Explanation:
Okay, so I need to find the value of this expression: (26² - 24² - 10)² - 10². Hmm, let me break this down step by step. First, I remember that there are some algebraic identities that might help simplify things here. Let me recall... Oh right! The difference of squares formula: a² - b² = (a - b)(a + b). Maybe I can use that somewhere here.

Looking at the expression inside the first square: 26² - 24². That looks like a difference of squares. Let me apply the formula there. So, 26² - 24² would be (26 - 24)(26 + 24). Let me calculate that. 26 minus 24 is 2, and 26 plus 24 is 50. So, multiplying those together: 2 * 50 = 100. Okay, so 26² - 24² is 100.

Wait, so then the expression inside the first square becomes 100 - 10. Let me write that down. So, inside the big parentheses, it's 100 - 10, which is 90. So now the entire expression simplifies to (90)² - 10².

Oh, another difference of squares! Because now we have something squared minus something else squared. Let me apply the formula again here. So, (90)² - (10)² = (90 - 10)(90 + 10). Calculating that: 90 - 10 is 80, and 90 + 10 is 100. Multiplying those together: 80 * 100.

Hmm, 80 times 100 is straightforward. That's 8000. So, putting it all together, the value of the original expression is 8000. Wait, let me just check my steps again to make sure I didn't make any mistakes.

First, 26² - 24²: yes, that's (26 - 24)(26 + 24) = 2*50 = 100. Then subtract 10: 100 - 10 = 90. Then square that: 90² is 8100. Then subtract 10² which is 100. So 8100 - 100 = 8000. Wait, hold on, I just realized that earlier I thought of (90)² - (10)² as (90 - 10)(90 + 10), which is 80*100 = 8000. But if I calculate 90² - 10² directly, that's 8100 - 100 = 8000. Yep, same answer. So both ways, it checks out.

But let me double-check the original problem to make sure I didn't misread anything. The problem is (26² - 24² - 10)² - 10². So first compute 26 squared, which is 676. Then 24 squared is 576. Then subtract those: 676 - 576 = 100. Then subtract 10: 100 - 10 = 90. Then square that: 90 squared is 8100. Then subtract 10 squared, which is 100. 8100 - 100 = 8000.

Yes, that's consistent. So both methods give me 8000. So I think that's the correct answer. Wait, but let me check my arithmetic once more.

26 squared: 26*26. Let me compute that. 20*20 = 400, 20*6 = 120, 6*20 = 120, 6*6 = 36. So (20 + 6)^2 = 20² + 2*20*6 + 6² = 400 + 240 + 36 = 676. Correct.

24 squared: 24*24. 20*20 = 400, 20*4 = 80, 4*20 = 80, 4*4 = 16. So (20 + 4)^2 = 400 + 160 + 16 = 576. Correct.

So 676 - 576 = 100. Then 100 - 10 = 90. Squared: 90*90. Let me do that. 9*9 = 81, so 90*90 = 8100. Then subtract 10^2 = 100. 8100 - 100 = 8000. Yep, still 8000.

So all my steps check out. Therefore, the value of the expression is 8000.

But just to be thorough, let me try expanding everything without using the difference of squares formula. Maybe that will help confirm.

Original expression: (26² - 24² - 10)² - 10².

First compute 26²: 676, 24²: 576. Then 676 - 576 is 100, as before. Then 100 - 10 is 90. Then 90 squared is 8100. Then subtract 10² which is 100. 8100 - 100 = 8000. Same result.

Alternatively, if I didn't notice the difference of squares, maybe I would expand the square term. Let me try that.

First, let me denote A = 26² - 24² - 10. Then the expression is A² - 10². But A is 100 - 10 = 90. So again, same thing.

Alternatively, suppose I didn't compute 26² - 24² first. Maybe expand (26² - 24² - 10)². But that seems more complicated. Let me see:

Let me write the inner part as (26² - 24²) - 10. As before, that's 100 - 10 = 90. So again, same thing.

Alternatively, if I didn't realize the difference of squares, maybe I could have done:

Compute 26² - 24² - 10. Let's compute 26² - 24² first. Which is (26 - 24)(26 + 24) = 2*50 = 100. So again, same steps.

Alternatively, someone might try to do it step by step without recognizing the difference of squares:

Compute 26²: 676.

Compute 24²: 576.

Subtract: 676 - 576 = 100.

Subtract 10: 100 - 10 = 90.

Square that: 90² = 8100.

Subtract 10²: 8100 - 100 = 8000.

Same answer again. So regardless of the method, I get 8000.

Alternatively, if someone made a mistake in calculation, maybe they could have messed up. Let me check if someone might have a different approach.

Wait, another way: Let me think about the entire expression as a difference of squares. Let me see:

The original expression is (something)² - (10)², which is (something - 10)(something + 10). But the "something" is 26² - 24² - 10. So if I write it as [(26² - 24² - 10) - 10][(26² - 24² - 10) + 10].

Wait, let me compute that. Let me denote X = 26² - 24² - 10. Then the expression is X² - 10² = (X - 10)(X + 10). So substituting X:

( (26² - 24² - 10) - 10 ) * ( (26² - 24² - 10) + 10 )

Simplify each part:

First part: 26² - 24² - 10 - 10 = 26² - 24² - 20

Second part: 26² - 24² - 10 + 10 = 26² - 24²

So the expression becomes (26² - 24² - 20) * (26² - 24²)

Earlier, we found 26² - 24² = 100. Therefore, substituting:

(100 - 20) * 100 = 80 * 100 = 8000.

Same answer again! So even if I approach it by factoring the entire expression as a difference of squares, I still get 8000.

Therefore, regardless of the method, the answer is 8000.

I think that's solid. All methods converge to 8000. Therefore, I can be confident that the value is 8000.

**Final Answer**
\boxed{8000}

To find the value of the expression:

$$
(26^2 - 24^2 - 10)^2 - 10^2
$$

we will evaluate it step by step using algebraic techniques and simplifications.

---

### Step 1: Recognize the Difference of Squares in the Inner Expression

The inner part $ 26^2 - 24^2 $ is a classic difference of squares. We can apply the identity:

$$
a^2 - b^2 = (a - b)(a + b)
$$

So:

$$
26^2 - 24^2 = (26 - 24)(26 + 24) = (2)(50) = 100
$$

---

### Step 2: Subtract 10 From the Result

Now, subtract 10 from the result of the above step:

$$
100 - 10 = 90
$$

---

### Step 3: Square the Result

Now square the value we just found:

$$
(90)^2 = 8100
$$

---

### Step 4: Subtract the Square of 10

Finally, subtract $10^2$:

$$
8100 - 100 = 8000
$$

---

### Final Answer

$$
\boxed{8000}
$$

[Assistant]
Explanation:
Okay, so I need to find the value of this expression: (26² - 24² - 10)² - 10². Hmm, let me break this down step by step. First, I remember that there are some algebraic identities that might help simplify things here. Let me recall... Oh right! The difference of squares formula: a² - b² = (a - b)(a + b). Maybe I can use that somewhere here.

Looking at the expression inside the first square: 26² - 24². That looks like a difference of squares. Let me apply the formula there. So, 26² - 24² would be (26 - 24)(26 + 24). Let me calculate that. 26 minus 24 is 2, and 26 plus 24 is 50. So, multiplying those together: 2 * 50 = 100. Okay, so 26² - 24² is 100.

Wait, so then the expression inside the first square becomes 100 - 10. Let me write that down. So, inside the big parentheses, it's 100 - 10, which is 90. So now the entire expression simplifies to (90)² - 10².

Oh, another difference of squares! Because now we have something squared minus something else squared. Let me apply the formula again here. So, (90)² - (10)² = (90 - 10)(90 + 10). Calculating that: 90 - 10 is 80, and 90 + 10 is 100. Multiplying those together: 80 * 100.

Hmm, 80 times 100 is straightforward. That's 8000. So, putting it all together, the value of the original expression is 8000. Wait, let me just check my steps again to make sure I didn't make any mistakes.

First, 26² - 24²: yes, that's (26 - 24)(26 + 24) = 2*50 = 100. Then subtract 10: 100 - 10 = 90. Then square that: 90² is 8100. Then subtract 10² which is 100. So 8100 - 100 = 8000. Wait, hold on, I just realized that earlier I thought of (90)² - (10)² as (90 - 10)(90 + 10), which is 80*100 = 8000. But if I calculate 90² - 10² directly, that's 8100 - 100 = 8000. Yep, same answer. So both ways, it checks out.

But let me double-check the original problem to make sure I didn't misread anything. The problem is (26² - 24² - 10)² - 10². So first compute 26 squared, which is 676. Then 24 squared is 576. Then subtract those: 676 - 576 = 100. Then subtract 10: 100 - 10 = 90. Then square that: 90 squared is 8100. Then subtract 10 squared, which is 100. 8100 - 100 = 8000.

Yes, that's consistent. So both methods give me 8000. So I think that's the correct answer. Wait, but let me check my arithmetic once more.

26 squared: 26*26. Let me compute that. 20*20 = 400, 20*6 = 120, 6*20 = 120, 6*6 = 36. So (20 + 6)^2 = 20² + 2*20*6 + 6² = 400 + 240 + 36 = 676. Correct.

24 squared: 24*24. 20*20 = 400, 20*4 = 80, 4*20 = 80, 4*4 = 16. So (20 + 4)^2 = 400 + 160 + 16 = 576. Correct.

So 676 - 576 = 100. Then 100 - 10 = 90. Squared: 90*90. Let me do that. 9*9 = 81, so 90*90 = 8100. Then subtract 10^2 = 100. 8100 - 100 = 8000. Yep, still 8000.

So all my steps check out. Therefore, the value of the expression is 8000.

But just to be thorough, let me try expanding everything without using the difference of squares formula. Maybe that will help confirm.

Original expression: (26² - 24² - 10)² - 10².

First compute 26²: 676, 24²: 576. Then 676 - 576 is 100, as before. Then 100 - 10 is 90. Then 90 squared is 8100. Then subtract 10² which is 100. 8100 - 100 = 8000. Same result.

Alternatively, if I didn't notice the difference of squares, maybe I would expand the square term. Let me try that.

First, let me denote A = 26² - 24² - 10. Then the expression is A² - 10². But A is 100 - 10 = 90. So again, same thing.

Alternatively, suppose I didn't compute 26² - 24² first. Maybe expand (26² - 24² - 10)². But that seems more complicated. Let me see:

Let me write the inner part as (26² - 24²) - 10. As before, that's 100 - 10 = 90. So again, same thing.

Alternatively, if I didn't realize the difference of squares, maybe I could have done:

Compute 26² - 24² - 10. Let's compute 26² - 24² first. Which is (26 - 24)(26 + 24) = 2*50 = 100. So again, same steps.

Alternatively, someone might try to do it step by step without recognizing the difference of squares:

Compute 26²: 676.

Compute 24²: 576.

Subtract: 676 - 576 = 100.

Subtract 10: 100 - 10 = 90.

Square that: 90² = 8100.

Subtract 10²: 8100 - 100 = 8000.

Same answer again. So regardless of the method, I get 8000.

Alternatively, if someone made a mistake in calculation, maybe they could have messed up. Let me check if someone might have a different approach.

Wait, another way: Let me think about the entire expression as a difference of squares. Let me see:

The original expression is (something)² - (10)², which is (something - 10)(something + 10). But the "something" is 26² - 24² - 10. So if I write it as [(26² - 24² - 10) - 10][(26² - 24² - 10) + 10].

Wait, let me compute that. Let me denote X = 26² - 24² - 10. Then the expression is X² - 10² = (X - 10)(X + 10). So substituting X:

( (26² - 24² - 10) - 10 ) * ( (26² - 24² - 10) + 10 )

Simplify each part:

First part: 26² - 24² - 10 - 10 = 26² - 24² - 20

Second part: 26² - 24² - 10 + 10 = 26² - 24²

So the expression becomes (26² - 24² - 20) * (26² - 24²)

Earlier, we found 26² - 24² = 100. Therefore, substituting:

(100 - 20) * 100 = 80 * 100 = 8000.

Same answer again! So even if I approach it by factoring the entire expression as a difference of squares, I still get 8000.

Therefore, regardless of the method, the answer is 8000.

I think that's solid. All methods converge to 8000. Therefore, I can be confident that the value is 8000.

**Final Answer**
\boxed{8000}

To find the value of the expression:

$$
(26^2 - 24^2 - 10)^2 - 10^2
$$

we will evaluate it step by step using algebraic techniques and simplifications.

---

### Step 1: Recognize the Difference of Squares in the Inner Expression

The inner part $ 26^2 - 24^2 $ is a classic difference of squares. We can apply the identity:

$$
a^2 - b^2 = (a - b)(a + b)
$$

So:

$$
26^2 - 24^2 = (26 - 24)(26 + 24) = (2)(50) = 100
$$

---

### Step 2: Subtract 10 From the Result

Now, subtract 10 from the result of the above step:

$$
100 - 10 = 90
$$

---

### Step 3: Square the Result

Now square the value we just found:

$$
(90)^2 = 8100
$$

---

### Step 4: Subtract the Square of 10

Finally, subtract $10^2$:

$$
8100 - 100 = 8000
$$

---

### Final Answer

$$
\boxed{8000}
$$

Solution: We know that $x^2 - y^2 = (x+y)(x-y)$. To begin with, let $x = 26^2 - 24^2 - 10$ and $y = 10$. Factoring $x^2 - y^2$ and substituting in gives $(26^2-24^2-10+10)(26^2-24^2-10-10)$. Now, let $x = 26$ and $y = 24$. Factoring $x^2 - y^2$ and substituting in gives $((26+24)(26-24)-10+10)((26+24)(26-24)-10-10)$. This simplifies to $(50\cdot 2)(50 \cdot 2 - 20)$, or $100 \cdot 80$. Thus, our final answer is $\boxed{8000}$.


Explanation:
Okay, so I need to figure out how many ways there are to arrange the letters of the word "ELLIPSE". Let me think about this step by step.

First, I remember that when you want to find the number of arrangements (permutations) of letters in a word, you usually start by considering if there are any repeating letters. If all the letters are unique, then it's just factorial of the number of letters. But if there are repeating letters, you have to divide by the factorial of the number of times each letter repeats.

So let me check the letters in "ELLIPSE". Let me write them down:

E, L, L, I, P, S, E.

Wait, let me count again. The word is E-L-L-I-P-S-E. So that's 7 letters total. Let me list them:

1. E
2. L
3. L
4. I
5. P
6. S
7. E

So there are 7 letters in total. Now, checking for duplicates. The letters E appears twice, L appears twice, and the rest are unique: I, P, S. So the letters E and L each have duplicates.

So in general, the formula for permutations of a multiset is:

Number of arrangements = n! / (n1! * n2! * ... * nk!)

where n is the total number of items, and n1, n2,...,nk are the frequencies of each distinct item.

In this case, n = 7. The duplicates are E (2 times) and L (2 times). So the formula would be:

7! divided by (2! * 2!) because there are two letters each repeated twice.

Let me confirm that. So if all letters were unique, it would be 7! But since there are duplicates, we have to divide by the factorials of their counts to account for the indistinguishability.

So that would be 7! / (2! * 2!). Let me compute that.

First, compute 7! which is 7 × 6 × 5 × 4 × 3 × 2 × 1. Let me compute that step by step.

7! = 5040.

Wait, let me verify:

7 × 6 = 42

42 × 5 = 210

210 × 4 = 840

840 × 3 = 2520

2520 × 2 = 5040

Yes, 7! is 5040.

Then, 2! is 2, so 2! * 2! is 2 * 2 = 4.

Therefore, the number of arrangements is 5040 / 4. Let me compute that.

5040 divided by 4.

Well, 5040 divided by 2 is 2520, divided by 2 again is 1260.

So 1260.

Therefore, is the answer 1260? Let me think again if I missed something.

Wait, are there only two letters that are repeated? Let me check again the letters in ELLIPSE.

Letters: E, L, L, I, P, S, E.

So E appears twice, L appears twice, and the rest (I, P, S) are once each. So yes, only two letters with duplicates, each appearing twice. So the formula is 7! divided by (2! * 2!).

But wait, another way to think: suppose I first calculate the total permutations without considering duplicates, which is 7!. Then, since the two E's are identical, we have overcounted by the number of ways the E's can be arranged among themselves, which is 2!, so we divide by 2!. Similarly, the two L's are also identical, so we divide by another 2!.

Therefore, total is 7! / (2! * 2!) = 5040 / 4 = 1260.

Therefore, the answer is 1260.

Wait, but let me check if there are more duplicates. For example, maybe I miscounted the letters. Let me write the letters again:

E, L, L, I, P, S, E.

So E is in positions 1 and 7. L is in positions 2 and 3. I, P, S are single. So yes, two E's and two L's. So that's correct.

Another way to think: Let's suppose all letters are distinct. Then the number of arrangements would be 7!. But since there are duplicates, each unique arrangement is counted multiple times. For example, swapping the two E's doesn't change the arrangement, so each unique arrangement is counted 2! times due to the E's. Similarly, swapping the two L's doesn't change the arrangement, so each unique arrangement is also counted 2! times due to the L's. Therefore, the total number of unique arrangements is 7! divided by (2! * 2!), which is again 1260.

Therefore, I think that's correct.

But let me try a smaller example to verify. For example, take the word "ELL". Which has letters E, L, L. The number of arrangements should be 3! / 2! = 3. Let's check:

Possible arrangements:

ELL, LEL, LLE. Yes, 3. So that works.

Similarly, for "EELL", which has 4 letters with two E's and two L's. The number of arrangements should be 4! / (2! * 2!) = 24 / 4 = 6. Let me list them:

EELL, ELEE, ELLE, LLEE, LELE, LLEL. Yes, that's 6.

So the formula works here. Therefore, applying the same logic to the original problem, the answer should be 7! / (2! * 2!) = 1260.

Wait, just to be thorough, let me check another example with three letters. Suppose the word is "AABB". Number of arrangements is 4! / (2! * 2!) = 6. Which is correct.

Therefore, I think my reasoning is sound.

Therefore, the number of ways to arrange the letters of the word ELLIPSE is 1260.

But just to make sure, let me check if there are any other letters that might be repeated. For example, maybe the letter 'P' or 'I' or 'S' are repeated? Let me check the original word again: ELLIPSE.

Breaking it down:

E, L, L, I, P, S, E.

So E is twice, L is twice, and I, P, S once each. So no, there are no other duplicates.

Therefore, the calculation is correct.

Alternatively, maybe I should calculate it step by step.

Total permutations: 7!

But since there are duplicates, we need to adjust.

First, consider the two E's. If all letters were unique, the number of arrangements would be 7!. But since the two E's are identical, each unique arrangement is counted 2! times (because swapping the E's doesn't create a new arrangement). Therefore, we divide by 2!.

Similarly, the two L's are also identical, so we have to divide by another 2! because swapping the L's doesn't change the arrangement.

Therefore, total number is 7! / (2! * 2!) = 5040 / 4 = 1260.

Yes, that seems consistent.

Alternatively, if I think of choosing positions for each letter.

First, choose positions for the E's. There are 7 positions. Choosing 2 positions for the E's: C(7, 2).

Then, from the remaining 5 positions, choose 2 positions for the L's: C(5, 2).

Then, the remaining 3 positions are for the letters I, P, S, which are all distinct. So they can be arranged in 3! ways.

Therefore, total number of arrangements is C(7, 2) * C(5, 2) * 3!

Let me compute that.

C(7, 2) is 21. C(5, 2) is 10. 3! is 6.

So total is 21 * 10 * 6.

21 * 10 is 210, 210 * 6 is 1260.

Same answer.

Therefore, this method also gives 1260.

Therefore, that confirms the answer.

Alternatively, think of it as permutations with repetition.

The formula is n! divided by the product of the factorials of the counts of each repeated element.

Which is exactly what we did.

Therefore, I think 1260 is correct.

Wait, just to make sure I didn't make an arithmetic mistake. Let me recompute 7! / (2! * 2!).

7! is 5040.

2! * 2! is 2 * 2 = 4.

5040 divided by 4: 5040 / 4. Let me do this division again.

Dividing 5040 by 4:

4 goes into 5 once (1), remainder 1.

Bring down 0: 10. 4 goes into 10 twice (2), remainder 2.

Bring down 4: 24. 4 goes into 24 six times (6).

Bring down 0: 0.

So the result is 1260.

Yes, that's correct.

Alternatively, 5040 divided by 2 is 2520, divided by 2 again is 1260.

Yes, same result.

Therefore, I feel confident that the answer is 1260.

I don't see any mistakes in my reasoning.

Another check: suppose we had a word with all letters distinct, say "ELLIPS" (but that's only 6 letters). Wait, no, the original word is 7 letters.

But just to think, if all letters were distinct, it would be 7! = 5040. Since we have duplicates, it's less. 1260 is less than 5040, which makes sense.

Alternatively, if there was only one pair of duplicates, say two E's, then it would be 7! / 2! = 2520. Which is more than 1260. So since there are two pairs of duplicates, we divide by another 2!, hence 2520 / 2 = 1260.

Yes, that's correct.

So all methods lead to the same answer. Therefore, I can be confident that the number of ways to arrange the letters of the word ELLIPSE is 1260.

**Final Answer**
The number of ways to arrange the letters of the word ELLIPSE is \boxed{1260}.

To determine the number of distinct ways to arrange the letters in the word **"ELLIPSE"**, we begin by analyzing the structure of the word.

---

### Step 1: Count the Letters

The word **"ELLIPSE"** consists of **7 letters**:

- **E** appears **2 times**
- **L** appears **2 times**
- **I**, **P**, and **S** each appear **1 time**

This tells us that the total number of **distinct permutations** is not simply 7! (which assumes all letters are unique), but must be adjusted for the repetitions.

---

### Step 2: Use the Formula for Multiset Permutations

When a set has repeated elements, the total number of distinct permutations is given by:

$$
\frac{n!}{n_1! \cdot n_2! \cdot \dots \cdot n_k!}
$$

where:
- $ n $ is the total number of items (letters),
- $ n_1, n_2, \dots, n_k $ are the frequencies of each repeated item.

In this case:
- $ n = 7 $ (total letters)
- $ n_1 = 2 $ (for E)
- $ n_2 = 2 $ (for L)
- $ n_3 = 1 $ (for I), $ n_4 = 1 $ (for P), $ n_5 = 1 $ (for S)

Thus, the formula becomes:

$$
\text{Number of arrangements} = \frac{7!}{2! \cdot 2!}
$$

---

### Step 3: Compute the Value

First, compute the factorials:

$$
7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040
$$
$$
2! = 2 \Rightarrow 2! \cdot 2! = 2 \cdot 2 = 4
$$

Now divide:

$$
\frac{5040}{4} = 1260
$$

---

### Step 4: Final Answer

Therefore, the number of ways to arrange the letters of the word **"ELLIPSE"** is:

$$
\boxed{1260}
$$

[Assistant]
Explanation:
Okay, so I need to figure out how many ways there are to arrange the letters of the word "ELLIPSE". Let me think about this step by step.

First, I remember that when you want to find the number of arrangements (permutations) of letters in a word, you usually start by considering if there are any repeating letters. If all the letters are unique, then it's just factorial of the number of letters. But if there are repeating letters, you have to divide by the factorial of the number of times each letter repeats.

So let me check the letters in "ELLIPSE". Let me write them down:

E, L, L, I, P, S, E.

Wait, let me count again. The word is E-L-L-I-P-S-E. So that's 7 letters total. Let me list them:

1. E
2. L
3. L
4. I
5. P
6. S
7. E

So there are 7 letters in total. Now, checking for duplicates. The letters E appears twice, L appears twice, and the rest are unique: I, P, S. So the letters E and L each have duplicates.

So in general, the formula for permutations of a multiset is:

Number of arrangements = n! / (n1! * n2! * ... * nk!)

where n is the total number of items, and n1, n2,...,nk are the frequencies of each distinct item.

In this case, n = 7. The duplicates are E (2 times) and L (2 times). So the formula would be:

7! divided by (2! * 2!) because there are two letters each repeated twice.

Let me confirm that. So if all letters were unique, it would be 7! But since there are duplicates, we have to divide by the factorials of their counts to account for the indistinguishability.

So that would be 7! / (2! * 2!). Let me compute that.

First, compute 7! which is 7 × 6 × 5 × 4 × 3 × 2 × 1. Let me compute that step by step.

7! = 5040.

Wait, let me verify:

7 × 6 = 42

42 × 5 = 210

210 × 4 = 840

840 × 3 = 2520

2520 × 2 = 5040

Yes, 7! is 5040.

Then, 2! is 2, so 2! * 2! is 2 * 2 = 4.

Therefore, the number of arrangements is 5040 / 4. Let me compute that.

5040 divided by 4.

Well, 5040 divided by 2 is 2520, divided by 2 again is 1260.

So 1260.

Therefore, is the answer 1260? Let me think again if I missed something.

Wait, are there only two letters that are repeated? Let me check again the letters in ELLIPSE.

Letters: E, L, L, I, P, S, E.

So E appears twice, L appears twice, and the rest (I, P, S) are once each. So yes, only two letters with duplicates, each appearing twice. So the formula is 7! divided by (2! * 2!).

But wait, another way to think: suppose I first calculate the total permutations without considering duplicates, which is 7!. Then, since the two E's are identical, we have overcounted by the number of ways the E's can be arranged among themselves, which is 2!, so we divide by 2!. Similarly, the two L's are also identical, so we divide by another 2!.

Therefore, total is 7! / (2! * 2!) = 5040 / 4 = 1260.

Therefore, the answer is 1260.

Wait, but let me check if there are more duplicates. For example, maybe I miscounted the letters. Let me write the letters again:

E, L, L, I, P, S, E.

So E is in positions 1 and 7. L is in positions 2 and 3. I, P, S are single. So yes, two E's and two L's. So that's correct.

Another way to think: Let's suppose all letters are distinct. Then the number of arrangements would be 7!. But since there are duplicates, each unique arrangement is counted multiple times. For example, swapping the two E's doesn't change the arrangement, so each unique arrangement is counted 2! times due to the E's. Similarly, swapping the two L's doesn't change the arrangement, so each unique arrangement is also counted 2! times due to the L's. Therefore, the total number of unique arrangements is 7! divided by (2! * 2!), which is again 1260.

Therefore, I think that's correct.

But let me try a smaller example to verify. For example, take the word "ELL". Which has letters E, L, L. The number of arrangements should be 3! / 2! = 3. Let's check:

Possible arrangements:

ELL, LEL, LLE. Yes, 3. So that works.

Similarly, for "EELL", which has 4 letters with two E's and two L's. The number of arrangements should be 4! / (2! * 2!) = 24 / 4 = 6. Let me list them:

EELL, ELEE, ELLE, LLEE, LELE, LLEL. Yes, that's 6.

So the formula works here. Therefore, applying the same logic to the original problem, the answer should be 7! / (2! * 2!) = 1260.

Wait, just to be thorough, let me check another example with three letters. Suppose the word is "AABB". Number of arrangements is 4! / (2! * 2!) = 6. Which is correct.

Therefore, I think my reasoning is sound.

Therefore, the number of ways to arrange the letters of the word ELLIPSE is 1260.

But just to make sure, let me check if there are any other letters that might be repeated. For example, maybe the letter 'P' or 'I' or 'S' are repeated? Let me check the original word again: ELLIPSE.

Breaking it down:

E, L, L, I, P, S, E.

So E is twice, L is twice, and I, P, S once each. So no, there are no other duplicates.

Therefore, the calculation is correct.

Alternatively, maybe I should calculate it step by step.

Total permutations: 7!

But since there are duplicates, we need to adjust.

First, consider the two E's. If all letters were unique, the number of arrangements would be 7!. But since the two E's are identical, each unique arrangement is counted 2! times (because swapping the E's doesn't create a new arrangement). Therefore, we divide by 2!.

Similarly, the two L's are also identical, so we have to divide by another 2! because swapping the L's doesn't change the arrangement.

Therefore, total number is 7! / (2! * 2!) = 5040 / 4 = 1260.

Yes, that seems consistent.

Alternatively, if I think of choosing positions for each letter.

First, choose positions for the E's. There are 7 positions. Choosing 2 positions for the E's: C(7, 2).

Then, from the remaining 5 positions, choose 2 positions for the L's: C(5, 2).

Then, the remaining 3 positions are for the letters I, P, S, which are all distinct. So they can be arranged in 3! ways.

Therefore, total number of arrangements is C(7, 2) * C(5, 2) * 3!

Let me compute that.

C(7, 2) is 21. C(5, 2) is 10. 3! is 6.

So total is 21 * 10 * 6.

21 * 10 is 210, 210 * 6 is 1260.

Same answer.

Therefore, this method also gives 1260.

Therefore, that confirms the answer.

Alternatively, think of it as permutations with repetition.

The formula is n! divided by the product of the factorials of the counts of each repeated element.

Which is exactly what we did.

Therefore, I think 1260 is correct.

Wait, just to make sure I didn't make an arithmetic mistake. Let me recompute 7! / (2! * 2!).

7! is 5040.

2! * 2! is 2 * 2 = 4.

5040 divided by 4: 5040 / 4. Let me do this division again.

Dividing 5040 by 4:

4 goes into 5 once (1), remainder 1.

Bring down 0: 10. 4 goes into 10 twice (2), remainder 2.

Bring down 4: 24. 4 goes into 24 six times (6).

Bring down 0: 0.

So the result is 1260.

Yes, that's correct.

Alternatively, 5040 divided by 2 is 2520, divided by 2 again is 1260.

Yes, same result.

Therefore, I feel confident that the answer is 1260.

I don't see any mistakes in my reasoning.

Another check: suppose we had a word with all letters distinct, say "ELLIPS" (but that's only 6 letters). Wait, no, the original word is 7 letters.

But just to think, if all letters were distinct, it would be 7! = 5040. Since we have duplicates, it's less. 1260 is less than 5040, which makes sense.

Alternatively, if there was only one pair of duplicates, say two E's, then it would be 7! / 2! = 2520. Which is more than 1260. So since there are two pairs of duplicates, we divide by another 2!, hence 2520 / 2 = 1260.

Yes, that's correct.

So all methods lead to the same answer. Therefore, I can be confident that the number of ways to arrange the letters of the word ELLIPSE is 1260.

**Final Answer**
The number of ways to arrange the letters of the word ELLIPSE is \boxed{1260}.

To determine the number of distinct ways to arrange the letters in the word **"ELLIPSE"**, we begin by analyzing the structure of the word.

---

### Step 1: Count the Letters

The word **"ELLIPSE"** consists of **7 letters**:

- **E** appears **2 times**
- **L** appears **2 times**
- **I**, **P**, and **S** each appear **1 time**

This tells us that the total number of **distinct permutations** is not simply 7! (which assumes all letters are unique), but must be adjusted for the repetitions.

---

### Step 2: Use the Formula for Multiset Permutations

When a set has repeated elements, the total number of distinct permutations is given by:

$$
\frac{n!}{n_1! \cdot n_2! \cdot \dots \cdot n_k!}
$$

where:
- $ n $ is the total number of items (letters),
- $ n_1, n_2, \dots, n_k $ are the frequencies of each repeated item.

In this case:
- $ n = 7 $ (total letters)
- $ n_1 = 2 $ (for E)
- $ n_2 = 2 $ (for L)
- $ n_3 = 1 $ (for I), $ n_4 = 1 $ (for P), $ n_5 = 1 $ (for S)

Thus, the formula becomes:

$$
\text{Number of arrangements} = \frac{7!}{2! \cdot 2!}
$$

---

### Step 3: Compute the Value

First, compute the factorials:

$$
7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040
$$
$$
2! = 2 \Rightarrow 2! \cdot 2! = 2 \cdot 2 = 4
$$

Now divide:

$$
\frac{5040}{4} = 1260
$$

---

### Step 4: Final Answer

Therefore, the number of ways to arrange the letters of the word **"ELLIPSE"** is:

$$
\boxed{1260}
$$

Solution: There are two E's, two L's, and seven total letters, so the answer is $\dfrac{7!}{2! \times 2!} = \boxed{1260}$.


Explanation:
Okay, so I need to solve the equation $2^{2x} = 256^{\frac{1}{2}}$. Hmm, let me think about how to approach this.

First, I remember that when dealing with exponential equations, it's often helpful to express both sides with the same base if possible. That way, I can set the exponents equal to each other. Let me check if 256 can be written as a power of 2.

I know that $2^8 = 256$, right? Let me verify that. $2^1 = 2$, $2^2 = 4$, $2^3 = 8$, $2^4 = 16$, $2^5 = 32$, $2^6 = 64$, $2^7 = 128$, and $2^8 = 256$. Yes, that's correct. So 256 is $2^8$.

So, substituting that into the equation, the right-hand side becomes $(2^8)^{\frac{1}{2}}$. Now, I need to simplify that. Remembering the exponent rule $(a^m)^n = a^{m \cdot n}$, so multiplying the exponents 8 and 1/2.

Calculating that, 8 times 1/2 is 4. Therefore, $(2^8)^{\frac{1}{2}} = 2^4$.

So now the equation simplifies to $2^{2x} = 2^4$. Since the bases are the same (both are base 2), I can set the exponents equal to each other. That gives me $2x = 4$.

Solving for x, I just need to divide both sides by 2. So, $x = 4 / 2 = 2$.

Wait, let me check if I did everything correctly. Let me verify by plugging x = 2 back into the original equation.

Left-hand side: $2^{2x}$ when x = 2 is $2^{2*2} = 2^4 = 16$.

Right-hand side: $256^{1/2}$ is the square root of 256, which is 16.

Yes, both sides equal 16. So that checks out.

But just to make sure I didn't skip any steps or make any mistakes in my reasoning, let me go through it again.

Original equation: $2^{2x} = 256^{1/2}$.

Express 256 as a power of 2: 256 is 2^8. So substituting, we have $2^{2x} = (2^8)^{1/2}$.

Apply exponent rule: $(2^8)^{1/2} = 2^{8*(1/2)} = 2^4$.

Therefore, equation becomes $2^{2x} = 2^4$.

Since the bases are equal, exponents must be equal: 2x = 4.

Solving for x: x = 4 / 2 = 2.

Yep, that seems right.

Alternatively, maybe there's another way to approach it? Let me think.

Another way could be to take the square root of both sides first. Wait, the original equation is $2^{2x} = 256^{1/2}$.

But 2^{2x} can be written as (2^x)^2. So maybe taking square roots on both sides? Let me try that.

Taking square roots of both sides: $\sqrt{2^{2x}} = \sqrt{256^{1/2}}$.

Simplifying the left side: $\sqrt{2^{2x}} = (2^{2x})^{1/2} = 2^{x}$.

On the right side: $\sqrt{256^{1/2}}$. Hmm, the square root of 256^{1/2} is the same as (256^{1/2})^{1/2} = 256^{1/4}.

But 256 is 2^8, so 256^{1/4} = (2^8)^{1/4} = 2^{8*(1/4)} = 2^2 = 4.

So now, the equation becomes 2^x = 4.

But 4 is 2^2, so 2^x = 2^2 implies x = 2.

Same result. So this method also gives x = 2.

So either way, whether I simplify the right-hand side first or take square roots, I end up with x = 2.

Alternatively, maybe I could use logarithms? Let me try that method as a check.

Starting with the original equation: $2^{2x} = 256^{1/2}$.

Take the natural logarithm of both sides: $\ln(2^{2x}) = \ln(256^{1/2})$.

Using the logarithm power rule: $2x \ln 2 = \frac{1}{2} \ln 256$.

Then, solving for x:

$x = \frac{\frac{1}{2} \ln 256}{2 \ln 2} = \frac{\ln 256}{4 \ln 2}$.

But 256 is 2^8, so $\ln 256 = \ln (2^8) = 8 \ln 2$.

Substituting back:

$x = \frac{8 \ln 2}{4 \ln 2} = \frac{8}{4} = 2$.

Same answer again.

So all three methods give me x = 2, so I feel confident that's the correct solution.

I think that covers all possible ways to approach this problem. It's a straightforward exponential equation once you express both sides with the same base.

Wait, just to ensure that there are no alternative interpretations of the original equation. Let me check the original problem again:

"Solve for x: $2^{2x} = 256^{\frac{1}{2}}$."

Yes, so 2 raised to the 2x power equals 256 raised to the 1/2 power.

No ambiguity there.

Another check: if I didn't realize that 256 is 2^8, could I still solve it? Let me try.

Suppose I didn't know 256 is a power of 2. Then maybe I could compute 256^(1/2) directly.

Well, 256 is 16 squared, right? Because 16*16 = 256. So square root of 256 is 16.

So the equation becomes 2^{2x} = 16.

Now, 16 is 2^4, so again 2^{2x} = 2^4, leading to 2x = 4, so x = 2.

Same answer.

So even if I didn't express 256 as a power of 2, but instead calculated the square root directly, I still get the same result.

Therefore, regardless of the method, x = 2 is the solution.

Is there a possibility that there could be another solution? For example, sometimes equations can have multiple solutions, but in this case, since the base of the exponential is 2, which is greater than 1, the function is strictly increasing, so the equation should have only one solution.

Also, since both sides are positive real numbers, and exponentials with base 2 are one-to-one functions, there is exactly one solution.

Therefore, x = 2 is the only solution.

So I think that's thorough. I considered multiple approaches, checked my work by substitution, verified the logic, and ensured that there are no alternative solutions.

**Final Answer**
The solution is \boxed{2}.

To solve the equation:

$$
2^{2x} = 256^{\frac{1}{2}}
$$

we begin by simplifying both sides of the equation to facilitate solving for $ x $.

---

### Step 1: Simplify the Right-Hand Side

We note that $ 256 $ is a power of 2:

$$
256 = 2^8
$$

Thus, replacing 256 in the equation:

$$
256^{\frac{1}{2}} = (2^8)^{\frac{1}{2}}
$$

Use the exponent rule $ (a^m)^n = a^{m \cdot n} $:

$$
(2^8)^{\frac{1}{2}} = 2^{8 \cdot \frac{1}{2}} = 2^4
$$

So the original equation now becomes:

$$
2^{2x} = 2^4
$$

---

### Step 2: Set Exponents Equal

Since both sides now have the same base (2), we can equate the exponents:

$$
2x = 4
$$

Solve for $ x $:

$$
x = \frac{4}{2} = 2
$$

---

### Step 3: Verification

To confirm the solution, substitute $ x = 2 $ back into the original equation:

Left-hand side:

$$
2^{2x} = 2^{2 \cdot 2} = 2^4 = 16
$$

Right-hand side:

$$
256^{\frac{1}{2}} = \sqrt{256} = 16
$$

Both sides are equal, so the solution is verified.

---

### Final Answer

The solution to the equation is:

$$
\boxed{2}
$$

[Assistant]
Explanation:
Okay, so I need to solve the equation $2^{2x} = 256^{\frac{1}{2}}$. Hmm, let me think about how to approach this.

First, I remember that when dealing with exponential equations, it's often helpful to express both sides with the same base if possible. That way, I can set the exponents equal to each other. Let me check if 256 can be written as a power of 2.

I know that $2^8 = 256$, right? Let me verify that. $2^1 = 2$, $2^2 = 4$, $2^3 = 8$, $2^4 = 16$, $2^5 = 32$, $2^6 = 64$, $2^7 = 128$, and $2^8 = 256$. Yes, that's correct. So 256 is $2^8$.

So, substituting that into the equation, the right-hand side becomes $(2^8)^{\frac{1}{2}}$. Now, I need to simplify that. Remembering the exponent rule $(a^m)^n = a^{m \cdot n}$, so multiplying the exponents 8 and 1/2.

Calculating that, 8 times 1/2 is 4. Therefore, $(2^8)^{\frac{1}{2}} = 2^4$.

So now the equation simplifies to $2^{2x} = 2^4$. Since the bases are the same (both are base 2), I can set the exponents equal to each other. That gives me $2x = 4$.

Solving for x, I just need to divide both sides by 2. So, $x = 4 / 2 = 2$.

Wait, let me check if I did everything correctly. Let me verify by plugging x = 2 back into the original equation.

Left-hand side: $2^{2x}$ when x = 2 is $2^{2*2} = 2^4 = 16$.

Right-hand side: $256^{1/2}$ is the square root of 256, which is 16.

Yes, both sides equal 16. So that checks out.

But just to make sure I didn't skip any steps or make any mistakes in my reasoning, let me go through it again.

Original equation: $2^{2x} = 256^{1/2}$.

Express 256 as a power of 2: 256 is 2^8. So substituting, we have $2^{2x} = (2^8)^{1/2}$.

Apply exponent rule: $(2^8)^{1/2} = 2^{8*(1/2)} = 2^4$.

Therefore, equation becomes $2^{2x} = 2^4$.

Since the bases are equal, exponents must be equal: 2x = 4.

Solving for x: x = 4 / 2 = 2.

Yep, that seems right.

Alternatively, maybe there's another way to approach it? Let me think.

Another way could be to take the square root of both sides first. Wait, the original equation is $2^{2x} = 256^{1/2}$.

But 2^{2x} can be written as (2^x)^2. So maybe taking square roots on both sides? Let me try that.

Taking square roots of both sides: $\sqrt{2^{2x}} = \sqrt{256^{1/2}}$.

Simplifying the left side: $\sqrt{2^{2x}} = (2^{2x})^{1/2} = 2^{x}$.

On the right side: $\sqrt{256^{1/2}}$. Hmm, the square root of 256^{1/2} is the same as (256^{1/2})^{1/2} = 256^{1/4}.

But 256 is 2^8, so 256^{1/4} = (2^8)^{1/4} = 2^{8*(1/4)} = 2^2 = 4.

So now, the equation becomes 2^x = 4.

But 4 is 2^2, so 2^x = 2^2 implies x = 2.

Same result. So this method also gives x = 2.

So either way, whether I simplify the right-hand side first or take square roots, I end up with x = 2.

Alternatively, maybe I could use logarithms? Let me try that method as a check.

Starting with the original equation: $2^{2x} = 256^{1/2}$.

Take the natural logarithm of both sides: $\ln(2^{2x}) = \ln(256^{1/2})$.

Using the logarithm power rule: $2x \ln 2 = \frac{1}{2} \ln 256$.

Then, solving for x:

$x = \frac{\frac{1}{2} \ln 256}{2 \ln 2} = \frac{\ln 256}{4 \ln 2}$.

But 256 is 2^8, so $\ln 256 = \ln (2^8) = 8 \ln 2$.

Substituting back:

$x = \frac{8 \ln 2}{4 \ln 2} = \frac{8}{4} = 2$.

Same answer again.

So all three methods give me x = 2, so I feel confident that's the correct solution.

I think that covers all possible ways to approach this problem. It's a straightforward exponential equation once you express both sides with the same base.

Wait, just to ensure that there are no alternative interpretations of the original equation. Let me check the original problem again:

"Solve for x: $2^{2x} = 256^{\frac{1}{2}}$."

Yes, so 2 raised to the 2x power equals 256 raised to the 1/2 power.

No ambiguity there.

Another check: if I didn't realize that 256 is 2^8, could I still solve it? Let me try.

Suppose I didn't know 256 is a power of 2. Then maybe I could compute 256^(1/2) directly.

Well, 256 is 16 squared, right? Because 16*16 = 256. So square root of 256 is 16.

So the equation becomes 2^{2x} = 16.

Now, 16 is 2^4, so again 2^{2x} = 2^4, leading to 2x = 4, so x = 2.

Same answer.

So even if I didn't express 256 as a power of 2, but instead calculated the square root directly, I still get the same result.

Therefore, regardless of the method, x = 2 is the solution.

Is there a possibility that there could be another solution? For example, sometimes equations can have multiple solutions, but in this case, since the base of the exponential is 2, which is greater than 1, the function is strictly increasing, so the equation should have only one solution.

Also, since both sides are positive real numbers, and exponentials with base 2 are one-to-one functions, there is exactly one solution.

Therefore, x = 2 is the only solution.

So I think that's thorough. I considered multiple approaches, checked my work by substitution, verified the logic, and ensured that there are no alternative solutions.

**Final Answer**
The solution is \boxed{2}.

To solve the equation:

$$
2^{2x} = 256^{\frac{1}{2}}
$$

we begin by simplifying both sides of the equation to facilitate solving for $ x $.

---

### Step 1: Simplify the Right-Hand Side

We note that $ 256 $ is a power of 2:

$$
256 = 2^8
$$

Thus, replacing 256 in the equation:

$$
256^{\frac{1}{2}} = (2^8)^{\frac{1}{2}}
$$

Use the exponent rule $ (a^m)^n = a^{m \cdot n} $:

$$
(2^8)^{\frac{1}{2}} = 2^{8 \cdot \frac{1}{2}} = 2^4
$$

So the original equation now becomes:

$$
2^{2x} = 2^4
$$

---

### Step 2: Set Exponents Equal

Since both sides now have the same base (2), we can equate the exponents:

$$
2x = 4
$$

Solve for $ x $:

$$
x = \frac{4}{2} = 2
$$

---

### Step 3: Verification

To confirm the solution, substitute $ x = 2 $ back into the original equation:

Left-hand side:

$$
2^{2x} = 2^{2 \cdot 2} = 2^4 = 16
$$

Right-hand side:

$$
256^{\frac{1}{2}} = \sqrt{256} = 16
$$

Both sides are equal, so the solution is verified.

---

### Final Answer

The solution to the equation is:

$$
\boxed{2}
$$

Solution:
\begin{align*}
2^{2x} & =256^{\frac{1}{2}} \\
2^{2x} & =(2^8)^{\frac{1}{2}} \\
2^{2x} & =(2^4) \\
2x & = 4 \\
x & = \boxed{2}
\end{align*}


Explanation:
Okay, so I have this equation here: √(3x - 5) = 2. I need to find all possible values of x. Hmm, let me think. Alright, square roots can sometimes be tricky because they only give the principal (non-negative) root, right? So, if I have a square root equal to a number, I have to make sure that the expression inside the square root is non-negative. Also, when I solve equations with square roots, I usually square both sides to eliminate the square root. Let me try that here.

First, let me write down the equation again to make sure I have it right:

√(3x - 5) = 2

Alright, so if I square both sides of the equation, that should get rid of the square root. Let me do that step by step. Squaring the left side: (√(3x - 5))² is just 3x - 5, because squaring a square root cancels them out. And squaring the right side: 2² is 4. So, after squaring both sides, the equation becomes:

3x - 5 = 4

Okay, that seems straightforward. Now I just need to solve for x. Let me add 5 to both sides to isolate the term with x. So:

3x - 5 + 5 = 4 + 5

Simplifying both sides:

3x = 9

Now, divide both sides by 3:

3x / 3 = 9 / 3

Which simplifies to:

x = 3

Wait, so x equals 3? Let me check if that's correct. Sometimes when you square both sides of an equation, you can introduce extraneous solutions, so I should verify that this solution actually works in the original equation.

Let me substitute x = 3 back into the original equation:

√(3*3 - 5) = √(9 - 5) = √4 = 2

Which is indeed equal to the right side of the equation, which is 2. So that works. So x = 3 is a valid solution. But the question asks for all possible values of x. Is there another solution?

Hmm, let me think. When you square both sides of an equation, sometimes you can get extra solutions, but in this case, since the square root function only outputs non-negative numbers, and we have √(3x - 5) equals 2, which is positive. So, squaring both sides should not introduce any extraneous solutions here because we already have the square root equal to a positive number. So, maybe there's only one solution here. Let me check if there's any other possibility.

Wait, could there be a negative solution? Let me think. Suppose someone might think, "Oh, maybe the square root could be negative, so maybe -2 is also a solution?" But no, the square root symbol specifically denotes the principal (non-negative) root. So √(something) is always non-negative. Therefore, even if we have an equation like √(something) = -2, that would have no solution because a square root can't be negative. In this case, though, the right side is 2, which is positive, so we don't have that problem.

So, going back, only x = 3 is the solution. Let me just check if the expression inside the square root is non-negative for x = 3. 3x - 5 when x = 3 is 9 - 5 = 4, which is positive. So that's okay. So there are no issues with the domain here. Therefore, x = 3 is the only solution.

But wait, the question says "all possible values of x." Maybe there's another value? Let me think again. Let me consider solving the equation step by step once more.

Original equation:

√(3x - 5) = 2

Square both sides:

(√(3x - 5))² = 2²

Which simplifies to:

3x - 5 = 4

Then, solving for x:

3x = 4 + 5

3x = 9

x = 3

So same result. There's no other solution here. Maybe if I think about the definition of the square root function, it's only defined for non-negative numbers. So 3x - 5 must be greater than or equal to zero. So 3x - 5 ≥ 0 ⇒ 3x ≥ 5 ⇒ x ≥ 5/3 ≈ 1.666...

But x = 3 is definitely in that domain. So there's no other solution. Therefore, the only possible value is 3.

Wait, but just to make sure, is there a case where squaring both sides could introduce another solution? For example, if the original equation had something like √(something) = something else, but here the right side is 2. Let me think of an example. Suppose if the equation was √(x) = x, then squaring both sides would give x = x², leading to x² - x = 0 ⇒ x(x - 1) = 0 ⇒ x = 0 or x = 1. But then you have to check both solutions. In this case, x = 0: √0 = 0, which is okay. x = 1: √1 = 1, which is also okay. So both are valid. But in our original problem, squaring only gave us one solution, which worked. So maybe in this case, there's only one solution.

Alternatively, if there was a ± symbol involved, but in this case, there's no ±. The original equation is √(something) equals 2, so we don't need to consider negative roots. Therefore, x = 3 is the only solution.

Therefore, I think the answer is x = 3. Let me check once more.

Original equation:

√(3x - 5) = 2

Plug in x = 3:

Left side: √(3*3 -5) = √(9 -5) = √4 = 2. Right side is 2. So yes, it works. Any other x?

Suppose someone says, what if 3x - 5 was another square that equals 4? But no, because we squared both sides and got 3x -5 = 4, so there's no other value for 3x -5. So 3x -5 is uniquely determined as 4. Therefore, x is uniquely determined as 3. Therefore, only one solution.

Therefore, the only possible value of x is 3.

**Final Answer**
The only possible value of $ x $ is \boxed{3}.

To solve the equation:

$$
\sqrt{3x - 5} = 2
$$

we begin by eliminating the square root through **squaring both sides** of the equation. This is a valid operation as long as we remember that the square root function is defined only for **non-negative** values and that squaring both sides may occasionally introduce extraneous solutions, which need to be verified.

---

### Step 1: Square Both Sides

$$
\left(\sqrt{3x - 5}\right)^2 = 2^2
$$

$$
3x - 5 = 4
$$

---

### Step 2: Solve for $ x $

Now solve the resulting linear equation:

$$
3x - 5 = 4
$$

Add 5 to both sides:

$$
3x = 9
$$

Divide both sides by 3:

$$
x = 3
$$

---

### Step 3: Verify the Solution

It's essential to verify that this value of $ x $ satisfies the original equation. Substitute $ x = 3 $ into the original expression:

$$
\sqrt{3(3) - 5} = \sqrt{9 - 5} = \sqrt{4} = 2
$$

This matches the right-hand side of the equation, confirming that $ x = 3 $ is indeed a valid solution.

---

### Step 4: Consider the Domain

The square root function is **only defined** when its argument is non-negative. Therefore, the expression under the square root must satisfy:

$$
3x - 5 \geq 0 \Rightarrow x \geq \frac{5}{3}
$$

Since $ x = 3 $ satisfies this condition, the solution lies within the valid domain.

---

### Final Conclusion

There are no other possible values of $ x $ that satisfy the original equation, because:

- Squaring both sides yields a **single unique solution**,
- The solution has been verified to satisfy the original equation,
- No other solution exists due to the behavior of the square root function and the nature of the equation.

---

### Final Answer

$$
\boxed{3}
$$

[Assistant]
Explanation:
Okay, so I have this equation here: √(3x - 5) = 2. I need to find all possible values of x. Hmm, let me think. Alright, square roots can sometimes be tricky because they only give the principal (non-negative) root, right? So, if I have a square root equal to a number, I have to make sure that the expression inside the square root is non-negative. Also, when I solve equations with square roots, I usually square both sides to eliminate the square root. Let me try that here.

First, let me write down the equation again to make sure I have it right:

√(3x - 5) = 2

Alright, so if I square both sides of the equation, that should get rid of the square root. Let me do that step by step. Squaring the left side: (√(3x - 5))² is just 3x - 5, because squaring a square root cancels them out. And squaring the right side: 2² is 4. So, after squaring both sides, the equation becomes:

3x - 5 = 4

Okay, that seems straightforward. Now I just need to solve for x. Let me add 5 to both sides to isolate the term with x. So:

3x - 5 + 5 = 4 + 5

Simplifying both sides:

3x = 9

Now, divide both sides by 3:

3x / 3 = 9 / 3

Which simplifies to:

x = 3

Wait, so x equals 3? Let me check if that's correct. Sometimes when you square both sides of an equation, you can introduce extraneous solutions, so I should verify that this solution actually works in the original equation.

Let me substitute x = 3 back into the original equation:

√(3*3 - 5) = √(9 - 5) = √4 = 2

Which is indeed equal to the right side of the equation, which is 2. So that works. So x = 3 is a valid solution. But the question asks for all possible values of x. Is there another solution?

Hmm, let me think. When you square both sides of an equation, sometimes you can get extra solutions, but in this case, since the square root function only outputs non-negative numbers, and we have √(3x - 5) equals 2, which is positive. So, squaring both sides should not introduce any extraneous solutions here because we already have the square root equal to a positive number. So, maybe there's only one solution here. Let me check if there's any other possibility.

Wait, could there be a negative solution? Let me think. Suppose someone might think, "Oh, maybe the square root could be negative, so maybe -2 is also a solution?" But no, the square root symbol specifically denotes the principal (non-negative) root. So √(something) is always non-negative. Therefore, even if we have an equation like √(something) = -2, that would have no solution because a square root can't be negative. In this case, though, the right side is 2, which is positive, so we don't have that problem.

So, going back, only x = 3 is the solution. Let me just check if the expression inside the square root is non-negative for x = 3. 3x - 5 when x = 3 is 9 - 5 = 4, which is positive. So that's okay. So there are no issues with the domain here. Therefore, x = 3 is the only solution.

But wait, the question says "all possible values of x." Maybe there's another value? Let me think again. Let me consider solving the equation step by step once more.

Original equation:

√(3x - 5) = 2

Square both sides:

(√(3x - 5))² = 2²

Which simplifies to:

3x - 5 = 4

Then, solving for x:

3x = 4 + 5

3x = 9

x = 3

So same result. There's no other solution here. Maybe if I think about the definition of the square root function, it's only defined for non-negative numbers. So 3x - 5 must be greater than or equal to zero. So 3x - 5 ≥ 0 ⇒ 3x ≥ 5 ⇒ x ≥ 5/3 ≈ 1.666...

But x = 3 is definitely in that domain. So there's no other solution. Therefore, the only possible value is 3.

Wait, but just to make sure, is there a case where squaring both sides could introduce another solution? For example, if the original equation had something like √(something) = something else, but here the right side is 2. Let me think of an example. Suppose if the equation was √(x) = x, then squaring both sides would give x = x², leading to x² - x = 0 ⇒ x(x - 1) = 0 ⇒ x = 0 or x = 1. But then you have to check both solutions. In this case, x = 0: √0 = 0, which is okay. x = 1: √1 = 1, which is also okay. So both are valid. But in our original problem, squaring only gave us one solution, which worked. So maybe in this case, there's only one solution.

Alternatively, if there was a ± symbol involved, but in this case, there's no ±. The original equation is √(something) equals 2, so we don't need to consider negative roots. Therefore, x = 3 is the only solution.

Therefore, I think the answer is x = 3. Let me check once more.

Original equation:

√(3x - 5) = 2

Plug in x = 3:

Left side: √(3*3 -5) = √(9 -5) = √4 = 2. Right side is 2. So yes, it works. Any other x?

Suppose someone says, what if 3x - 5 was another square that equals 4? But no, because we squared both sides and got 3x -5 = 4, so there's no other value for 3x -5. So 3x -5 is uniquely determined as 4. Therefore, x is uniquely determined as 3. Therefore, only one solution.

Therefore, the only possible value of x is 3.

**Final Answer**
The only possible value of $ x $ is \boxed{3}.

To solve the equation:

$$
\sqrt{3x - 5} = 2
$$

we begin by eliminating the square root through **squaring both sides** of the equation. This is a valid operation as long as we remember that the square root function is defined only for **non-negative** values and that squaring both sides may occasionally introduce extraneous solutions, which need to be verified.

---

### Step 1: Square Both Sides

$$
\left(\sqrt{3x - 5}\right)^2 = 2^2
$$

$$
3x - 5 = 4
$$

---

### Step 2: Solve for $ x $

Now solve the resulting linear equation:

$$
3x - 5 = 4
$$

Add 5 to both sides:

$$
3x = 9
$$

Divide both sides by 3:

$$
x = 3
$$

---

### Step 3: Verify the Solution

It's essential to verify that this value of $ x $ satisfies the original equation. Substitute $ x = 3 $ into the original expression:

$$
\sqrt{3(3) - 5} = \sqrt{9 - 5} = \sqrt{4} = 2
$$

This matches the right-hand side of the equation, confirming that $ x = 3 $ is indeed a valid solution.

---

### Step 4: Consider the Domain

The square root function is **only defined** when its argument is non-negative. Therefore, the expression under the square root must satisfy:

$$
3x - 5 \geq 0 \Rightarrow x \geq \frac{5}{3}
$$

Since $ x = 3 $ satisfies this condition, the solution lies within the valid domain.

---

### Final Conclusion

There are no other possible values of $ x $ that satisfy the original equation, because:

- Squaring both sides yields a **single unique solution**,
- The solution has been verified to satisfy the original equation,
- No other solution exists due to the behavior of the square root function and the nature of the equation.

---

### Final Answer

$$
\boxed{3}
$$

Solution:
First, we start by squaring both sides of the equation \begin{align*} (\sqrt{3x-5})^2& =(2)^2
\\ \Rightarrow\qquad 3x-5& =4
\\\Rightarrow\qquad 3x& =9
\\\Rightarrow\qquad x& =\boxed{3}.
\end{align*}Testing, we find that this value of $x$ does indeed satisfy the equation.


Explanation:
Okay, let me try to figure out whether the composition of these functions, f(f(g(f(g(f(x)))))) is even, odd, or neither. Alright, let's start by recalling what even and odd functions are.

An even function satisfies the condition that g(-x) = g(x) for all x in its domain. On the other hand, an odd function satisfies f(-x) = -f(x) for all x. So, given that f is odd and g is even, I need to analyze how these properties propagate through the composition of functions.

The function we need to check is f(f(g(f(g(f(x)))))). That's a bit of a complex composition. Let me write it down step by step to see if I can track the transformations. Let me denote each step:

Let me define:

Let’s denote the innermost function first. Let’s start from the inside and work our way outwards. Let me think about how each function affects the input when we replace x with -x.

Let me consider the entire composition function, let's call it H(x) = f(f(g(f(g(f(x)))))).

We need to find H(-x) and compare it to H(x) and -H(x) to determine if it's even, odd, or neither.

So, H(-x) = f(f(g(f(g(f(-x)))))).

Now, let me analyze each layer step by step. Let's start with the innermost function, which is f(x). Let me denote each composition step:

Let me define:

Let’s break down H(x):

Starting from the inside:

First, we have f(x). Then, we apply g to that result: g(f(x)). Then, apply f to that result: f(g(f(x))). Then, apply g to that: g(f(g(f(x)))). Then apply f to that: f(g(f(g(f(x))))). Then apply f again to that: f(f(g(f(g(f(x)))))).

Wait, but maybe breaking it down step by step is too abstract. Let me try to substitute step by step.

Let me start by computing H(-x):

H(-x) = f(f(g(f(g(f(-x)))))).

Now, let's work from the inside out. Let's start with f(-x). Since f is odd, f(-x) = -f(x). So, replacing f(-x) with -f(x):

So, f(-x) = -f(x). Therefore, f(-x) is -f(x). Let me write that down.

So, H(-x) = f(f(g(f(g(-f(x)))))).

Wait, because the innermost function is f(-x) = -f(x). Then, moving out, the next layer is g applied to that result. Let me write each step:

Let me denote:

Let’s define step by step:

Let me define:

Let’s call the innermost function as A = f(x). Then, next layer is B = g(A) = g(f(x)). Then, next is C = f(B) = f(g(f(x))). Then, next is D = g(C) = g(f(g(f(x)))). Then, next is E = f(D) = f(g(f(g(f(x))))). Then, finally, H(x) = f(E) = f(f(g(f(g(f(x)))))).

Similarly, for H(-x), we start with f(-x) = -f(x). Let's do the same steps but replacing x with -x:

Starting with f(-x) = -f(x). Then, the next layer is g applied to that: g(-f(x)). Now, since g is even, g(-y) = g(y) for any y. Therefore, g(-f(x)) = g(f(x)). Therefore, that layer becomes g(f(x)).

So, proceeding:

H(-x) = f(f(g(f(g(f(-x))))) = f(f(g(f(g(-f(x))))) = f(f(g(f(g(-f(x)))))

Now, let me go step by step again:

First layer: f(-x) = -f(x)

Second layer: g(f(-x)) = g(-f(x)) = g(f(x)) because g is even.

Third layer: f(g(f(-x))) = f(g(f(x))).

Fourth layer: g(f(g(f(-x)))) = g(f(g(f(x)))) because we just saw that f(g(f(-x))) is f(g(f(x))).

Fifth layer: f(g(f(g(f(-x))))) = f(g(f(g(f(x))))).

Therefore, H(-x) is f(f(g(f(g(f(x))))) = H(x)?

Wait, that seems like H(-x) = H(x)? So that would mean H(x) is even?

Wait, hold on. Wait, let me check again step by step. Maybe I missed a step.

Wait, H(-x) is f(f(g(f(g(f(-x)))))).

First, f(-x) is -f(x). Then, inside the next function, we have f(g(f(-x)))?

Wait, let me write each step more carefully.

Let me define:

Let me start with the innermost function: f(-x) = -f(x).

Now, moving outwards, the next function is g applied to that: g(f(-x)) = g(-f(x)).

Since g is even, g(-f(x)) = g(f(x)).

Next, moving outwards, we apply f to that result: f(g(f(-x))) = f(g(f(x))).

Next, moving outwards, we apply g to that result: g(f(g(f(-x)))) = g(f(g(f(x)))).

Next, moving outwards, we apply f to that result: f(g(f(g(f(-x))))) = f(g(f(g(f(x))))).

Finally, the outermost function is f applied to that: f(f(g(f(g(f(-x)))))) = f(f(g(f(g(f(x)))))).

Which is exactly H(x). Therefore, H(-x) = H(x). Therefore, H is even?

Wait, that would mean that the entire composition is even? But let me check again, since this seems surprising. Let me go through each step again, perhaps I made a mistake.

Let me try again:

H(-x) = f(f(g(f(g(f(-x)))))).

Starting from the inside:

First, f(-x) = -f(x).

Then, the next layer is f(-x) inside g: g(f(-x)) = g(-f(x)).

Since g is even, g(-f(x)) = g(f(x)).

Then, the next layer is f(g(f(-x))) = f(g(f(x))).

Then, the next layer is g(f(g(f(-x)))) = g(f(g(f(x)))).

Then, the next layer is f(g(f(g(f(-x))))) = f(g(f(g(f(x))))).

Then, the outermost f: f(f(g(f(g(f(x))))) = H(x).

Therefore, H(-x) = H(x). Therefore, H(x) is even?

But let me check with an example. Maybe take specific functions for f and g to test.

Let me take f(x) as an odd function. Let's say f(x) = x, which is odd. And g(x) as an even function, say g(x) = x².

Then let's compute H(x) = f(f(g(f(g(f(x)))))).

Let me compute H(x):

First compute f(x): x.

Then g(f(x)) = g(x) = x².

Then f(g(f(x))) = f(x²) = x².

Then g(f(g(f(x)))) = g(x²) = (x²)² = x⁴.

Then f(g(f(g(f(x))))) = f(x⁴) = x⁴.

Then H(x) = f(f(g(f(g(f(x))))) = f(x⁴) = x⁴.

So H(x) = x⁴, which is even. Therefore, in this case, H(x) is even. So that matches our previous conclusion.

But let me try another example to check.

Take f(x) = x³ (odd), and g(x) = x² (even).

Compute H(x) = f(f(g(f(g(f(x)))))).

Compute step by step:

Start with f(x) = x³.

Then g(f(x)) = g(x³) = (x³)² = x⁶.

Then f(g(f(x))) = f(x⁶) = (x⁶)³ = x¹⁸.

Then g(f(g(f(x)))) = g(x¹⁸) = (x¹⁸)² = x³⁶.

Then f(g(f(g(f(x))))) = f(x³⁶) = (x³⁶)³ = x¹⁰⁸.

Then H(x) = f(f(g(f(g(f(x))))) = f(x¹⁰⁸) = (x¹⁰⁸)³ = x³²⁴.

Which is x³²⁴. Which is even, since (-x)³²⁴ = x³²⁴. Therefore, again H(x) is even.

Hmm, interesting. So both examples give me even functions. So maybe H(x) is even? Wait, but maybe I should test a case where the composition could be odd?

Wait, but maybe the answer is even. But let me think again. Let me check with another example where maybe f is a different odd function.

Take f(x) = sin(x), which is odd, and g(x) = cos(x), which is even. Let me compute H(x) = f(f(g(f(g(f(x)))))).

Let me compute step by step:

Start with f(x) = sin(x).

g(f(x)) = cos(sin(x)).

f(g(f(x))) = sin(cos(sin(x))).

g(f(g(f(x)))) = cos(sin(cos(sin(x)))).

f(g(f(g(f(x))))) = sin(cos(sin(cos(sin(x))))).

H(x) = f(f(g(f(g(f(x))))) = sin(sin(cos(sin(cos(sin(x)))))).

Now, let's compute H(-x):

H(-x) = sin(sin(cos(sin(cos(sin(-x)))))).

But sin(-x) = -sin(x). Let me substitute step by step:

Start with f(-x) = sin(-x) = -sin(x).

Then, g(f(-x)) = cos(-sin(x)) = cos(sin(x)) because cos is even.

Then, f(g(f(-x))) = sin(cos(sin(x))).

Then, g(f(g(f(-x)))) = cos(sin(cos(sin(x)))).

Then, f(g(f(g(f(-x))))) = sin(cos(sin(cos(sin(x))))).

Finally, H(-x) = f(f(g(f(g(f(-x))))) = sin(sin(cos(sin(cos(sin(x)))))) = H(x).

Therefore, H(-x) = H(x), hence H(x) is even.

So even in this case, H(x) is even. Therefore, maybe regardless of the specific odd and even functions, the composition is even?

Wait, but let me check if there is a case where it's odd?

Wait, let me think again. Suppose there is a function where the composition becomes odd. But according to the previous reasoning, H(-x) = H(x). Therefore, it's even.

But let me think again about the general case. Let me try to formalize the steps.

Given that f is odd and g is even, let me analyze the composition.

Let me consider H(x) = f(f(g(f(g(f(x)))))).

Let me try to compute H(-x):

H(-x) = f(f(g(f(g(f(-x))))) )

As f is odd, f(-x) = -f(x). Therefore, f(-x) = -f(x).

So, substitute into H(-x):

H(-x) = f(f(g(f(g(-f(x))))) )

Now, since g is even, g(-y) = g(y) for any y. Therefore, g(-f(x)) = g(f(x)).

Therefore:

H(-x) = f(f(g(f(g(f(x))))) )

But notice that the expression inside is f(g(f(g(f(x))))) which is the same as the inner part of H(x). Wait, H(x) is f(f(g(f(g(f(x)))))).

Wait, let me check:

H(x) = f(f(g(f(g(f(x)))))).

H(-x) = f(f(g(f(g(-f(x))))) ) = f(f(g(f(g(f(x))))) ). So yes, H(-x) = H(x). Therefore, H is even.

Therefore, regardless of the specific odd and even functions, this composition is even. Therefore, the answer is even.

Wait, but let me check once again for errors in reasoning. Maybe I missed a step?

Let me try to think about the number of f's and g's. Let me count. The composition is f composed with f composed with g composed with f composed with g composed with f. So:

Starting from the inside:

f(x) → then g(f(x)) → then f(g(f(x))) → then g(f(g(f(x)))) → then f(g(f(g(f(x))))) → then f(f(g(f(g(f(x)))))).

So, total steps: starting with f(x), then alternating between g and f?

Wait, let me think in terms of the number of function applications. Let me consider how each function affects the parity.

Let me recall that composing functions can affect the parity. For example, composing an even function with another function can sometimes preserve or change the parity. Similarly with odd functions.

But maybe we can think about the composition step by step. Let me think about the number of times we apply odd and even functions.

But perhaps another way is to consider the effect of replacing x with -x. Let me track how the parity changes through each function.

Starting with H(x) = f(f(g(f(g(f(x)))))).

Let me define each layer:

Let me denote:

Let’s define:

Let’s let’s denote:

Let’s define the innermost function as:

Layer 1: f(x)

Layer 2: g(Layer 1) = g(f(x))

Layer 3: f(Layer 2) = f(g(f(x)))

Layer 4: g(Layer 3) = g(f(g(f(x))))

Layer 5: f(Layer 4) = f(g(f(g(f(x)))))

Layer 6: f(Layer 5) = f(f(g(f(g(f(x)))))) = H(x)

So, each layer is:

Layer 1: f(x) [odd]

Layer 2: g(Layer 1) [even applied to odd: even * odd? Wait, not sure. But since g is even, and Layer 1 is f(x). Let me think about the parity of each layer.

But maybe instead of parity, think about how each layer transforms when x is replaced by -x.

But perhaps this is complicating. Alternatively, since we already did the substitution and found that H(-x) = H(x), maybe this is sufficient?

Alternatively, let me think about the parity of each layer.

Layer 1: f(x) is odd. So, Layer1(-x) = -Layer1(x).

Layer 2: g(Layer1(x)). Since g is even, Layer2(-x) = g(Layer1(-x)) = g(-Layer1(x)) = g(Layer1(x)) = Layer2(x). Therefore, Layer2 is even.

Layer3: f(Layer2(x)). Since Layer2 is even, then Layer3(-x) = f(Layer2(-x)) = f(Layer2(x)) = Layer3(x). Therefore, Layer3 is even? Wait, no: Wait, if Layer2 is even, then f(Layer2(x)) is f applied to an even function. Since f is odd, what is the parity of f applied to an even function?

Let me think. Suppose we have a function h(x) = f(k(x)), where k(x) is even. Then h(-x) = f(k(-x)) = f(k(x)) = h(x). Therefore, h(x) is even. Because applying f to an even function gives an even function? Wait, no, not necessarily. Wait, if k(x) is even, then k(-x) = k(x). So, h(-x) = f(k(-x)) = f(k(x)) = h(x). Therefore, h(x) is even. Therefore, Layer3 is even.

Layer4: g(Layer3(x)). Since Layer3 is even, and g is even, then Layer4(-x) = g(Layer3(-x)) = g(Layer3(x)) = Layer4(x). Therefore, Layer4 is even.

Layer5: f(Layer4(x)). Since Layer4 is even, then Layer5(-x) = f(Layer4(-x)) = f(Layer4(x)) = -f(Layer4(x))? Wait, no. Wait, f is odd. If Layer4(x) is even, then f(Layer4(x)) is f applied to an even function. Let me denote m(x) = f(k(x)), where k(x) is even. Then m(-x) = f(k(-x)) = f(k(x)) = m(x). Wait, no: Wait, f is odd, so m(-x) = f(k(-x)) = f(k(x)) because k is even. So m(-x) = f(k(x)) = m(x). Therefore, m(x) is even? Wait, that's conflicting with my previous thought. Wait, if m(x) = f(k(x)), where k is even, then m(-x) = f(k(-x)) = f(k(x)) = m(x). Therefore, m(x) is even. Therefore, Layer5 is even?

Wait, but then Layer5 is even. Therefore, Layer5(-x) = Layer5(x). Then Layer6 is f(Layer5(x)). So, Layer6(-x) = f(Layer5(-x)) = f(Layer5(x)) = Layer6(x). Wait, but f is odd. So if Layer5 is even, then f(Layer5(x)) is f applied to an even function. Which, as we saw, would be even? Wait, no. Wait, if you have an even function m(x), then f(m(x)) would be... Let's check:

Let me take m(x) even, then f(m(-x)) = f(m(x)) because m is even. But since f is odd, f(m(-x)) = -f(-m(x))? Wait, no. Wait, f is odd, so f(-y) = -f(y). Therefore, f(m(-x)) = f(m(x)) because m(-x) = m(x). Therefore, f(m(-x)) = f(m(x)), which is the same as f(m(x)). Therefore, f(m(x)) is even? Because f(m(-x)) = f(m(x)), which is the definition of even function. Therefore, f(m(x)) is even. Therefore, Layer6 is even? Therefore, H(x) = Layer6 is even. Therefore, H(-x) = H(x). Therefore, H is even.

Therefore, regardless of the functions, as long as f is odd and g is even, the composition is even. Therefore, the answer should be even.

But let me check again with the previous layers. Let me check again:

Layer1: odd

Layer2: even

Layer3: f applied to even: even

Layer4: g applied to even: even

Layer5: f applied to even: even

Layer6: f applied to even: even. Therefore, H(x) is even.

Therefore, yes, regardless of the number of layers, as long as we start with layer1 odd, and then each subsequent layer alternates between g (even) and f (odd). Wait, but in our case, the layers are: f, g, f, g, f, f. Wait, let me check:

Wait, H(x) is f(f(g(f(g(f(x)))))). So, starting from innermost:

Start with f(x) [layer1: f]

Then g(f(x)) [layer2: g]

Then f(g(f(x))) [layer3: f]

Then g(f(g(f(x)))) [layer4: g]

Then f(g(f(g(f(x))))) [layer5: f]

Then f(f(g(f(g(f(x))))) [layer6: f]

So, layers: f, g, f, g, f, f.

So, starting with f (odd), then g (even), f (odd), g (even), f (odd), f (odd). So, the last layer is f again. However, through the previous analysis, each subsequent layer's parity was determined based on previous.

But as we saw, layer1: odd

layer2: even

layer3: even (because f applied to even)

layer4: even (g applied to even)

layer5: even (f applied to even)

layer6: even (f applied to even). So yes, layer6 is even. Therefore, H(x) is even.

Therefore, the composition is even.

Therefore, the answer is even.

But let me check once more with a different approach. Let me think about the entire composition's parity.

We can think of the composition as a series of operations. Let me note that:

- Composing an even function with another function preserves or changes the parity based on the function's parity.

But since we are dealing with multiple compositions, maybe we can track the number of odd functions.

But here's a thought: each time we compose an odd function with another function, it can invert the parity, while even functions preserve it. However, since we are composing multiple functions, it's a bit more complex.

Alternatively, since we have already shown through substitution that H(-x) = H(x), which is the definition of even function, and verified with examples, perhaps that's sufficient.

Alternatively, maybe think of the entire composition as follows:

Start with x. Apply f (odd), then g (even), then f (odd), then g (even), then f (odd), then f (odd). Let me think about the total number of odd functions. Each odd function can flip the sign, but since they are nested, it might cancel out.

But let me think in terms of the overall operation:

Let me think about replacing x with -x and track the sign changes.

Let me denote:

Start with x. Replace with -x.

First, the innermost f(x) becomes -f(x).

Then, the next layer is g(f(x)) which becomes g(-f(x)) = g(f(x)) since g is even.

Then, the next layer is f(g(f(x))) which becomes f(g(f(x))) because the previous layer was even.

Then, the next layer is g(f(g(f(x)))) which becomes g(f(g(f(x)))) since the previous layer was even.

Then, the next layer is f(g(f(g(f(x))))) which becomes f(g(f(g(f(x))))) because the previous layer was even.

Finally, the outermost f applied to that: f(...) becomes -f(...) because f is odd. Wait, but hold on:

Wait, hold on. Wait, let me track again carefully:

Starting with replacing x by -x:

1. Start with f(x) → becomes -f(x)

2. Then, apply g to that: g(-f(x)) = g(f(x))

3. Then apply f to that: f(g(f(x))) → which is the same as before, since the input is g(f(x)), which is same as before. Wait, but when replacing x by -x, we need to track how each step affects.

Wait, perhaps my previous reasoning was wrong? Let me think again step by step.

Let me think of H(x) as a composition of 6 functions:

Let me write H(x) = f_6(f_5(f_4(f_3(f_2(f_1(x))))))

Where f_1(x) = f(x)

f_2(x) = g(f(x))

f_3(x) = f(g(f(x)))

f_4(x) = g(f(g(f(x))))

f_5(x) = f(g(f(g(f(x)))))

f_6(x) = f(f(g(f(g(f(x)))))) = H(x)

But when we replace x with -x, we have:

H(-x) = f_6(f_5(f_4(f_3(f_2(f_1(-x))))))

Now, let me compute each step:

Start from the inside:

f_1(-x) = f(-x) = -f(x)

Then f_2(-x) = g(f_1(-x)) = g(-f(x)) = g(f(x)) because g is even

Then f_3(-x) = f(f_2(-x)) = f(g(f(x))) = f_3(x)

Then f_4(-x) = g(f_3(-x)) = g(f_3(x)) = f_4(x)

Then f_5(-x) = f(f_4(-x)) = f(f_4(x)) = f_5(x)

Then f_6(-x) = f(f_5(-x)) = f(f_5(x)) = f_6(x)

Therefore, H(-x) = H(x). Therefore, H is even.

Yes, this is the same conclusion as before. Therefore, regardless of the specific functions, as long as f is odd and g is even, the composition H is even.

Therefore, the answer should be even.

But just to make sure, let me think if there is any scenario where this might not hold. For example, if the number of odd functions is odd, but in this case, we have multiple odd functions. Wait, but each application of an odd function depends on the previous result. However, due to the even functions being in between, they might have canceled out the sign changes.

In the above step-by-step analysis, every time we applied an odd function (f), it was applied to an even function, which didn't change the sign. Therefore, the sign flip from f was negated by the even functions. For example, when we had f applied to an even function, which gave us an even function again. Wait, but in the final step, we have f applied to an even function (f_5(x)). Since f is odd, f applied to an even function is even? Wait, earlier analysis said that f applied to an even function is even.

Yes, for example, suppose m(x) is even, then f(m(x)) is even because f(m(-x)) = f(m(x)). Hence, f(m(x)) is even. Therefore, even though f is odd, when applied to an even function, the result is even. Therefore, all layers after the first become even functions, and the final application of f is to an even function, resulting in an even function. Therefore, H(x) is even.

Therefore, I'm pretty confident the answer is even.

**Final Answer**
\boxed{even}

To determine whether the function
$$
f(f(g(f(g(f(x))))))
$$
is **even**, **odd**, or **neither**, we start by understanding the key properties of the given functions:

- $ f(x) $ is an **odd** function, so we have:
  $$
  f(-x) = -f(x)
  $$
- $ g(x) $ is an **even** function, so we have:
  $$
  g(-x) = g(x)
  $$

Our goal is to evaluate the **parity** of the composite function
$$
H(x) = f(f(g(f(g(f(x))))))
$$
by computing $ H(-x) $ and comparing it with $ H(x) $ and $ -H(x) $.

---

### Step-by-step Evaluation of $ H(-x) $

We proceed from the **innermost** to the **outermost** function in the domain:

1. **Innermost:**
   $$
   f(-x) = -f(x)
   $$

2. **Apply $ g $:**
   $$
   g(f(-x)) = g(-f(x)) = g(f(x)) \quad \text{(since $ g $ is even)}
   $$

3. **Apply $ f $:**
   $$
   f(g(f(-x))) = f(g(f(x)))
   $$

4. **Apply $ g $:**
   $$
   g(f(g(f(-x)))) = g(f(g(f(x))))
   $$

5. **Apply $ f $:**
   $$
   f(g(f(g(f(-x))))) = f(g(f(g(f(x)))))
   $$

6. **Apply final $ f $:**
   $$
   H(-x) = f(f(g(f(g(f(-x))))) = f(f(g(f(g(f(x))))) = H(x)
   $$

---

### Conclusion

We have shown that the expression $ H(-x) $ simplifies to $ H(x) $. That is:

$$
H(-x) = H(x)
$$

This **defines an even function**. Therefore, the composition
$$
f(f(g(f(g(f(x))))))
$$
is **even**.

---

### Final Answer:
$$
\boxed{even}
$$

[Assistant]
Explanation:
Okay, let me try to figure out whether the composition of these functions, f(f(g(f(g(f(x)))))) is even, odd, or neither. Alright, let's start by recalling what even and odd functions are.

An even function satisfies the condition that g(-x) = g(x) for all x in its domain. On the other hand, an odd function satisfies f(-x) = -f(x) for all x. So, given that f is odd and g is even, I need to analyze how these properties propagate through the composition of functions.

The function we need to check is f(f(g(f(g(f(x)))))). That's a bit of a complex composition. Let me write it down step by step to see if I can track the transformations. Let me denote each step:

Let me define:

Let’s denote the innermost function first. Let’s start from the inside and work our way outwards. Let me think about how each function affects the input when we replace x with -x.

Let me consider the entire composition function, let's call it H(x) = f(f(g(f(g(f(x)))))).

We need to find H(-x) and compare it to H(x) and -H(x) to determine if it's even, odd, or neither.

So, H(-x) = f(f(g(f(g(f(-x)))))).

Now, let me analyze each layer step by step. Let's start with the innermost function, which is f(x). Let me denote each composition step:

Let me define:

Let’s break down H(x):

Starting from the inside:

First, we have f(x). Then, we apply g to that result: g(f(x)). Then, apply f to that result: f(g(f(x))). Then, apply g to that: g(f(g(f(x)))). Then apply f to that: f(g(f(g(f(x))))). Then apply f again to that: f(f(g(f(g(f(x)))))).

Wait, but maybe breaking it down step by step is too abstract. Let me try to substitute step by step.

Let me start by computing H(-x):

H(-x) = f(f(g(f(g(f(-x)))))).

Now, let's work from the inside out. Let's start with f(-x). Since f is odd, f(-x) = -f(x). So, replacing f(-x) with -f(x):

So, f(-x) = -f(x). Therefore, f(-x) is -f(x). Let me write that down.

So, H(-x) = f(f(g(f(g(-f(x)))))).

Wait, because the innermost function is f(-x) = -f(x). Then, moving out, the next layer is g applied to that result. Let me write each step:

Let me denote:

Let’s define step by step:

Let me define:

Let’s call the innermost function as A = f(x). Then, next layer is B = g(A) = g(f(x)). Then, next is C = f(B) = f(g(f(x))). Then, next is D = g(C) = g(f(g(f(x)))). Then, next is E = f(D) = f(g(f(g(f(x))))). Then, finally, H(x) = f(E) = f(f(g(f(g(f(x)))))).

Similarly, for H(-x), we start with f(-x) = -f(x). Let's do the same steps but replacing x with -x:

Starting with f(-x) = -f(x). Then, the next layer is g applied to that: g(-f(x)). Now, since g is even, g(-y) = g(y) for any y. Therefore, g(-f(x)) = g(f(x)). Therefore, that layer becomes g(f(x)).

So, proceeding:

H(-x) = f(f(g(f(g(f(-x))))) = f(f(g(f(g(-f(x))))) = f(f(g(f(g(-f(x)))))

Now, let me go step by step again:

First layer: f(-x) = -f(x)

Second layer: g(f(-x)) = g(-f(x)) = g(f(x)) because g is even.

Third layer: f(g(f(-x))) = f(g(f(x))).

Fourth layer: g(f(g(f(-x)))) = g(f(g(f(x)))) because we just saw that f(g(f(-x))) is f(g(f(x))).

Fifth layer: f(g(f(g(f(-x))))) = f(g(f(g(f(x))))).

Therefore, H(-x) is f(f(g(f(g(f(x))))) = H(x)?

Wait, that seems like H(-x) = H(x)? So that would mean H(x) is even?

Wait, hold on. Wait, let me check again step by step. Maybe I missed a step.

Wait, H(-x) is f(f(g(f(g(f(-x)))))).

First, f(-x) is -f(x). Then, inside the next function, we have f(g(f(-x)))?

Wait, let me write each step more carefully.

Let me define:

Let me start with the innermost function: f(-x) = -f(x).

Now, moving outwards, the next function is g applied to that: g(f(-x)) = g(-f(x)).

Since g is even, g(-f(x)) = g(f(x)).

Next, moving outwards, we apply f to that result: f(g(f(-x))) = f(g(f(x))).

Next, moving outwards, we apply g to that result: g(f(g(f(-x)))) = g(f(g(f(x)))).

Next, moving outwards, we apply f to that result: f(g(f(g(f(-x))))) = f(g(f(g(f(x))))).

Finally, the outermost function is f applied to that: f(f(g(f(g(f(-x)))))) = f(f(g(f(g(f(x)))))).

Which is exactly H(x). Therefore, H(-x) = H(x). Therefore, H is even?

Wait, that would mean that the entire composition is even? But let me check again, since this seems surprising. Let me go through each step again, perhaps I made a mistake.

Let me try again:

H(-x) = f(f(g(f(g(f(-x)))))).

Starting from the inside:

First, f(-x) = -f(x).

Then, the next layer is f(-x) inside g: g(f(-x)) = g(-f(x)).

Since g is even, g(-f(x)) = g(f(x)).

Then, the next layer is f(g(f(-x))) = f(g(f(x))).

Then, the next layer is g(f(g(f(-x)))) = g(f(g(f(x)))).

Then, the next layer is f(g(f(g(f(-x))))) = f(g(f(g(f(x))))).

Then, the outermost f: f(f(g(f(g(f(x))))) = H(x).

Therefore, H(-x) = H(x). Therefore, H(x) is even?

But let me check with an example. Maybe take specific functions for f and g to test.

Let me take f(x) as an odd function. Let's say f(x) = x, which is odd. And g(x) as an even function, say g(x) = x².

Then let's compute H(x) = f(f(g(f(g(f(x)))))).

Let me compute H(x):

First compute f(x): x.

Then g(f(x)) = g(x) = x².

Then f(g(f(x))) = f(x²) = x².

Then g(f(g(f(x)))) = g(x²) = (x²)² = x⁴.

Then f(g(f(g(f(x))))) = f(x⁴) = x⁴.

Then H(x) = f(f(g(f(g(f(x))))) = f(x⁴) = x⁴.

So H(x) = x⁴, which is even. Therefore, in this case, H(x) is even. So that matches our previous conclusion.

But let me try another example to check.

Take f(x) = x³ (odd), and g(x) = x² (even).

Compute H(x) = f(f(g(f(g(f(x)))))).

Compute step by step:

Start with f(x) = x³.

Then g(f(x)) = g(x³) = (x³)² = x⁶.

Then f(g(f(x))) = f(x⁶) = (x⁶)³ = x¹⁸.

Then g(f(g(f(x)))) = g(x¹⁸) = (x¹⁸)² = x³⁶.

Then f(g(f(g(f(x))))) = f(x³⁶) = (x³⁶)³ = x¹⁰⁸.

Then H(x) = f(f(g(f(g(f(x))))) = f(x¹⁰⁸) = (x¹⁰⁸)³ = x³²⁴.

Which is x³²⁴. Which is even, since (-x)³²⁴ = x³²⁴. Therefore, again H(x) is even.

Hmm, interesting. So both examples give me even functions. So maybe H(x) is even? Wait, but maybe I should test a case where the composition could be odd?

Wait, but maybe the answer is even. But let me think again. Let me check with another example where maybe f is a different odd function.

Take f(x) = sin(x), which is odd, and g(x) = cos(x), which is even. Let me compute H(x) = f(f(g(f(g(f(x)))))).

Let me compute step by step:

Start with f(x) = sin(x).

g(f(x)) = cos(sin(x)).

f(g(f(x))) = sin(cos(sin(x))).

g(f(g(f(x)))) = cos(sin(cos(sin(x)))).

f(g(f(g(f(x))))) = sin(cos(sin(cos(sin(x))))).

H(x) = f(f(g(f(g(f(x))))) = sin(sin(cos(sin(cos(sin(x)))))).

Now, let's compute H(-x):

H(-x) = sin(sin(cos(sin(cos(sin(-x)))))).

But sin(-x) = -sin(x). Let me substitute step by step:

Start with f(-x) = sin(-x) = -sin(x).

Then, g(f(-x)) = cos(-sin(x)) = cos(sin(x)) because cos is even.

Then, f(g(f(-x))) = sin(cos(sin(x))).

Then, g(f(g(f(-x)))) = cos(sin(cos(sin(x)))).

Then, f(g(f(g(f(-x))))) = sin(cos(sin(cos(sin(x))))).

Finally, H(-x) = f(f(g(f(g(f(-x))))) = sin(sin(cos(sin(cos(sin(x)))))) = H(x).

Therefore, H(-x) = H(x), hence H(x) is even.

So even in this case, H(x) is even. Therefore, maybe regardless of the specific odd and even functions, the composition is even?

Wait, but let me check if there is a case where it's odd?

Wait, let me think again. Suppose there is a function where the composition becomes odd. But according to the previous reasoning, H(-x) = H(x). Therefore, it's even.

But let me think again about the general case. Let me try to formalize the steps.

Given that f is odd and g is even, let me analyze the composition.

Let me consider H(x) = f(f(g(f(g(f(x)))))).

Let me try to compute H(-x):

H(-x) = f(f(g(f(g(f(-x))))) )

As f is odd, f(-x) = -f(x). Therefore, f(-x) = -f(x).

So, substitute into H(-x):

H(-x) = f(f(g(f(g(-f(x))))) )

Now, since g is even, g(-y) = g(y) for any y. Therefore, g(-f(x)) = g(f(x)).

Therefore:

H(-x) = f(f(g(f(g(f(x))))) )

But notice that the expression inside is f(g(f(g(f(x))))) which is the same as the inner part of H(x). Wait, H(x) is f(f(g(f(g(f(x)))))).

Wait, let me check:

H(x) = f(f(g(f(g(f(x)))))).

H(-x) = f(f(g(f(g(-f(x))))) ) = f(f(g(f(g(f(x))))) ). So yes, H(-x) = H(x). Therefore, H is even.

Therefore, regardless of the specific odd and even functions, this composition is even. Therefore, the answer is even.

Wait, but let me check once again for errors in reasoning. Maybe I missed a step?

Let me try to think about the number of f's and g's. Let me count. The composition is f composed with f composed with g composed with f composed with g composed with f. So:

Starting from the inside:

f(x) → then g(f(x)) → then f(g(f(x))) → then g(f(g(f(x)))) → then f(g(f(g(f(x))))) → then f(f(g(f(g(f(x)))))).

So, total steps: starting with f(x), then alternating between g and f?

Wait, let me think in terms of the number of function applications. Let me consider how each function affects the parity.

Let me recall that composing functions can affect the parity. For example, composing an even function with another function can sometimes preserve or change the parity. Similarly with odd functions.

But maybe we can think about the composition step by step. Let me think about the number of times we apply odd and even functions.

But perhaps another way is to consider the effect of replacing x with -x. Let me track how the parity changes through each function.

Starting with H(x) = f(f(g(f(g(f(x)))))).

Let me define each layer:

Let me denote:

Let’s define:

Let’s let’s denote:

Let’s define the innermost function as:

Layer 1: f(x)

Layer 2: g(Layer 1) = g(f(x))

Layer 3: f(Layer 2) = f(g(f(x)))

Layer 4: g(Layer 3) = g(f(g(f(x))))

Layer 5: f(Layer 4) = f(g(f(g(f(x)))))

Layer 6: f(Layer 5) = f(f(g(f(g(f(x)))))) = H(x)

So, each layer is:

Layer 1: f(x) [odd]

Layer 2: g(Layer 1) [even applied to odd: even * odd? Wait, not sure. But since g is even, and Layer 1 is f(x). Let me think about the parity of each layer.

But maybe instead of parity, think about how each layer transforms when x is replaced by -x.

But perhaps this is complicating. Alternatively, since we already did the substitution and found that H(-x) = H(x), maybe this is sufficient?

Alternatively, let me think about the parity of each layer.

Layer 1: f(x) is odd. So, Layer1(-x) = -Layer1(x).

Layer 2: g(Layer1(x)). Since g is even, Layer2(-x) = g(Layer1(-x)) = g(-Layer1(x)) = g(Layer1(x)) = Layer2(x). Therefore, Layer2 is even.

Layer3: f(Layer2(x)). Since Layer2 is even, then Layer3(-x) = f(Layer2(-x)) = f(Layer2(x)) = Layer3(x). Therefore, Layer3 is even? Wait, no: Wait, if Layer2 is even, then f(Layer2(x)) is f applied to an even function. Since f is odd, what is the parity of f applied to an even function?

Let me think. Suppose we have a function h(x) = f(k(x)), where k(x) is even. Then h(-x) = f(k(-x)) = f(k(x)) = h(x). Therefore, h(x) is even. Because applying f to an even function gives an even function? Wait, no, not necessarily. Wait, if k(x) is even, then k(-x) = k(x). So, h(-x) = f(k(-x)) = f(k(x)) = h(x). Therefore, h(x) is even. Therefore, Layer3 is even.

Layer4: g(Layer3(x)). Since Layer3 is even, and g is even, then Layer4(-x) = g(Layer3(-x)) = g(Layer3(x)) = Layer4(x). Therefore, Layer4 is even.

Layer5: f(Layer4(x)). Since Layer4 is even, then Layer5(-x) = f(Layer4(-x)) = f(Layer4(x)) = -f(Layer4(x))? Wait, no. Wait, f is odd. If Layer4(x) is even, then f(Layer4(x)) is f applied to an even function. Let me denote m(x) = f(k(x)), where k(x) is even. Then m(-x) = f(k(-x)) = f(k(x)) = m(x). Wait, no: Wait, f is odd, so m(-x) = f(k(-x)) = f(k(x)) because k is even. So m(-x) = f(k(x)) = m(x). Therefore, m(x) is even? Wait, that's conflicting with my previous thought. Wait, if m(x) = f(k(x)), where k is even, then m(-x) = f(k(-x)) = f(k(x)) = m(x). Therefore, m(x) is even. Therefore, Layer5 is even?

Wait, but then Layer5 is even. Therefore, Layer5(-x) = Layer5(x). Then Layer6 is f(Layer5(x)). So, Layer6(-x) = f(Layer5(-x)) = f(Layer5(x)) = Layer6(x). Wait, but f is odd. So if Layer5 is even, then f(Layer5(x)) is f applied to an even function. Which, as we saw, would be even? Wait, no. Wait, if you have an even function m(x), then f(m(x)) would be... Let's check:

Let me take m(x) even, then f(m(-x)) = f(m(x)) because m is even. But since f is odd, f(m(-x)) = -f(-m(x))? Wait, no. Wait, f is odd, so f(-y) = -f(y). Therefore, f(m(-x)) = f(m(x)) because m(-x) = m(x). Therefore, f(m(-x)) = f(m(x)), which is the same as f(m(x)). Therefore, f(m(x)) is even? Because f(m(-x)) = f(m(x)), which is the definition of even function. Therefore, f(m(x)) is even. Therefore, Layer6 is even? Therefore, H(x) = Layer6 is even. Therefore, H(-x) = H(x). Therefore, H is even.

Therefore, regardless of the functions, as long as f is odd and g is even, the composition is even. Therefore, the answer should be even.

But let me check again with the previous layers. Let me check again:

Layer1: odd

Layer2: even

Layer3: f applied to even: even

Layer4: g applied to even: even

Layer5: f applied to even: even

Layer6: f applied to even: even. Therefore, H(x) is even.

Therefore, yes, regardless of the number of layers, as long as we start with layer1 odd, and then each subsequent layer alternates between g (even) and f (odd). Wait, but in our case, the layers are: f, g, f, g, f, f. Wait, let me check:

Wait, H(x) is f(f(g(f(g(f(x)))))). So, starting from innermost:

Start with f(x) [layer1: f]

Then g(f(x)) [layer2: g]

Then f(g(f(x))) [layer3: f]

Then g(f(g(f(x)))) [layer4: g]

Then f(g(f(g(f(x))))) [layer5: f]

Then f(f(g(f(g(f(x))))) [layer6: f]

So, layers: f, g, f, g, f, f.

So, starting with f (odd), then g (even), f (odd), g (even), f (odd), f (odd). So, the last layer is f again. However, through the previous analysis, each subsequent layer's parity was determined based on previous.

But as we saw, layer1: odd

layer2: even

layer3: even (because f applied to even)

layer4: even (g applied to even)

layer5: even (f applied to even)

layer6: even (f applied to even). So yes, layer6 is even. Therefore, H(x) is even.

Therefore, the composition is even.

Therefore, the answer is even.

But let me check once more with a different approach. Let me think about the entire composition's parity.

We can think of the composition as a series of operations. Let me note that:

- Composing an even function with another function preserves or changes the parity based on the function's parity.

But since we are dealing with multiple compositions, maybe we can track the number of odd functions.

But here's a thought: each time we compose an odd function with another function, it can invert the parity, while even functions preserve it. However, since we are composing multiple functions, it's a bit more complex.

Alternatively, since we have already shown through substitution that H(-x) = H(x), which is the definition of even function, and verified with examples, perhaps that's sufficient.

Alternatively, maybe think of the entire composition as follows:

Start with x. Apply f (odd), then g (even), then f (odd), then g (even), then f (odd), then f (odd). Let me think about the total number of odd functions. Each odd function can flip the sign, but since they are nested, it might cancel out.

But let me think in terms of the overall operation:

Let me think about replacing x with -x and track the sign changes.

Let me denote:

Start with x. Replace with -x.

First, the innermost f(x) becomes -f(x).

Then, the next layer is g(f(x)) which becomes g(-f(x)) = g(f(x)) since g is even.

Then, the next layer is f(g(f(x))) which becomes f(g(f(x))) because the previous layer was even.

Then, the next layer is g(f(g(f(x)))) which becomes g(f(g(f(x)))) since the previous layer was even.

Then, the next layer is f(g(f(g(f(x))))) which becomes f(g(f(g(f(x))))) because the previous layer was even.

Finally, the outermost f applied to that: f(...) becomes -f(...) because f is odd. Wait, but hold on:

Wait, hold on. Wait, let me track again carefully:

Starting with replacing x by -x:

1. Start with f(x) → becomes -f(x)

2. Then, apply g to that: g(-f(x)) = g(f(x))

3. Then apply f to that: f(g(f(x))) → which is the same as before, since the input is g(f(x)), which is same as before. Wait, but when replacing x by -x, we need to track how each step affects.

Wait, perhaps my previous reasoning was wrong? Let me think again step by step.

Let me think of H(x) as a composition of 6 functions:

Let me write H(x) = f_6(f_5(f_4(f_3(f_2(f_1(x))))))

Where f_1(x) = f(x)

f_2(x) = g(f(x))

f_3(x) = f(g(f(x)))

f_4(x) = g(f(g(f(x))))

f_5(x) = f(g(f(g(f(x)))))

f_6(x) = f(f(g(f(g(f(x)))))) = H(x)

But when we replace x with -x, we have:

H(-x) = f_6(f_5(f_4(f_3(f_2(f_1(-x))))))

Now, let me compute each step:

Start from the inside:

f_1(-x) = f(-x) = -f(x)

Then f_2(-x) = g(f_1(-x)) = g(-f(x)) = g(f(x)) because g is even

Then f_3(-x) = f(f_2(-x)) = f(g(f(x))) = f_3(x)

Then f_4(-x) = g(f_3(-x)) = g(f_3(x)) = f_4(x)

Then f_5(-x) = f(f_4(-x)) = f(f_4(x)) = f_5(x)

Then f_6(-x) = f(f_5(-x)) = f(f_5(x)) = f_6(x)

Therefore, H(-x) = H(x). Therefore, H is even.

Yes, this is the same conclusion as before. Therefore, regardless of the specific functions, as long as f is odd and g is even, the composition H is even.

Therefore, the answer should be even.

But just to make sure, let me think if there is any scenario where this might not hold. For example, if the number of odd functions is odd, but in this case, we have multiple odd functions. Wait, but each application of an odd function depends on the previous result. However, due to the even functions being in between, they might have canceled out the sign changes.

In the above step-by-step analysis, every time we applied an odd function (f), it was applied to an even function, which didn't change the sign. Therefore, the sign flip from f was negated by the even functions. For example, when we had f applied to an even function, which gave us an even function again. Wait, but in the final step, we have f applied to an even function (f_5(x)). Since f is odd, f applied to an even function is even? Wait, earlier analysis said that f applied to an even function is even.

Yes, for example, suppose m(x) is even, then f(m(x)) is even because f(m(-x)) = f(m(x)). Hence, f(m(x)) is even. Therefore, even though f is odd, when applied to an even function, the result is even. Therefore, all layers after the first become even functions, and the final application of f is to an even function, resulting in an even function. Therefore, H(x) is even.

Therefore, I'm pretty confident the answer is even.

**Final Answer**
\boxed{even}

To determine whether the function
$$
f(f(g(f(g(f(x))))))
$$
is **even**, **odd**, or **neither**, we start by understanding the key properties of the given functions:

- $ f(x) $ is an **odd** function, so we have:
  $$
  f(-x) = -f(x)
  $$
- $ g(x) $ is an **even** function, so we have:
  $$
  g(-x) = g(x)
  $$

Our goal is to evaluate the **parity** of the composite function
$$
H(x) = f(f(g(f(g(f(x))))))
$$
by computing $ H(-x) $ and comparing it with $ H(x) $ and $ -H(x) $.

---

### Step-by-step Evaluation of $ H(-x) $

We proceed from the **innermost** to the **outermost** function in the domain:

1. **Innermost:**
   $$
   f(-x) = -f(x)
   $$

2. **Apply $ g $:**
   $$
   g(f(-x)) = g(-f(x)) = g(f(x)) \quad \text{(since $ g $ is even)}
   $$

3. **Apply $ f $:**
   $$
   f(g(f(-x))) = f(g(f(x)))
   $$

4. **Apply $ g $:**
   $$
   g(f(g(f(-x)))) = g(f(g(f(x))))
   $$

5. **Apply $ f $:**
   $$
   f(g(f(g(f(-x))))) = f(g(f(g(f(x)))))
   $$

6. **Apply final $ f $:**
   $$
   H(-x) = f(f(g(f(g(f(-x))))) = f(f(g(f(g(f(x))))) = H(x)
   $$

---

### Conclusion

We have shown that the expression $ H(-x) $ simplifies to $ H(x) $. That is:

$$
H(-x) = H(x)
$$

This **defines an even function**. Therefore, the composition
$$
f(f(g(f(g(f(x))))))
$$
is **even**.

---

### Final Answer:
$$
\boxed{even}
$$

Solution:
We have that
\[f(f(g(f(g(f(-x)))))) = f(f(g(f(g(-f(x)))))) = f(f(g(f(g(f(x)))))),\]so the function is $\boxed{\text{even}}.$

More generally, if we have a composition of functions, and at least one of the functions is even, then the whole composition of functions is even.