Explanation:
Okay, so I need to figure out how many ways there are to arrange the letters of the word "ELLIPSE". Let me start by recalling what I know about permutations of letters with repeated characters.

First, let me write down the word and count the letters. The word is ELLIPSE. Let me spell it out: E, L, L, I, P, S, E. Wait, let me count again. E is the first letter, then L, L, I, P, S, E. So that's 7 letters in total. Let me check again: E (1), L (2), L (3), I (4), P (5), S (6), E (7). Yep, 7 letters.

Now, I need to check if there are any repeating letters because that affects the number of unique permutations. Let me list out each letter and their counts.

Looking at the letters: E, L, L, I, P, S, E.

So, E appears twice, L appears twice, and the other letters I, P, S each appear once. Let me confirm:

- E: 2 times
- L: 2 times
- I: 1 time
- P: 1 time
- S: 1 time

Yes, that's correct. So total letters: 7. Repeated letters: E and L each twice.

Now, the formula for permutations of a multiset is n! divided by the product of the factorials of the counts of each repeated element. So in general, if you have n items where there are duplicates, the number of distinct arrangements is n! divided by (k1! * k2! * ... * km!), where ki is the number of times each repeated element occurs.

So applying that here, n is 7. The duplicates are E (2 times) and L (2 times). So the formula would be 7! divided by (2! * 2!).

Let me compute that. First, 7! is 7 factorial. Let me calculate that.

7! = 7 × 6 × 5 × 4 × 3 × 2 × 1. Let me compute step by step:

7 × 6 = 42

42 × 5 = 210

210 × 4 = 840

840 × 3 = 2520

2520 × 2 = 5040

5040 × 1 = 5040. So 7! is 5040.

Then, the denominator is 2! * 2! which is 2 * 2 = 4.

Therefore, the number of distinct arrangements is 5040 divided by 4. Let me compute that.

5040 ÷ 4. Well, 5040 divided by 2 is 2520, divided by 2 again is 1260. So 5040 / 4 = 1260.

Therefore, the number of distinct arrangements is 1260.

Wait, but let me make sure I didn't miss any other repeated letters. Let me check again. The letters in ELLIPSE are E, L, L, I, P, S, E. So E is twice, L is twice, and the rest are unique. So yes, only two letters are repeated, each twice. So the formula applies here.

Alternatively, maybe I can think of it as first arranging all 7 letters, which would be 7! ways, but then dividing by the duplicates. Since the two E's are indistinct, swapping them doesn't create a new arrangement, so we divide by 2! for the E's. Similarly, the two L's are indistinct, so divide by another 2! for the L's.

So that gives 7! / (2! * 2!) = 5040 / 4 = 1260.

Is there another way to think about this? Maybe by choosing positions for the repeated letters? Let me try that approach to verify.

Suppose I first choose positions for the E's. There are 7 letters, so the number of ways to choose positions for the two E's is C(7, 2). Then, after placing the E's, there are 5 remaining positions. Then, choose positions for the two L's, which would be 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.

So total number of arrangements would be C(7,2) * C(5,2) * 3!.

Let me compute that.

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

So 21 * 10 * 6. Let me compute that.

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

Same result! So that confirms that the answer is 1260.

Alternatively, another way: suppose I think of all permutations, then adjust for overcounting.

But I think the two methods I used are sufficient.

Wait, let me check if there's any other possible mistake. For example, maybe I miscounted the letters? Let me check again.

ELLIPSE: E, L, L, I, P, S, E. So letters are E, L, L, I, P, S, E. So E is twice, L is twice, and I, P, S once each. So total letters: 2 + 2 + 1 + 1 + 1 = 7. Correct.

Another possible mistake: maybe I forgot that there are two different repeated letters? But in the formula, we divide by 2! for each repeated letter. So since there are two letters each repeated twice, we divide by 2! * 2! as I did.

Alternatively, if there was a letter repeated three times, we would divide by 3! etc.

So, seems correct.

Alternatively, maybe I can think of the total permutations as 7! and then divide by the permutations of the identical letters. Since the two E's can be arranged among themselves in 2! ways, but since they are identical, those arrangements are not unique. Similarly for the two L's. Therefore, total unique permutations are 7! / (2! * 2!).

Yes, that's the same as before.

So, all methods lead to 1260.

Therefore, I think the answer is 1260.

But just to be thorough, let me try a smaller example to see if my reasoning holds.

Suppose I have the word "AABB". How many arrangements?

Using the formula: 4! / (2! * 2!) = 24 / 4 = 6.

Listing them out: AABB, ABAB, ABBA, BAAB, BABA, BBAA. Yep, 6.

Alternatively, using the other method: choose positions for A's: C(4,2) = 6, then the B's are in the remaining positions. So 6, which matches.

So that works.

Another example: "AABC". Letters A, A, B, C. Number of arrangements: 4! / 2! = 24 / 2 = 12.

Alternatively, choose positions for A's: C(4,2) = 6, then arrange B and C in the remaining 2 positions: 2! = 2. Total 6 * 2 = 12. Correct.

So that seems to work.

Therefore, applying the same logic to ELLIPSE, which has two pairs of duplicates, the formula is 7! / (2! * 2!) = 1260.

Therefore, I think that is the correct answer.

I don't see any mistakes in my reasoning. So I think the number of distinct arrangements 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 of the word **"ELLIPSE"**, we begin by analyzing the composition of the word.

---

### Step 1: Count the Total Letters and Identify Repeats

The word **ELLIPSE** consists of the following letters:

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

So, the total number of letters is:

$$
2 + 2 + 1 + 1 + 1 = 7
$$

---

### Step 2: Apply the Permutations of a Multiset Formula

When arranging letters where some are repeated, the number of **distinct permutations** is given by the formula:

$$
\frac{n!}{k_1! \cdot k_2! \cdot \ldots \cdot k_m!}
$$

Where:
- $ n $ is the total number of letters
- $ k_1, k_2, \ldots, k_m $ are the frequencies of each repeated letter

In this case:
- $ n = 7 $
- $ k_1 = 2 $ (for E)
- $ k_2 = 2 $ (for L)
- The other letters (I, P, S) appear only once, so their factorials are $ 1! $, which do not affect the result

So the number of distinct arrangements is:

$$
\frac{7!}{2! \cdot 2!} = \frac{5040}{2 \cdot 2} = \frac{5040}{4} = 1260
$$

---

### Step 3: Verification via Alternative Method

We can also verify this by considering the process of choosing positions for the repeated letters:

1. Choose 2 positions out of 7 for the E's: $ \binom{7}{2} = 21 $
2. From the remaining 5 positions, choose 2 for the L's: $ \binom{5}{2} = 10 $
3. The remaining 3 positions are for the distinct letters I, P, and S, which can be arranged in $ 3! = 6 $ ways

Total number of arrangements:

$$
21 \times 10 \times 6 = 1260
$$

This confirms the result.

---

### Final Answer

$$
\boxed{1260}
$$

[Assistant]
Explanation:
Okay, so I need to figure out how many ways there are to arrange the letters of the word "ELLIPSE". Let me start by recalling what I know about permutations of letters with repeated characters.

First, let me write down the word and count the letters. The word is ELLIPSE. Let me spell it out: E, L, L, I, P, S, E. Wait, let me count again. E is the first letter, then L, L, I, P, S, E. So that's 7 letters in total. Let me check again: E (1), L (2), L (3), I (4), P (5), S (6), E (7). Yep, 7 letters.

Now, I need to check if there are any repeating letters because that affects the number of unique permutations. Let me list out each letter and their counts.

Looking at the letters: E, L, L, I, P, S, E.

So, E appears twice, L appears twice, and the other letters I, P, S each appear once. Let me confirm:

- E: 2 times
- L: 2 times
- I: 1 time
- P: 1 time
- S: 1 time

Yes, that's correct. So total letters: 7. Repeated letters: E and L each twice.

Now, the formula for permutations of a multiset is n! divided by the product of the factorials of the counts of each repeated element. So in general, if you have n items where there are duplicates, the number of distinct arrangements is n! divided by (k1! * k2! * ... * km!), where ki is the number of times each repeated element occurs.

So applying that here, n is 7. The duplicates are E (2 times) and L (2 times). So the formula would be 7! divided by (2! * 2!).

Let me compute that. First, 7! is 7 factorial. Let me calculate that.

7! = 7 × 6 × 5 × 4 × 3 × 2 × 1. Let me compute step by step:

7 × 6 = 42

42 × 5 = 210

210 × 4 = 840

840 × 3 = 2520

2520 × 2 = 5040

5040 × 1 = 5040. So 7! is 5040.

Then, the denominator is 2! * 2! which is 2 * 2 = 4.

Therefore, the number of distinct arrangements is 5040 divided by 4. Let me compute that.

5040 ÷ 4. Well, 5040 divided by 2 is 2520, divided by 2 again is 1260. So 5040 / 4 = 1260.

Therefore, the number of distinct arrangements is 1260.

Wait, but let me make sure I didn't miss any other repeated letters. Let me check again. The letters in ELLIPSE are E, L, L, I, P, S, E. So E is twice, L is twice, and the rest are unique. So yes, only two letters are repeated, each twice. So the formula applies here.

Alternatively, maybe I can think of it as first arranging all 7 letters, which would be 7! ways, but then dividing by the duplicates. Since the two E's are indistinct, swapping them doesn't create a new arrangement, so we divide by 2! for the E's. Similarly, the two L's are indistinct, so divide by another 2! for the L's.

So that gives 7! / (2! * 2!) = 5040 / 4 = 1260.

Is there another way to think about this? Maybe by choosing positions for the repeated letters? Let me try that approach to verify.

Suppose I first choose positions for the E's. There are 7 letters, so the number of ways to choose positions for the two E's is C(7, 2). Then, after placing the E's, there are 5 remaining positions. Then, choose positions for the two L's, which would be 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.

So total number of arrangements would be C(7,2) * C(5,2) * 3!.

Let me compute that.

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

So 21 * 10 * 6. Let me compute that.

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

Same result! So that confirms that the answer is 1260.

Alternatively, another way: suppose I think of all permutations, then adjust for overcounting.

But I think the two methods I used are sufficient.

Wait, let me check if there's any other possible mistake. For example, maybe I miscounted the letters? Let me check again.

ELLIPSE: E, L, L, I, P, S, E. So letters are E, L, L, I, P, S, E. So E is twice, L is twice, and I, P, S once each. So total letters: 2 + 2 + 1 + 1 + 1 = 7. Correct.

Another possible mistake: maybe I forgot that there are two different repeated letters? But in the formula, we divide by 2! for each repeated letter. So since there are two letters each repeated twice, we divide by 2! * 2! as I did.

Alternatively, if there was a letter repeated three times, we would divide by 3! etc.

So, seems correct.

Alternatively, maybe I can think of the total permutations as 7! and then divide by the permutations of the identical letters. Since the two E's can be arranged among themselves in 2! ways, but since they are identical, those arrangements are not unique. Similarly for the two L's. Therefore, total unique permutations are 7! / (2! * 2!).

Yes, that's the same as before.

So, all methods lead to 1260.

Therefore, I think the answer is 1260.

But just to be thorough, let me try a smaller example to see if my reasoning holds.

Suppose I have the word "AABB". How many arrangements?

Using the formula: 4! / (2! * 2!) = 24 / 4 = 6.

Listing them out: AABB, ABAB, ABBA, BAAB, BABA, BBAA. Yep, 6.

Alternatively, using the other method: choose positions for A's: C(4,2) = 6, then the B's are in the remaining positions. So 6, which matches.

So that works.

Another example: "AABC". Letters A, A, B, C. Number of arrangements: 4! / 2! = 24 / 2 = 12.

Alternatively, choose positions for A's: C(4,2) = 6, then arrange B and C in the remaining 2 positions: 2! = 2. Total 6 * 2 = 12. Correct.

So that seems to work.

Therefore, applying the same logic to ELLIPSE, which has two pairs of duplicates, the formula is 7! / (2! * 2!) = 1260.

Therefore, I think that is the correct answer.

I don't see any mistakes in my reasoning. So I think the number of distinct arrangements 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 of the word **"ELLIPSE"**, we begin by analyzing the composition of the word.

---

### Step 1: Count the Total Letters and Identify Repeats

The word **ELLIPSE** consists of the following letters:

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

So, the total number of letters is:

$$
2 + 2 + 1 + 1 + 1 = 7
$$

---

### Step 2: Apply the Permutations of a Multiset Formula

When arranging letters where some are repeated, the number of **distinct permutations** is given by the formula:

$$
\frac{n!}{k_1! \cdot k_2! \cdot \ldots \cdot k_m!}
$$

Where:
- $ n $ is the total number of letters
- $ k_1, k_2, \ldots, k_m $ are the frequencies of each repeated letter

In this case:
- $ n = 7 $
- $ k_1 = 2 $ (for E)
- $ k_2 = 2 $ (for L)
- The other letters (I, P, S) appear only once, so their factorials are $ 1! $, which do not affect the result

So the number of distinct arrangements is:

$$
\frac{7!}{2! \cdot 2!} = \frac{5040}{2 \cdot 2} = \frac{5040}{4} = 1260
$$

---

### Step 3: Verification via Alternative Method

We can also verify this by considering the process of choosing positions for the repeated letters:

1. Choose 2 positions out of 7 for the E's: $ \binom{7}{2} = 21 $
2. From the remaining 5 positions, choose 2 for the L's: $ \binom{5}{2} = 10 $
3. The remaining 3 positions are for the distinct letters I, P, and S, which can be arranged in $ 3! = 6 $ ways

Total number of arrangements:

$$
21 \times 10 \times 6 = 1260
$$

This confirms the result.

---

### Final Answer

$$
\boxed{1260}
$$

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 $2^8 = 256$ because $2^10 is 1024, so 2^8 is 256. Let me verify that: 2^1 is 2, 2^2 is 4, 2^3 is 8, 2^4 is 16, 2^5 is 32, 2^6 is 64, 2^7 is 128, and 2^8 is 256. Yep, that's right. So 256 is $2^8$.

So substituting that into the original equation, the right-hand side becomes $(2^8)^{\frac{1}{2}}$. Now, I remember that when you raise a power to another power, you multiply the exponents. So $(2^8)^{\frac{1}{2}} = 2^{8 \times \frac{1}{2}} = 2^4$.

Therefore, 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 $x = \frac{4}{2} = 2$.

Wait, let me double-check my steps to make sure I didn't make a mistake. Starting with the original equation: $2^{2x} = 256^{1/2}$.

Expressed as powers of 2, 256 is indeed 2^8, so the right side is (2^8)^{1/2} which is 2^{4}. Then, 2^{2x} equals 2^4, so exponents must be equal: 2x = 4, so x = 2.

Let me plug x = 2 back into the original equation to verify.

Left side: $2^{2*2} = 2^4 = 16$.

Right side: $256^{1/2} = \sqrt{256} = 16$.

Yes, both sides equal 16. So x = 2 is the correct solution.

Alternatively, maybe I can approach this problem using logarithms? Let me try that method to confirm.

Starting with $2^{2x} = 256^{1/2}$. Taking the logarithm base 2 of both sides:

$\log_2(2^{2x}) = \log_2(256^{1/2})$.

Using the logarithm power rule, $\log_b(a^c) = c \log_b a$, so:

Left side becomes $2x \log_2 2 = 2x * 1 = 2x$.

Right side becomes $\frac{1}{2} \log_2 256$.

Since 256 is 2^8, $\log_2 256 = 8$, so the right side is $\frac{1}{2} * 8 = 4$.

Therefore, 2x = 4, so x = 2. Same result.

Another way: Maybe express both sides with exponents and solve.

Alternatively, maybe I can compute 256^{1/2} first. The square root of 256 is 16, as I did before. Then the equation becomes $2^{2x} = 16$.

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

All methods lead to the same answer. So I think x = 2 is definitely the solution.

Is there any other way someone might approach this problem? Maybe by trial and error? Let's see. If I guess x = 2, then 2x = 4, so 2^4 = 16. And sqrt(256) is 16. So that works. If I tried x = 1, then left side is 2^2 = 4, right side is 16, which is not equal. If x = 3, left side is 2^6 = 64, which is bigger than 16. So x=2 is the only integer that works.

Alternatively, if someone didn't remember that 256 is 2^8, they might try to compute 256^(1/2) directly. Since 256 is 16 squared, because 16*16=256. So sqrt(256)=16. Then, 2^{2x}=16. Then, since 16 is 2^4, so again 2x=4, x=2.

So regardless of the method, the answer is x=2.

I think that's thorough. I considered multiple approaches: expressing both sides with the same base, using logarithms, direct computation, and even trial and error. All lead to the same conclusion. Therefore, I'm confident that x=2 is the correct solution.

**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.

---

### Step 1: Express 256 as a power of 2

We know that:

$$
256 = 2^8
$$

So, we can rewrite the right-hand side of the equation as:

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

Using the exponent rule $(a^m)^n = a^{m \cdot n}$, we simplify:

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

---

### Step 2: Rewrite the equation

Now the equation becomes:

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

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

$$
2x = 4
$$

---

### Step 3: Solve for $x$

Divide both sides by 2:

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

---

### Step 4: Verify 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, confirming that the solution is correct.

---

### Final Answer

$$
\boxed{2}
$$

[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 $2^8 = 256$ because $2^10 is 1024, so 2^8 is 256. Let me verify that: 2^1 is 2, 2^2 is 4, 2^3 is 8, 2^4 is 16, 2^5 is 32, 2^6 is 64, 2^7 is 128, and 2^8 is 256. Yep, that's right. So 256 is $2^8$.

So substituting that into the original equation, the right-hand side becomes $(2^8)^{\frac{1}{2}}$. Now, I remember that when you raise a power to another power, you multiply the exponents. So $(2^8)^{\frac{1}{2}} = 2^{8 \times \frac{1}{2}} = 2^4$.

Therefore, 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 $x = \frac{4}{2} = 2$.

Wait, let me double-check my steps to make sure I didn't make a mistake. Starting with the original equation: $2^{2x} = 256^{1/2}$.

Expressed as powers of 2, 256 is indeed 2^8, so the right side is (2^8)^{1/2} which is 2^{4}. Then, 2^{2x} equals 2^4, so exponents must be equal: 2x = 4, so x = 2.

Let me plug x = 2 back into the original equation to verify.

Left side: $2^{2*2} = 2^4 = 16$.

Right side: $256^{1/2} = \sqrt{256} = 16$.

Yes, both sides equal 16. So x = 2 is the correct solution.

Alternatively, maybe I can approach this problem using logarithms? Let me try that method to confirm.

Starting with $2^{2x} = 256^{1/2}$. Taking the logarithm base 2 of both sides:

$\log_2(2^{2x}) = \log_2(256^{1/2})$.

Using the logarithm power rule, $\log_b(a^c) = c \log_b a$, so:

Left side becomes $2x \log_2 2 = 2x * 1 = 2x$.

Right side becomes $\frac{1}{2} \log_2 256$.

Since 256 is 2^8, $\log_2 256 = 8$, so the right side is $\frac{1}{2} * 8 = 4$.

Therefore, 2x = 4, so x = 2. Same result.

Another way: Maybe express both sides with exponents and solve.

Alternatively, maybe I can compute 256^{1/2} first. The square root of 256 is 16, as I did before. Then the equation becomes $2^{2x} = 16$.

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

All methods lead to the same answer. So I think x = 2 is definitely the solution.

Is there any other way someone might approach this problem? Maybe by trial and error? Let's see. If I guess x = 2, then 2x = 4, so 2^4 = 16. And sqrt(256) is 16. So that works. If I tried x = 1, then left side is 2^2 = 4, right side is 16, which is not equal. If x = 3, left side is 2^6 = 64, which is bigger than 16. So x=2 is the only integer that works.

Alternatively, if someone didn't remember that 256 is 2^8, they might try to compute 256^(1/2) directly. Since 256 is 16 squared, because 16*16=256. So sqrt(256)=16. Then, 2^{2x}=16. Then, since 16 is 2^4, so again 2x=4, x=2.

So regardless of the method, the answer is x=2.

I think that's thorough. I considered multiple approaches: expressing both sides with the same base, using logarithms, direct computation, and even trial and error. All lead to the same conclusion. Therefore, I'm confident that x=2 is the correct solution.

**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.

---

### Step 1: Express 256 as a power of 2

We know that:

$$
256 = 2^8
$$

So, we can rewrite the right-hand side of the equation as:

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

Using the exponent rule $(a^m)^n = a^{m \cdot n}$, we simplify:

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

---

### Step 2: Rewrite the equation

Now the equation becomes:

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

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

$$
2x = 4
$$

---

### Step 3: Solve for $x$

Divide both sides by 2:

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

---

### Step 4: Verify 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, confirming that the solution is correct.

---

### Final Answer

$$
\boxed{2}
$$

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 need to solve the equation √(3x - 5) = 2. Hmm, let me think. First, I remember that when there's a square root involved, I should probably get rid of it by squaring both sides of the equation. That way, I can solve for x more easily. Let me try that.

So, if I square both sides, the left side becomes (√(3x - 5))², which should just be 3x - 5, right? Because squaring a square root cancels them out. And the right side is 2 squared, which is 4. So now the equation is 3x - 5 = 4.

Alright, now I need to solve for x. Let me add 5 to both sides to get rid of the -5. That would give me 3x = 4 + 5, which is 3x = 9. Then, dividing both sides by 3, I get x = 9/3, so x = 3.

Wait, but hold on. I remember that sometimes when you square both sides of an equation, you might introduce extraneous solutions. So I should check if this solution actually works in the original equation. Let me plug x = 3 back into the original equation to verify.

Original equation: √(3x - 5) = 2. Plugging in x = 3, we get √(3*3 - 5) = √(9 - 5) = √4 = 2. Yep, that's equal to 2, which matches the right side. So x = 3 is a valid solution.

But the question says "find all possible values of x." Is there a possibility of another solution? Let me think. The square root function only gives the principal (non-negative) root, so √(something) is always non-negative. In this case, the right side is 2, which is positive, so there's no issue there.

But maybe I should consider if there's another way to approach this? For example, could there be a negative solution? Let me see. If I had, say, √(something) = -2, that would have no solution because square roots are always non-negative. But in our case, the right side is positive, so we don't have to worry about negative solutions.

Therefore, squaring both sides gave me x = 3, and checking it shows it works. So there's only one solution here.

Wait, but maybe I should check if there are any restrictions on the domain of the original equation? The expression under the square root, 3x - 5, must be greater than or equal to zero. So 3x - 5 ≥ 0. Solving that, 3x ≥ 5, so x ≥ 5/3. Since 5/3 is approximately 1.666..., and our solution x = 3 is greater than that, so it's within the domain.

Therefore, there are no other solutions because the equation only has one solution that satisfies both the equation and the domain restriction.

Let me recap to make sure I didn't miss anything. Start with √(3x - 5) = 2. Square both sides: 3x - 5 = 4. Solve for x: 3x = 9, so x = 3. Check the solution in the original equation: works. Check domain: x = 3 is allowed. Therefore, the only solution is x = 3.

I think that's all. I don't see any other possible solutions. Maybe if there was a different equation, but in this case, it seems straightforward.

Another way to think about it: the square root function is one-to-one for non-negative numbers, so if √(A) = B, then A = B², provided that B is non-negative. Since 2 is non-negative, this is valid. Therefore, the equation is equivalent to 3x - 5 = 4, leading to x = 3.

Yep, that's consistent with what I did before. So I think I'm confident that x = 3 is the only solution.

**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. The most straightforward method is to **square both sides** of the equation, which removes the square root and simplifies the expression.

---

### Step 1: Square Both Sides

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

$$
3x - 5 = 4
$$

---

### 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

We must check that the solution satisfies the **original equation** and that the expression under the square root is **non-negative** (since square roots are only defined for non-negative values in the real number system).

Substitute $ x = 3 $ into the original equation:

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

This matches the right-hand side of the original equation, so the solution is valid.

Also, check the domain of the square root expression:

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

Since $ x = 3 $ satisfies this inequality, it is within the domain.

---

### Step 4: Consider for Additional Solutions

The square root function is **non-negative**, and the right-hand side of the original equation is also positive (2). Therefore, there is **no possibility** of a negative solution or any other solution that would satisfy the equation.

---

### Final Answer

$$
\boxed{3}
$$

[Assistant]
Explanation:
Okay, so I need to solve the equation √(3x - 5) = 2. Hmm, let me think. First, I remember that when there's a square root involved, I should probably get rid of it by squaring both sides of the equation. That way, I can solve for x more easily. Let me try that.

So, if I square both sides, the left side becomes (√(3x - 5))², which should just be 3x - 5, right? Because squaring a square root cancels them out. And the right side is 2 squared, which is 4. So now the equation is 3x - 5 = 4.

Alright, now I need to solve for x. Let me add 5 to both sides to get rid of the -5. That would give me 3x = 4 + 5, which is 3x = 9. Then, dividing both sides by 3, I get x = 9/3, so x = 3.

Wait, but hold on. I remember that sometimes when you square both sides of an equation, you might introduce extraneous solutions. So I should check if this solution actually works in the original equation. Let me plug x = 3 back into the original equation to verify.

Original equation: √(3x - 5) = 2. Plugging in x = 3, we get √(3*3 - 5) = √(9 - 5) = √4 = 2. Yep, that's equal to 2, which matches the right side. So x = 3 is a valid solution.

But the question says "find all possible values of x." Is there a possibility of another solution? Let me think. The square root function only gives the principal (non-negative) root, so √(something) is always non-negative. In this case, the right side is 2, which is positive, so there's no issue there.

But maybe I should consider if there's another way to approach this? For example, could there be a negative solution? Let me see. If I had, say, √(something) = -2, that would have no solution because square roots are always non-negative. But in our case, the right side is positive, so we don't have to worry about negative solutions.

Therefore, squaring both sides gave me x = 3, and checking it shows it works. So there's only one solution here.

Wait, but maybe I should check if there are any restrictions on the domain of the original equation? The expression under the square root, 3x - 5, must be greater than or equal to zero. So 3x - 5 ≥ 0. Solving that, 3x ≥ 5, so x ≥ 5/3. Since 5/3 is approximately 1.666..., and our solution x = 3 is greater than that, so it's within the domain.

Therefore, there are no other solutions because the equation only has one solution that satisfies both the equation and the domain restriction.

Let me recap to make sure I didn't miss anything. Start with √(3x - 5) = 2. Square both sides: 3x - 5 = 4. Solve for x: 3x = 9, so x = 3. Check the solution in the original equation: works. Check domain: x = 3 is allowed. Therefore, the only solution is x = 3.

I think that's all. I don't see any other possible solutions. Maybe if there was a different equation, but in this case, it seems straightforward.

Another way to think about it: the square root function is one-to-one for non-negative numbers, so if √(A) = B, then A = B², provided that B is non-negative. Since 2 is non-negative, this is valid. Therefore, the equation is equivalent to 3x - 5 = 4, leading to x = 3.

Yep, that's consistent with what I did before. So I think I'm confident that x = 3 is the only solution.

**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. The most straightforward method is to **square both sides** of the equation, which removes the square root and simplifies the expression.

---

### Step 1: Square Both Sides

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

$$
3x - 5 = 4
$$

---

### 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

We must check that the solution satisfies the **original equation** and that the expression under the square root is **non-negative** (since square roots are only defined for non-negative values in the real number system).

Substitute $ x = 3 $ into the original equation:

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

This matches the right-hand side of the original equation, so the solution is valid.

Also, check the domain of the square root expression:

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

Since $ x = 3 $ satisfies this inequality, it is within the domain.

---

### Step 4: Consider for Additional Solutions

The square root function is **non-negative**, and the right-hand side of the original equation is also positive (2). Therefore, there is **no possibility** of a negative solution or any other solution that would satisfy 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, so I need to figure out whether the composition of these functions, f(f(g(f(g(f(x)))))) is even, odd, or neither. Let me 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 h(-x) = -h(x) for all x. So, given that f is odd and g is even, I need to analyze the composition step by step.

Let me write down the composition: f(f(g(f(g(f(x)))))). That's a lot of functions nested inside each other. Maybe I should work from the inside out, applying the properties of even and odd functions as I go. Let me try to break it down step by step.

First, let me denote the innermost function and work my way out. Let me define some intermediate functions to make this manageable. Let me start with the innermost function, which is f(x). Let me call that step 1: f(x). Then, moving outward, the next layer is g(f(x)). Let me call that step 2: g(f(x)). Then, the next layer is f(g(f(x))), which is step 3. Then, step 4 would be g(f(g(f(x)))), and step 5 is f(g(f(g(f(x))))), and finally step 6 is f(f(g(f(g(f(x)))))). Wait, but the original composition is f(f(g(f(g(f(x)))))). Let me check again:

Original composition: f(f(g(f(g(f(x)))))). Let me count the layers:

Starting from the inside:

1. f(x)
2. g(f(x))
3. f(g(f(x)))
4. g(f(g(f(x))))
5. f(g(f(g(f(x)))))
6. f(f(g(f(g(f(x))))))

So, there are six layers. Let me try to analyze each step's parity (evenness or oddness) as I go.

Starting with step 1: f(x). Since f is odd, we know that f(-x) = -f(x).

Step 2: g(f(x)). Since g is even, we know that g(-y) = g(y) for any y. Let me substitute y = f(x). So, g(-f(x)) = g(f(x)). But what is the expression here? The function at step 2 is g(f(x)). Let me consider replacing x with -x in step 2. Let me compute g(f(-x)). Since f is odd, f(-x) = -f(x). Therefore, g(f(-x)) = g(-f(x)). Since g is even, this is equal to g(f(x)). Therefore, step 2, which is g(f(x)), is even. Because replacing x with -x gives the same value. Therefore, step 2 is even.

So, step 2: even function.

Moving to step 3: f(g(f(x))). Let me denote this as f(step2). Since step2 is even, let me call it h(x) = g(f(x)), which is even. Therefore, step3 is f(h(x)). Now, since f is odd, let me check if f(h(x)) is odd or even. Let me compute f(h(-x)). Since h(x) is even, h(-x) = h(x). Therefore, f(h(-x)) = f(h(x)). But since f is odd, f(-y) = -f(y). However, here we have f(h(x)), and h(-x) = h(x). Therefore, f(h(-x)) = f(h(x)). Therefore, the function f(h(x)) is even? Wait, no. Wait, if I have a function k(x) = f(h(x)), then k(-x) = f(h(-x)) = f(h(x)) = k(x). Therefore, k(x) is even. Therefore, step3 is even. Wait, but f is odd and h is even. So, composing an odd function with an even function... Let me recall that the composition of an odd function and an even function is odd? Wait, no. Let me think. Suppose we have f(g(x)), where f is odd and g is even. Then f(g(-x)) = f(g(x)) because g(-x) = g(x). Therefore, f(g(-x)) = f(g(x)), so f(g(x)) is even. Therefore, the composition of an odd function and an even function is even. Therefore, step3 is even. So step3 is even.

So step3: even.

Moving to step4: g(f(g(f(x)))). Let me denote step3 as h(x) = f(g(f(x))) which is even. Therefore, step4 is g(h(x)). Since g is even, let me check if g(h(x)) is even or odd. Let me compute g(h(-x)). Since h(x) is even, h(-x) = h(x). Therefore, g(h(-x)) = g(h(x)). Therefore, step4 is even. Therefore, step4 is even.

So step4: even.

Moving to step5: f(g(f(g(f(x))))). Let me denote step4 as k(x) = g(f(g(f(x)))), which is even. Therefore, step5 is f(k(x)). Since f is odd, let me check if f(k(x)) is odd or even. Let me compute f(k(-x)). Since k(x) is even, k(-x) = k(x). Therefore, f(k(-x)) = f(k(x)). But since f is odd, f(-y) = -f(y). However, here we have f(k(x)), and k(-x) = k(x). Therefore, f(k(-x)) = f(k(x)). Therefore, the function f(k(x)) is even? Wait, similar to before. If we have a function m(x) = f(k(x)), then m(-x) = f(k(-x)) = f(k(x)) = m(x). Therefore, m(x) is even. Therefore, step5 is even.

Wait, but hold on. If f is odd and k(x) is even, then f(k(x)) is even? Let me check with an example. Suppose f is odd, say f(x) = x, and k(x) is even, say k(x) = x². Then f(k(x)) = x², which is even. So yes, that works. So, if f is odd and k is even, then f(k(x)) is even. Therefore, step5 is even.

So step5: even.

Now step6: f(f(g(f(g(f(x)))))) which is f(step5). Since step5 is even, let me denote step5 as m(x) = f(g(f(g(f(x))))) which is even. Therefore, step6 is f(m(x)). Now, since f is odd, let me check if f(m(x)) is odd or even. Let me compute f(m(-x)). Since m(x) is even, m(-x) = m(x). Therefore, f(m(-x)) = f(m(x)). But since f is odd, f(-y) = -f(y). However, here we have f(m(x)), and m(-x) = m(x). Therefore, f(m(-x)) = f(m(x)). Therefore, the function f(m(x)) is even? Wait, same as before. If we have a function n(x) = f(m(x)), then n(-x) = f(m(-x)) = f(m(x)) = n(x). Therefore, n(x) is even. Therefore, step6 is even.

Wait, but hold on. If f is odd and m(x) is even, then f(m(x)) is even? Let me check with an example again. Let f(x) = -x (which is odd) and m(x) = x² (even). Then f(m(x)) = -x², which is even. Because -x² is even. So yes, even. So even though f is odd, composing it with an even function gives an even function. Therefore, step6 is even.

But wait, the original question is about f(f(g(f(g(f(x)))))). According to this step-by-step analysis, it's even. But let me check again. Wait, maybe I made a mistake here. Let me go through each step again carefully.

Starting with f(x) is odd. Then g(f(x)) is even. Then f(g(f(x))) is even? Wait, earlier I thought that composing an odd function with an even function gives even. Wait, let me think again. Suppose f is odd and h is even. Then f(h(x)) is even? Let me check. Let me compute f(h(-x)) = f(h(x)) because h is even. Therefore, f(h(-x)) = f(h(x)), so yes, f(h(x)) is even. Therefore, step3 is even. Then step4 is g(step3) which is even. Then step5 is f(step4) which is even. Then step6 is f(step5) which is even. Therefore, the entire composition is even? But wait, let me check with specific functions. Let me take concrete examples for f and g.

Let me take f(x) as an odd function, say f(x) = x. Let me take g(x) as an even function, say g(x) = x². Then let's compute the composition f(f(g(f(g(f(x)))))).

Let me compute step by step:

Start with f(x) = x.

Then, innermost is f(x) = x.

Then, next layer: g(f(x)) = g(x) = x².

Then, next layer: f(g(f(x))) = f(x²) = x².

Then, next layer: g(f(g(f(x)))) = g(x²) = (x²)² = x⁴.

Then, next layer: f(g(f(g(f(x))))) = f(x⁴) = x⁴.

Then, the outermost layer: f(f(g(f(g(f(x)))))) = f(x⁴) = x⁴.

Now, x⁴ is an even function, since (-x)^4 = x^4. Therefore, in this case, the composition is even.

But let me try another example to check. Let me take f(x) = x^3 (which is odd) and g(x) = x² (even). Then compute the composition:

Start with f(x) = x³.

Then, g(f(x)) = (x³)² = x^6.

Then, f(g(f(x))) = f(x^6) = (x^6)^3 = x^{18}.

Then, g(f(g(f(x)))) = g(x^{18}) = (x^{18})² = x^{36}.

Then, f(g(f(g(f(x))))) = f(x^{36}) = (x^{36})³ = x^{108}.

Then, f(f(g(f(g(f(x)))))) = f(x^{108}) = (x^{108})³ = x^{324}.

Which is even, since (-x)^{324} = x^{324}. So again, even.

Hmm, so in both examples, the composition is even. But let me try another example where maybe the result is different? Wait, but maybe it's always even? Let me think.

Alternatively, maybe I made a mistake in assuming that all steps are even. Let me check again. Let me try with different functions. Let me take f(x) = -x (odd), and g(x) = x² (even). Then compute the composition:

Start with f(x) = -x.

Then, g(f(x)) = g(-x) = (-x)^2 = x².

Then, f(g(f(x))) = f(x²) = -x².

Then, g(f(g(f(x)))) = g(-x²) = (-x²)^2 = x^4.

Then, f(g(f(g(f(x))))) = f(x^4) = -x^4.

Then, f(f(g(f(g(f(x)))))) = f(-x^4) = -(-x^4) = x^4.

Which is even. So again, even.

Wait, so in this case, the final result is x^4, which is even. So again, even.

But let me try with another function. Let me take f(x) = x^3 (odd) and g(x) = |x| (even). Let me compute the composition:

Start with f(x) = x³.

g(f(x)) = |x³|. Since x³ is odd, |x³| is even? Wait, |x³| = |x|³, which is even? Let me check: |(-x)³| = |-x³| = |x³| = |x|³. So yes, it's even. So g(f(x)) is even.

Then f(g(f(x))) = f(|x³|) = (|x³|)^3 = |x|⁹. Which is even, since |x|⁹ is even.

Then g(f(g(f(x)))) = g(|x|⁹) = ||x|⁹| = |x|⁹, which is even.

Then f(g(f(g(f(x))))) = f(|x|⁹) = (|x|⁹)^3 = |x|²⁷, which is even.

Then f(f(g(f(g(f(x)))))) = f(|x|²⁷) = (|x|²⁷)^3 = |x|⁸¹, which is even.

So again, even.

Hmm. So in all these examples, the composition ends up being even. Therefore, maybe the answer is even? But let me think again.

Wait, but let me try to think more abstractly. Let me try to analyze the composition step by step, using the properties of even and odd functions.

Let me denote the entire composition as H(x) = f(f(g(f(g(f(x)))))). Let me compute H(-x) and see if it equals H(x) or -H(x) or neither.

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

Let me compute step by step:

Start with f(-x). Since f is odd, f(-x) = -f(x).

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

Then, compute f(g(f(-x))) = f(g(f(x))). Since g(f(x)) is some value, let me denote it as y. Then f(y) is just f(y). But since we have already computed that g(f(-x)) = g(f(x)), then f(g(f(-x))) = f(g(f(x))).

Then, compute g(f(g(f(-x)))) = g(f(g(f(x)))). Let me denote z = g(f(x)), then f(g(f(x))) = f(z). Then, g(f(z)) = g(f(z)). But since g is even, and f(z) is some value. Wait, but z is g(f(x)), which is even? Wait, earlier steps showed that g(f(x)) is even. Wait, but let me check again.

Wait, let me track each step:

Starting with H(-x):

1. f(-x) = -f(x) [since f is odd]
2. g(f(-x)) = g(-f(x)) = g(f(x)) [since g is even]
3. f(g(f(-x))) = f(g(f(x))) [from step 2]
4. g(f(g(f(-x)))) = g(f(g(f(x)))) [from step 3]
5. f(g(f(g(f(-x))))) = f(g(f(g(f(x))))) [from step 4]
6. f(f(g(f(g(f(-x)))))) = f(f(g(f(g(f(x))))) [from step 5]

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

Wait, so according to this, H(-x) = H(x), so H is even. Therefore, the composition is even.

But let me check again. Let me write this more carefully:

Let me compute H(-x):

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

Let me compute f(-x) = -f(x) [odd]

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

Then, compute f(g(f(-x))) = f(g(f(x))).

Then, compute g(f(g(f(-x)))) = g(f(g(f(x)))).

Then, compute f(g(f(g(f(-x))))) = f(g(f(g(f(x))))).

Then, compute f(f(g(f(g(f(-x)))))) = f(f(g(f(g(f(x)))))) = H(x).

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

Therefore, regardless of the specific functions f and g (as long as f is odd and g is even), the composition H(x) is even.

But wait, let me check if there's a case where this might not hold. Suppose f is odd and g is even. Let me think of another example where maybe the composition could be different? Wait, but according to the algebra above, H(-x) = H(x). Therefore, regardless of the functions, as long as f is odd and g is even, the composition is even.

But let me think again. Suppose f is odd and g is even. Let me consider the composition step by step. Let me think about the number of times f and g are applied.

Starting from the inside:

Let me denote the innermost function as f(x). Then, moving outward:

1. f(x) is odd.
2. g(f(x)) is even (as g is even and f(x) is odd, but composition of even and odd is even).
3. f(g(f(x))) is even (as f is odd and g(f(x)) is even, composition of odd and even is even).
4. g(f(g(f(x)))) is even (g is even and f(g(f(x))) is even, composition of even and even is even).
5. f(g(f(g(f(x))))) is even (f is odd and g(...) is even, composition of odd and even is even).
6. f(f(g(f(g(f(x)))))) is even (f is odd and f(...) is even, composition of odd and even is even).

Therefore, each time we apply f or g, depending on the previous function's parity, we can determine the parity of the next step.

So, starting with odd, then even, then even, then even, then even, then even. Therefore, the final composition is even.

Therefore, the answer should be even.

But let me check again with another example where maybe the functions are more complex. Let me take f(x) = sin(x) (odd) and g(x) = cos(x) (even). Let me compute H(x) = f(f(g(f(g(f(x)))))).

Compute step by step:

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

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

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))))).

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

Now, let me check if this function is even. Let me compute H(-x):

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

Compute sin(-x) = -sin(x). Then:

sin(-x) = -sin(x)

cos(sin(-x)) = cos(-sin(x)) = cos(sin(x)) because cos is even.

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

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

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

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

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

Therefore, even in this trigonometric example, the composition is even.

Therefore, it seems that regardless of the specific odd and even functions, the composition ends up being even.

But let me think again. Suppose that there is some cancellation or something. Wait, but according to the algebra, every time we apply f or g, the parity is determined. Let me think of the number of times f is applied.

The composition is f(f(g(f(g(f(x)))))). Let me count the number of f and g functions:

Starting from the inside:

- f(x)
- g(f(x))
- f(g(f(x)))
- g(f(g(f(x))))
- f(g(f(g(f(x)))))
- f(f(g(f(g(f(x))))))

So, there are 2 f's and 3 g's? Wait, no. Let me count:

Original composition: f(f(g(f(g(f(x)))))).

Breaking down:

- The innermost is 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 number of f's: Let's see:

First f: f(x)

Then, inside the first g: f(x)

Then, inside the first f: g(f(x))

Then, inside the second g: f(g(f(x)))

Then, inside the second f: g(f(g(f(x))))

Then, inside the second f: f(g(f(g(f(x)))))

Wait, maybe it's better to count the number of f and g functions:

The composition is f composed with f composed with g composed with f composed with g composed with f. So, the order is:

Starting from the innermost:

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, the number of f's is 3? Wait, no. Let me count:

Each time we have a function, starting from the innermost:

1. f(x)
2. g(f(x))
3. f(g(f(x)))
4. g(f(g(f(x))))
5. f(g(f(g(f(x)))))
6. f(f(g(f(g(f(x))))))

So, the total number of f's is 3? Wait, no. Let me count how many times f is applied:

- The first f is applied once (f(x))
- Then, the second f is applied in step 3: f(g(f(x)))
- Then, the third f is applied in step 5: f(g(f(g(f(x)))))
- Then, the fourth f is applied in step 6: f(f(...))

Wait, actually, the composition is f(f(g(f(g(f(x)))))) which is f composed with f composed with g composed with f composed with g composed with f. So, the order of composition is:

Starting from the innermost:

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, the number of f's is 4? Wait, no. Let me think of the composition as:

Let me write it as f1(f2(g1(f3(g2(f4(x)))))) where f1, f2, f3, f4 are f's and g1, g2 are g's. Wait, maybe not. Let me think:

The composition is f(f(g(f(g(f(x)))))).

So, starting from the inside:

Start with f(x). Then, apply g to that: g(f(x)). Then apply f to that: 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 to that: f(f(g(f(g(f(x)))))).

So, the number of f's is 4? Wait, no. Let me count:

- The first f is f(x)
- Then, the second f is in f(g(f(x)))
- The third f is in g(f(g(f(x))))
Wait, no. Wait, the functions are:

Starting from the innermost:

1. f(x)
2. g(f(x)) → g applied to f(x)
3. f(g(f(x))) → f applied to g(f(x))
4. g(f(g(f(x)))) → g applied to f(g(f(x)))
5. f(g(f(g(f(x))))) → f applied to g(f(g(f(x))))
6. f(f(g(f(g(f(x)))))) → f applied to f(g(f(g(f(x)))))

Therefore, the number of f's is 4? Wait, no. Let me count how many times f is used:

- The first f is in step 1: f(x)
- The second f is in step 3: f(g(f(x)))
- The third f is in step 5: f(g(f(g(f(x)))))
- The fourth f is in step 6: f(f(...))

So, total of four f's? Wait, but the original composition is f(f(g(f(g(f(x)))))) which is f composed with f composed with g composed with f composed with g composed with f. So, the number of f's is 4? Wait, no. Let me think of the composition as follows:

Let me denote the functions as follows:

Let me define:

Let’s say we have functions:

Start with f(x).

Then, apply g: g(f(x)).

Then apply f: f(g(f(x))).

Then apply g: g(f(g(f(x)))).

Then apply f: f(g(f(g(f(x))))).

Then apply f: f(f(g(f(g(f(x)))))).

So, the total number of f's is 4? Wait, no. Let me count:

- The first f is f(x)
- The second f is in f(g(f(x)))
- The third f is in f(g(f(g(f(x)))))
- The fourth f is in f(f(...))

So, yes, four f's. But the original composition is f(f(g(f(g(f(x)))))) which is f composed with f composed with g composed with f composed with g composed with f. So, the number of f's is 4? Wait, no. Let me think of the composition as:

The outermost function is f, then another f, then g, then f, then g, then f. So, total of four f's and two g's? Wait, no. Let me count:

Starting from the outside:

1. f
2. f
3. g
4. f
5. g
6. f

So, total of four f's and two g's? Wait, no. Wait, the composition is f(f(g(f(g(f(x)))))) which is f composed with f composed with g composed with f composed with g composed with f. So, the order is f, f, g, f, g, f. Therefore, total of four f's and two g's? Wait, no. Let me count:

- The first f is the outermost function.
- Then another f.
- Then a g.
- Then an f.
- Then a g.
- Then an f.

So, total of four f's and two g's? Wait, no. Wait, the composition is f(f(g(f(g(f(x)))))) which is f applied to [f applied to [g applied to [f applied to [g applied to [f applied to x]]]]]. So, the number of f's is 4? Let me count:

- The first f is the outermost.
- The second f is inside the first f.
- The third f is inside the second f.
- The fourth f is inside the third f.

Wait, no. Let me think recursively. The innermost is f(x). Then, the next layer is g(f(x)). Then, the next layer is f(g(f(x))). Then, the next layer is g(f(g(f(x)))). Then, the next layer is f(g(f(g(f(x))))). Then, the outermost layer is f(f(g(f(g(f(x)))))). Therefore, the number of f's is 4? Wait, no. Let me count:

- The innermost is f(x): 1st f
- Then, g(f(x)): 1st g
- Then, f(g(f(x))): 2nd f
- Then, g(f(g(f(x)))): 2nd g
- Then, f(g(f(g(f(x))))): 3rd f
- Then, f(f(g(f(g(f(x)))))): 4th f

Yes, so there are four f's and two g's. Therefore, the composition is f(f(g(f(g(f(x)))))) which is four f's and two g's.

But regardless of the number, the key is that each time we apply f or g, depending on the previous function's parity, we can determine the parity of the next function.

But according to the previous analysis, regardless of the number of applications, the final result is even.

But let me think again. Suppose that instead of f being odd and g being even, we have different numbers of f and g. But in this case, the composition ends up being even.

Alternatively, maybe there's a general rule here. Let me think about the parity of the composition.

Let me consider that composing an odd function with another function: if the inner function is even, then the composition is odd? Wait, no. Wait, if you have an odd function f and an even function h, then f(h(x)) is even? Because f(h(-x)) = f(h(x)) since h is even. Therefore, yes, it's even.

If you have an even function h and an odd function f, then h(f(x)) is even? Wait, h is even, f is odd. Then h(f(-x)) = h(-f(x)) = h(f(x)) because h is even. Therefore, h(f(x)) is even.

Therefore, composing an even function with an odd function gives an even function.

Similarly, composing an odd function with an odd function gives an even function? Wait, no. If you have two odd functions, f and g, then f(g(-x)) = f(-g(x)) = -f(g(x)) because f is odd. Therefore, f(g(-x)) = -f(g(x)), so the composition is odd.

Similarly, composing two even functions gives an even function.

So, the key is that when you compose functions, the parity depends on the number of odd functions in the composition.

But in our case, the composition is f(f(g(f(g(f(x)))))). Let me think of the entire composition as a series of operations. Let me track the parity at each step:

Start with f(x): odd.

Then, apply g: g(f(x)) is even (even function composed with odd function).

Then, apply f: f(g(f(x))) is odd composed with even function, which is even.

Then, apply g: g(f(g(f(x)))) is even composed with even function, which is even.

Then, apply f: f(g(f(g(f(x))))) is odd composed with even function, which is even.

Then, apply f: f(f(g(f(g(f(x)))))) is odd composed with even function, which is even.

Therefore, each time, after the first step, the parity is even, and then applying f or g preserves the evenness.

Therefore, the final result is even.

Therefore, regardless of the specific 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 once more with a different example where maybe the functions are not polynomials. Let me take f(x) = x^3 (odd) and g(x) = x^2 (even). Then compute H(x) = f(f(g(f(g(f(x)))))).

Compute step by step:

Start with f(x) = x³.

g(f(x)) = (x³)² = x⁶.

f(g(f(x))) = (x⁶)³ = x¹⁸.

g(f(g(f(x)))) = (x¹⁸)² = x³⁶.

f(g(f(g(f(x))))) = (x³⁶)³ = x¹⁰⁸.

f(f(g(f(g(f(x)))))) = (x¹⁰⁸)³ = x³²⁴.

Which is even, as (-x)^324 = x^324.

Another example: f(x) = -x (odd), g(x) = x² (even). Then:

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

Compute:

f(x) = -x.

g(f(x)) = (-x)² = x².

f(g(f(x))) = -x².

g(f(g(f(x)))) = (-x²)² = x⁴.

f(g(f(g(f(x))))) = -x⁴.

f(f(g(f(g(f(x)))))) = -(-x⁴) = x⁴, which is even.

Therefore, again, even.

Therefore, I think I can be confident that the composition is even.

But just to make sure, let me think if there's any possible way that the composition could be odd. Suppose that the number of odd functions in the composition is odd? Wait, but in this case, the composition is f(f(g(f(g(f(x)))))). Let me count the number of f's. There are four f's. Each f is odd. So, composing four odd functions.

But composing odd functions:

- Odd * Odd = Even (since f(g(x)) where f and g are odd: f(g(-x)) = f(-g(x)) = -f(g(x)) = -f(g(x)), so it's odd? Wait, no. Wait, if you have two odd functions, their composition is odd. Wait, let me check:

Let f and g be odd. Then, f(g(-x)) = f(-g(x)) = -f(g(x)) because f is odd. Therefore, f(g(-x)) = -f(g(x)), so the composition is odd.

Therefore, composing two odd functions gives an odd function.

But in our case, the composition is f(f(g(f(g(f(x)))))). Let me think of it as:

Start with f(x) (odd).

Then, apply g (even): g(f(x)) is even.

Then, apply f (odd): f(g(f(x))) is even (odd composed with even is even).

Then, apply g (even): g(f(g(f(x)))) is even (even composed with even is even).

Then, apply f (odd): f(g(f(g(f(x))))) is even (odd composed with even is even).

Then, apply f (odd): f(f(g(f(g(f(x)))))) is odd composed with even, which is even.

Therefore, even.

So, even though there are four f's (odd functions), the composition ends up being even because the number of times we compose odd functions with even functions is such that the overall parity is even.

Therefore, the answer is even.

Therefore, after all these steps and examples, I think the composition is even.

**Final Answer**
\boxed{even}

To determine whether the composition
$$
f(f(g(f(g(f(x))))))
$$
is **even**, **odd**, or **neither**, we analyze the behavior of the functions under the transformation $ x \to -x $, using the definitions of **even** and **odd** functions:

- A function $ h(x) $ is **even** if $ h(-x) = h(x) $ for all $ x $.
- A function $ h(x) $ is **odd** if $ h(-x) = -h(x) $ for all $ x $.

We are given:
- $ f(x) $ is **odd**, so $ f(-x) = -f(x) $.
- $ g(x) $ is **even**, so $ g(-x) = g(x) $.

---

### Step-by-Step Analysis

Let’s evaluate the composition step by step, starting from the innermost function and moving outward.

#### Step 1: $ f(x) $
- $ f(x) $ is **odd**.

#### Step 2: $ g(f(x)) $
- Since $ f(x) $ is odd, $ f(-x) = -f(x) $.
- Then $ g(f(-x)) = g(-f(x)) = g(f(x)) $, because $ g $ is even.
- So, $ g(f(x)) $ is **even**.

#### Step 3: $ f(g(f(x))) $
- $ g(f(x)) $ is even, so $ g(f(-x)) = g(f(x)) $.
- Then $ f(g(f(-x))) = f(g(f(x))) $, because $ f $ is odd and $ g(f(x)) $ is even.
- So, $ f(g(f(x))) $ is **even**.

#### Step 4: $ g(f(g(f(x)))) $
- $ f(g(f(x))) $ is even, so $ f(g(f(-x))) = f(g(f(x))) $.
- Then $ g(f(g(f(-x)))) = g(f(g(f(x)))) $, because $ g $ is even.
- So, $ g(f(g(f(x)))) $ is **even**.

#### Step 5: $ f(g(f(g(f(x))))) $
- $ g(f(g(f(x)))) $ is even, so $ f(g(f(g(f(-x))))) = f(g(f(g(f(x))))) $.
- So, $ f(g(f(g(f(x))))) $ is **even**.

#### Step 6: $ f(f(g(f(g(f(x)))))) $
- $ f(g(f(g(f(x))))) $ is even, so $ f(f(g(f(g(f(-x))))) = f(f(g(f(g(f(x))))) $.
- So, the entire composition is **even**.

---

### General Rule

We can summarize the pattern:
- Composing an **odd** function with an **even** function results in an **even** function.
- Composing an **even** function with an **even** function results in an **even** function.
- Composing an **odd** function with an **odd** function results in an **odd** function.

In our case, the composition involves:
- 4 applications of $ f $ (odd),
- 2 applications of $ g $ (even).

Each time an **odd** function is applied to an **even** function, the result is **even**. Since the final function is built from such compositions, the result is **even**.

---

### Final Answer

$$
\boxed{even}
$$

[Assistant]
Explanation:
Okay, so I need to figure out whether the composition of these functions, f(f(g(f(g(f(x)))))) is even, odd, or neither. Let me 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 h(-x) = -h(x) for all x. So, given that f is odd and g is even, I need to analyze the composition step by step.

Let me write down the composition: f(f(g(f(g(f(x)))))). That's a lot of functions nested inside each other. Maybe I should work from the inside out, applying the properties of even and odd functions as I go. Let me try to break it down step by step.

First, let me denote the innermost function and work my way out. Let me define some intermediate functions to make this manageable. Let me start with the innermost function, which is f(x). Let me call that step 1: f(x). Then, moving outward, the next layer is g(f(x)). Let me call that step 2: g(f(x)). Then, the next layer is f(g(f(x))), which is step 3. Then, step 4 would be g(f(g(f(x)))), and step 5 is f(g(f(g(f(x))))), and finally step 6 is f(f(g(f(g(f(x)))))). Wait, but the original composition is f(f(g(f(g(f(x)))))). Let me check again:

Original composition: f(f(g(f(g(f(x)))))). Let me count the layers:

Starting from the inside:

1. f(x)
2. g(f(x))
3. f(g(f(x)))
4. g(f(g(f(x))))
5. f(g(f(g(f(x)))))
6. f(f(g(f(g(f(x))))))

So, there are six layers. Let me try to analyze each step's parity (evenness or oddness) as I go.

Starting with step 1: f(x). Since f is odd, we know that f(-x) = -f(x).

Step 2: g(f(x)). Since g is even, we know that g(-y) = g(y) for any y. Let me substitute y = f(x). So, g(-f(x)) = g(f(x)). But what is the expression here? The function at step 2 is g(f(x)). Let me consider replacing x with -x in step 2. Let me compute g(f(-x)). Since f is odd, f(-x) = -f(x). Therefore, g(f(-x)) = g(-f(x)). Since g is even, this is equal to g(f(x)). Therefore, step 2, which is g(f(x)), is even. Because replacing x with -x gives the same value. Therefore, step 2 is even.

So, step 2: even function.

Moving to step 3: f(g(f(x))). Let me denote this as f(step2). Since step2 is even, let me call it h(x) = g(f(x)), which is even. Therefore, step3 is f(h(x)). Now, since f is odd, let me check if f(h(x)) is odd or even. Let me compute f(h(-x)). Since h(x) is even, h(-x) = h(x). Therefore, f(h(-x)) = f(h(x)). But since f is odd, f(-y) = -f(y). However, here we have f(h(x)), and h(-x) = h(x). Therefore, f(h(-x)) = f(h(x)). Therefore, the function f(h(x)) is even? Wait, no. Wait, if I have a function k(x) = f(h(x)), then k(-x) = f(h(-x)) = f(h(x)) = k(x). Therefore, k(x) is even. Therefore, step3 is even. Wait, but f is odd and h is even. So, composing an odd function with an even function... Let me recall that the composition of an odd function and an even function is odd? Wait, no. Let me think. Suppose we have f(g(x)), where f is odd and g is even. Then f(g(-x)) = f(g(x)) because g(-x) = g(x). Therefore, f(g(-x)) = f(g(x)), so f(g(x)) is even. Therefore, the composition of an odd function and an even function is even. Therefore, step3 is even. So step3 is even.

So step3: even.

Moving to step4: g(f(g(f(x)))). Let me denote step3 as h(x) = f(g(f(x))) which is even. Therefore, step4 is g(h(x)). Since g is even, let me check if g(h(x)) is even or odd. Let me compute g(h(-x)). Since h(x) is even, h(-x) = h(x). Therefore, g(h(-x)) = g(h(x)). Therefore, step4 is even. Therefore, step4 is even.

So step4: even.

Moving to step5: f(g(f(g(f(x))))). Let me denote step4 as k(x) = g(f(g(f(x)))), which is even. Therefore, step5 is f(k(x)). Since f is odd, let me check if f(k(x)) is odd or even. Let me compute f(k(-x)). Since k(x) is even, k(-x) = k(x). Therefore, f(k(-x)) = f(k(x)). But since f is odd, f(-y) = -f(y). However, here we have f(k(x)), and k(-x) = k(x). Therefore, f(k(-x)) = f(k(x)). Therefore, the function f(k(x)) is even? Wait, similar to before. If we have a function m(x) = f(k(x)), then m(-x) = f(k(-x)) = f(k(x)) = m(x). Therefore, m(x) is even. Therefore, step5 is even.

Wait, but hold on. If f is odd and k(x) is even, then f(k(x)) is even? Let me check with an example. Suppose f is odd, say f(x) = x, and k(x) is even, say k(x) = x². Then f(k(x)) = x², which is even. So yes, that works. So, if f is odd and k is even, then f(k(x)) is even. Therefore, step5 is even.

So step5: even.

Now step6: f(f(g(f(g(f(x)))))) which is f(step5). Since step5 is even, let me denote step5 as m(x) = f(g(f(g(f(x))))) which is even. Therefore, step6 is f(m(x)). Now, since f is odd, let me check if f(m(x)) is odd or even. Let me compute f(m(-x)). Since m(x) is even, m(-x) = m(x). Therefore, f(m(-x)) = f(m(x)). But since f is odd, f(-y) = -f(y). However, here we have f(m(x)), and m(-x) = m(x). Therefore, f(m(-x)) = f(m(x)). Therefore, the function f(m(x)) is even? Wait, same as before. If we have a function n(x) = f(m(x)), then n(-x) = f(m(-x)) = f(m(x)) = n(x). Therefore, n(x) is even. Therefore, step6 is even.

Wait, but hold on. If f is odd and m(x) is even, then f(m(x)) is even? Let me check with an example again. Let f(x) = -x (which is odd) and m(x) = x² (even). Then f(m(x)) = -x², which is even. Because -x² is even. So yes, even. So even though f is odd, composing it with an even function gives an even function. Therefore, step6 is even.

But wait, the original question is about f(f(g(f(g(f(x)))))). According to this step-by-step analysis, it's even. But let me check again. Wait, maybe I made a mistake here. Let me go through each step again carefully.

Starting with f(x) is odd. Then g(f(x)) is even. Then f(g(f(x))) is even? Wait, earlier I thought that composing an odd function with an even function gives even. Wait, let me think again. Suppose f is odd and h is even. Then f(h(x)) is even? Let me check. Let me compute f(h(-x)) = f(h(x)) because h is even. Therefore, f(h(-x)) = f(h(x)), so yes, f(h(x)) is even. Therefore, step3 is even. Then step4 is g(step3) which is even. Then step5 is f(step4) which is even. Then step6 is f(step5) which is even. Therefore, the entire composition is even? But wait, let me check with specific functions. Let me take concrete examples for f and g.

Let me take f(x) as an odd function, say f(x) = x. Let me take g(x) as an even function, say g(x) = x². Then let's compute the composition f(f(g(f(g(f(x)))))).

Let me compute step by step:

Start with f(x) = x.

Then, innermost is f(x) = x.

Then, next layer: g(f(x)) = g(x) = x².

Then, next layer: f(g(f(x))) = f(x²) = x².

Then, next layer: g(f(g(f(x)))) = g(x²) = (x²)² = x⁴.

Then, next layer: f(g(f(g(f(x))))) = f(x⁴) = x⁴.

Then, the outermost layer: f(f(g(f(g(f(x)))))) = f(x⁴) = x⁴.

Now, x⁴ is an even function, since (-x)^4 = x^4. Therefore, in this case, the composition is even.

But let me try another example to check. Let me take f(x) = x^3 (which is odd) and g(x) = x² (even). Then compute the composition:

Start with f(x) = x³.

Then, g(f(x)) = (x³)² = x^6.

Then, f(g(f(x))) = f(x^6) = (x^6)^3 = x^{18}.

Then, g(f(g(f(x)))) = g(x^{18}) = (x^{18})² = x^{36}.

Then, f(g(f(g(f(x))))) = f(x^{36}) = (x^{36})³ = x^{108}.

Then, f(f(g(f(g(f(x)))))) = f(x^{108}) = (x^{108})³ = x^{324}.

Which is even, since (-x)^{324} = x^{324}. So again, even.

Hmm, so in both examples, the composition is even. But let me try another example where maybe the result is different? Wait, but maybe it's always even? Let me think.

Alternatively, maybe I made a mistake in assuming that all steps are even. Let me check again. Let me try with different functions. Let me take f(x) = -x (odd), and g(x) = x² (even). Then compute the composition:

Start with f(x) = -x.

Then, g(f(x)) = g(-x) = (-x)^2 = x².

Then, f(g(f(x))) = f(x²) = -x².

Then, g(f(g(f(x)))) = g(-x²) = (-x²)^2 = x^4.

Then, f(g(f(g(f(x))))) = f(x^4) = -x^4.

Then, f(f(g(f(g(f(x)))))) = f(-x^4) = -(-x^4) = x^4.

Which is even. So again, even.

Wait, so in this case, the final result is x^4, which is even. So again, even.

But let me try with another function. Let me take f(x) = x^3 (odd) and g(x) = |x| (even). Let me compute the composition:

Start with f(x) = x³.

g(f(x)) = |x³|. Since x³ is odd, |x³| is even? Wait, |x³| = |x|³, which is even? Let me check: |(-x)³| = |-x³| = |x³| = |x|³. So yes, it's even. So g(f(x)) is even.

Then f(g(f(x))) = f(|x³|) = (|x³|)^3 = |x|⁹. Which is even, since |x|⁹ is even.

Then g(f(g(f(x)))) = g(|x|⁹) = ||x|⁹| = |x|⁹, which is even.

Then f(g(f(g(f(x))))) = f(|x|⁹) = (|x|⁹)^3 = |x|²⁷, which is even.

Then f(f(g(f(g(f(x)))))) = f(|x|²⁷) = (|x|²⁷)^3 = |x|⁸¹, which is even.

So again, even.

Hmm. So in all these examples, the composition ends up being even. Therefore, maybe the answer is even? But let me think again.

Wait, but let me try to think more abstractly. Let me try to analyze the composition step by step, using the properties of even and odd functions.

Let me denote the entire composition as H(x) = f(f(g(f(g(f(x)))))). Let me compute H(-x) and see if it equals H(x) or -H(x) or neither.

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

Let me compute step by step:

Start with f(-x). Since f is odd, f(-x) = -f(x).

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

Then, compute f(g(f(-x))) = f(g(f(x))). Since g(f(x)) is some value, let me denote it as y. Then f(y) is just f(y). But since we have already computed that g(f(-x)) = g(f(x)), then f(g(f(-x))) = f(g(f(x))).

Then, compute g(f(g(f(-x)))) = g(f(g(f(x)))). Let me denote z = g(f(x)), then f(g(f(x))) = f(z). Then, g(f(z)) = g(f(z)). But since g is even, and f(z) is some value. Wait, but z is g(f(x)), which is even? Wait, earlier steps showed that g(f(x)) is even. Wait, but let me check again.

Wait, let me track each step:

Starting with H(-x):

1. f(-x) = -f(x) [since f is odd]
2. g(f(-x)) = g(-f(x)) = g(f(x)) [since g is even]
3. f(g(f(-x))) = f(g(f(x))) [from step 2]
4. g(f(g(f(-x)))) = g(f(g(f(x)))) [from step 3]
5. f(g(f(g(f(-x))))) = f(g(f(g(f(x))))) [from step 4]
6. f(f(g(f(g(f(-x)))))) = f(f(g(f(g(f(x))))) [from step 5]

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

Wait, so according to this, H(-x) = H(x), so H is even. Therefore, the composition is even.

But let me check again. Let me write this more carefully:

Let me compute H(-x):

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

Let me compute f(-x) = -f(x) [odd]

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

Then, compute f(g(f(-x))) = f(g(f(x))).

Then, compute g(f(g(f(-x)))) = g(f(g(f(x)))).

Then, compute f(g(f(g(f(-x))))) = f(g(f(g(f(x))))).

Then, compute f(f(g(f(g(f(-x)))))) = f(f(g(f(g(f(x)))))) = H(x).

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

Therefore, regardless of the specific functions f and g (as long as f is odd and g is even), the composition H(x) is even.

But wait, let me check if there's a case where this might not hold. Suppose f is odd and g is even. Let me think of another example where maybe the composition could be different? Wait, but according to the algebra above, H(-x) = H(x). Therefore, regardless of the functions, as long as f is odd and g is even, the composition is even.

But let me think again. Suppose f is odd and g is even. Let me consider the composition step by step. Let me think about the number of times f and g are applied.

Starting from the inside:

Let me denote the innermost function as f(x). Then, moving outward:

1. f(x) is odd.
2. g(f(x)) is even (as g is even and f(x) is odd, but composition of even and odd is even).
3. f(g(f(x))) is even (as f is odd and g(f(x)) is even, composition of odd and even is even).
4. g(f(g(f(x)))) is even (g is even and f(g(f(x))) is even, composition of even and even is even).
5. f(g(f(g(f(x))))) is even (f is odd and g(...) is even, composition of odd and even is even).
6. f(f(g(f(g(f(x)))))) is even (f is odd and f(...) is even, composition of odd and even is even).

Therefore, each time we apply f or g, depending on the previous function's parity, we can determine the parity of the next step.

So, starting with odd, then even, then even, then even, then even, then even. Therefore, the final composition is even.

Therefore, the answer should be even.

But let me check again with another example where maybe the functions are more complex. Let me take f(x) = sin(x) (odd) and g(x) = cos(x) (even). Let me compute H(x) = f(f(g(f(g(f(x)))))).

Compute step by step:

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

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

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))))).

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

Now, let me check if this function is even. Let me compute H(-x):

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

Compute sin(-x) = -sin(x). Then:

sin(-x) = -sin(x)

cos(sin(-x)) = cos(-sin(x)) = cos(sin(x)) because cos is even.

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

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

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

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

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

Therefore, even in this trigonometric example, the composition is even.

Therefore, it seems that regardless of the specific odd and even functions, the composition ends up being even.

But let me think again. Suppose that there is some cancellation or something. Wait, but according to the algebra, every time we apply f or g, the parity is determined. Let me think of the number of times f is applied.

The composition is f(f(g(f(g(f(x)))))). Let me count the number of f and g functions:

Starting from the inside:

- f(x)
- g(f(x))
- f(g(f(x)))
- g(f(g(f(x))))
- f(g(f(g(f(x)))))
- f(f(g(f(g(f(x))))))

So, there are 2 f's and 3 g's? Wait, no. Let me count:

Original composition: f(f(g(f(g(f(x)))))).

Breaking down:

- The innermost is 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 number of f's: Let's see:

First f: f(x)

Then, inside the first g: f(x)

Then, inside the first f: g(f(x))

Then, inside the second g: f(g(f(x)))

Then, inside the second f: g(f(g(f(x))))

Then, inside the second f: f(g(f(g(f(x)))))

Wait, maybe it's better to count the number of f and g functions:

The composition is f composed with f composed with g composed with f composed with g composed with f. So, the order is:

Starting from the innermost:

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, the number of f's is 3? Wait, no. Let me count:

Each time we have a function, starting from the innermost:

1. f(x)
2. g(f(x))
3. f(g(f(x)))
4. g(f(g(f(x))))
5. f(g(f(g(f(x)))))
6. f(f(g(f(g(f(x))))))

So, the total number of f's is 3? Wait, no. Let me count how many times f is applied:

- The first f is applied once (f(x))
- Then, the second f is applied in step 3: f(g(f(x)))
- Then, the third f is applied in step 5: f(g(f(g(f(x)))))
- Then, the fourth f is applied in step 6: f(f(...))

Wait, actually, the composition is f(f(g(f(g(f(x)))))) which is f composed with f composed with g composed with f composed with g composed with f. So, the order of composition is:

Starting from the innermost:

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, the number of f's is 4? Wait, no. Let me think of the composition as:

Let me write it as f1(f2(g1(f3(g2(f4(x)))))) where f1, f2, f3, f4 are f's and g1, g2 are g's. Wait, maybe not. Let me think:

The composition is f(f(g(f(g(f(x)))))).

So, starting from the inside:

Start with f(x). Then, apply g to that: g(f(x)). Then apply f to that: 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 to that: f(f(g(f(g(f(x)))))).

So, the number of f's is 4? Wait, no. Let me count:

- The first f is f(x)
- Then, the second f is in f(g(f(x)))
- The third f is in g(f(g(f(x))))
Wait, no. Wait, the functions are:

Starting from the innermost:

1. f(x)
2. g(f(x)) → g applied to f(x)
3. f(g(f(x))) → f applied to g(f(x))
4. g(f(g(f(x)))) → g applied to f(g(f(x)))
5. f(g(f(g(f(x))))) → f applied to g(f(g(f(x))))
6. f(f(g(f(g(f(x)))))) → f applied to f(g(f(g(f(x)))))

Therefore, the number of f's is 4? Wait, no. Let me count how many times f is used:

- The first f is in step 1: f(x)
- The second f is in step 3: f(g(f(x)))
- The third f is in step 5: f(g(f(g(f(x)))))
- The fourth f is in step 6: f(f(...))

So, total of four f's? Wait, but the original composition is f(f(g(f(g(f(x)))))) which is f composed with f composed with g composed with f composed with g composed with f. So, the number of f's is 4? Wait, no. Let me think of the composition as follows:

Let me denote the functions as follows:

Let me define:

Let’s say we have functions:

Start with f(x).

Then, apply g: g(f(x)).

Then apply f: f(g(f(x))).

Then apply g: g(f(g(f(x)))).

Then apply f: f(g(f(g(f(x))))).

Then apply f: f(f(g(f(g(f(x)))))).

So, the total number of f's is 4? Wait, no. Let me count:

- The first f is f(x)
- The second f is in f(g(f(x)))
- The third f is in f(g(f(g(f(x)))))
- The fourth f is in f(f(...))

So, yes, four f's. But the original composition is f(f(g(f(g(f(x)))))) which is f composed with f composed with g composed with f composed with g composed with f. So, the number of f's is 4? Wait, no. Let me think of the composition as:

The outermost function is f, then another f, then g, then f, then g, then f. So, total of four f's and two g's? Wait, no. Let me count:

Starting from the outside:

1. f
2. f
3. g
4. f
5. g
6. f

So, total of four f's and two g's? Wait, no. Wait, the composition is f(f(g(f(g(f(x)))))) which is f composed with f composed with g composed with f composed with g composed with f. So, the order is f, f, g, f, g, f. Therefore, total of four f's and two g's? Wait, no. Let me count:

- The first f is the outermost function.
- Then another f.
- Then a g.
- Then an f.
- Then a g.
- Then an f.

So, total of four f's and two g's? Wait, no. Wait, the composition is f(f(g(f(g(f(x)))))) which is f applied to [f applied to [g applied to [f applied to [g applied to [f applied to x]]]]]. So, the number of f's is 4? Let me count:

- The first f is the outermost.
- The second f is inside the first f.
- The third f is inside the second f.
- The fourth f is inside the third f.

Wait, no. Let me think recursively. The innermost is f(x). Then, the next layer is g(f(x)). Then, the next layer is f(g(f(x))). Then, the next layer is g(f(g(f(x)))). Then, the next layer is f(g(f(g(f(x))))). Then, the outermost layer is f(f(g(f(g(f(x)))))). Therefore, the number of f's is 4? Wait, no. Let me count:

- The innermost is f(x): 1st f
- Then, g(f(x)): 1st g
- Then, f(g(f(x))): 2nd f
- Then, g(f(g(f(x)))): 2nd g
- Then, f(g(f(g(f(x))))): 3rd f
- Then, f(f(g(f(g(f(x)))))): 4th f

Yes, so there are four f's and two g's. Therefore, the composition is f(f(g(f(g(f(x)))))) which is four f's and two g's.

But regardless of the number, the key is that each time we apply f or g, depending on the previous function's parity, we can determine the parity of the next function.

But according to the previous analysis, regardless of the number of applications, the final result is even.

But let me think again. Suppose that instead of f being odd and g being even, we have different numbers of f and g. But in this case, the composition ends up being even.

Alternatively, maybe there's a general rule here. Let me think about the parity of the composition.

Let me consider that composing an odd function with another function: if the inner function is even, then the composition is odd? Wait, no. Wait, if you have an odd function f and an even function h, then f(h(x)) is even? Because f(h(-x)) = f(h(x)) since h is even. Therefore, yes, it's even.

If you have an even function h and an odd function f, then h(f(x)) is even? Wait, h is even, f is odd. Then h(f(-x)) = h(-f(x)) = h(f(x)) because h is even. Therefore, h(f(x)) is even.

Therefore, composing an even function with an odd function gives an even function.

Similarly, composing an odd function with an odd function gives an even function? Wait, no. If you have two odd functions, f and g, then f(g(-x)) = f(-g(x)) = -f(g(x)) because f is odd. Therefore, f(g(-x)) = -f(g(x)), so the composition is odd.

Similarly, composing two even functions gives an even function.

So, the key is that when you compose functions, the parity depends on the number of odd functions in the composition.

But in our case, the composition is f(f(g(f(g(f(x)))))). Let me think of the entire composition as a series of operations. Let me track the parity at each step:

Start with f(x): odd.

Then, apply g: g(f(x)) is even (even function composed with odd function).

Then, apply f: f(g(f(x))) is odd composed with even function, which is even.

Then, apply g: g(f(g(f(x)))) is even composed with even function, which is even.

Then, apply f: f(g(f(g(f(x))))) is odd composed with even function, which is even.

Then, apply f: f(f(g(f(g(f(x)))))) is odd composed with even function, which is even.

Therefore, each time, after the first step, the parity is even, and then applying f or g preserves the evenness.

Therefore, the final result is even.

Therefore, regardless of the specific 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 once more with a different example where maybe the functions are not polynomials. Let me take f(x) = x^3 (odd) and g(x) = x^2 (even). Then compute H(x) = f(f(g(f(g(f(x)))))).

Compute step by step:

Start with f(x) = x³.

g(f(x)) = (x³)² = x⁶.

f(g(f(x))) = (x⁶)³ = x¹⁸.

g(f(g(f(x)))) = (x¹⁸)² = x³⁶.

f(g(f(g(f(x))))) = (x³⁶)³ = x¹⁰⁸.

f(f(g(f(g(f(x)))))) = (x¹⁰⁸)³ = x³²⁴.

Which is even, as (-x)^324 = x^324.

Another example: f(x) = -x (odd), g(x) = x² (even). Then:

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

Compute:

f(x) = -x.

g(f(x)) = (-x)² = x².

f(g(f(x))) = -x².

g(f(g(f(x)))) = (-x²)² = x⁴.

f(g(f(g(f(x))))) = -x⁴.

f(f(g(f(g(f(x)))))) = -(-x⁴) = x⁴, which is even.

Therefore, again, even.

Therefore, I think I can be confident that the composition is even.

But just to make sure, let me think if there's any possible way that the composition could be odd. Suppose that the number of odd functions in the composition is odd? Wait, but in this case, the composition is f(f(g(f(g(f(x)))))). Let me count the number of f's. There are four f's. Each f is odd. So, composing four odd functions.

But composing odd functions:

- Odd * Odd = Even (since f(g(x)) where f and g are odd: f(g(-x)) = f(-g(x)) = -f(g(x)) = -f(g(x)), so it's odd? Wait, no. Wait, if you have two odd functions, their composition is odd. Wait, let me check:

Let f and g be odd. Then, f(g(-x)) = f(-g(x)) = -f(g(x)) because f is odd. Therefore, f(g(-x)) = -f(g(x)), so the composition is odd.

Therefore, composing two odd functions gives an odd function.

But in our case, the composition is f(f(g(f(g(f(x)))))). Let me think of it as:

Start with f(x) (odd).

Then, apply g (even): g(f(x)) is even.

Then, apply f (odd): f(g(f(x))) is even (odd composed with even is even).

Then, apply g (even): g(f(g(f(x)))) is even (even composed with even is even).

Then, apply f (odd): f(g(f(g(f(x))))) is even (odd composed with even is even).

Then, apply f (odd): f(f(g(f(g(f(x)))))) is odd composed with even, which is even.

Therefore, even.

So, even though there are four f's (odd functions), the composition ends up being even because the number of times we compose odd functions with even functions is such that the overall parity is even.

Therefore, the answer is even.

Therefore, after all these steps and examples, I think the composition is even.

**Final Answer**
\boxed{even}

To determine whether the composition
$$
f(f(g(f(g(f(x))))))
$$
is **even**, **odd**, or **neither**, we analyze the behavior of the functions under the transformation $ x \to -x $, using the definitions of **even** and **odd** functions:

- A function $ h(x) $ is **even** if $ h(-x) = h(x) $ for all $ x $.
- A function $ h(x) $ is **odd** if $ h(-x) = -h(x) $ for all $ x $.

We are given:
- $ f(x) $ is **odd**, so $ f(-x) = -f(x) $.
- $ g(x) $ is **even**, so $ g(-x) = g(x) $.

---

### Step-by-Step Analysis

Let’s evaluate the composition step by step, starting from the innermost function and moving outward.

#### Step 1: $ f(x) $
- $ f(x) $ is **odd**.

#### Step 2: $ g(f(x)) $
- Since $ f(x) $ is odd, $ f(-x) = -f(x) $.
- Then $ g(f(-x)) = g(-f(x)) = g(f(x)) $, because $ g $ is even.
- So, $ g(f(x)) $ is **even**.

#### Step 3: $ f(g(f(x))) $
- $ g(f(x)) $ is even, so $ g(f(-x)) = g(f(x)) $.
- Then $ f(g(f(-x))) = f(g(f(x))) $, because $ f $ is odd and $ g(f(x)) $ is even.
- So, $ f(g(f(x))) $ is **even**.

#### Step 4: $ g(f(g(f(x)))) $
- $ f(g(f(x))) $ is even, so $ f(g(f(-x))) = f(g(f(x))) $.
- Then $ g(f(g(f(-x)))) = g(f(g(f(x)))) $, because $ g $ is even.
- So, $ g(f(g(f(x)))) $ is **even**.

#### Step 5: $ f(g(f(g(f(x))))) $
- $ g(f(g(f(x)))) $ is even, so $ f(g(f(g(f(-x))))) = f(g(f(g(f(x))))) $.
- So, $ f(g(f(g(f(x))))) $ is **even**.

#### Step 6: $ f(f(g(f(g(f(x)))))) $
- $ f(g(f(g(f(x))))) $ is even, so $ f(f(g(f(g(f(-x))))) = f(f(g(f(g(f(x))))) $.
- So, the entire composition is **even**.

---

### General Rule

We can summarize the pattern:
- Composing an **odd** function with an **even** function results in an **even** function.
- Composing an **even** function with an **even** function results in an **even** function.
- Composing an **odd** function with an **odd** function results in an **odd** function.

In our case, the composition involves:
- 4 applications of $ f $ (odd),
- 2 applications of $ g $ (even).

Each time an **odd** function is applied to an **even** function, the result is **even**. Since the final function is built from such compositions, the result 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.


Explanation:
Okay, so I need to solve the equation (x/2) + (x/3) = 5. Hmm, let me think about how to approach this. I remember that when you have fractions in an equation, it's often helpful to eliminate them by finding a common denominator or multiplying both sides by the least common multiple of the denominators. Let me try that.

First, let me identify the denominators here. They are 2 and 3. The least common multiple of 2 and 3 is 6, right? So if I multiply both sides of the equation by 6, that should eliminate the fractions. Let me write that down:

6 * [(x/2) + (x/3)] = 6 * 5

Now, distributing the 6 to each term inside the brackets:

6*(x/2) + 6*(x/3) = 30

Calculating each term:

For the first term, 6 divided by 2 is 3, so 3x. For the second term, 6 divided by 3 is 2, so 2x. So the equation becomes:

3x + 2x = 30

Combining like terms:

3x + 2x is 5x, so:

5x = 30

Now, to solve for x, I just need to divide both sides by 5:

x = 30 / 5

Which simplifies to:

x = 6

Wait, let me check if that makes sense. Let me substitute x = 6 back into the original equation to verify.

Original equation: (6/2) + (6/3) = 5

Calculating each term:

6 divided by 2 is 3, and 6 divided by 3 is 2. Adding them together: 3 + 2 = 5. Which matches the right side of the equation. So that checks out.

But just to make sure I didn't make any mistakes in my steps, let me go through them again. Starting with multiplying both sides by 6. That's a valid step because 6 is not zero, so it's allowed. Then distributing the 6 correctly: 6*(x/2) is indeed 3x and 6*(x/3) is 2x. Adding those gives 5x, which equals 30. Dividing by 5 gives x = 6. Yep, that all seems correct.

Alternatively, maybe I can solve it another way to confirm. Let me try combining the fractions first before multiplying. The original equation is (x/2) + (x/3) = 5. To combine the fractions, I need a common denominator, which is 6 again. So converting each term:

x/2 is equivalent to 3x/6, and x/3 is equivalent to 2x/6. Adding them together:

3x/6 + 2x/6 = (3x + 2x)/6 = 5x/6

So the equation becomes:

5x/6 = 5

Then, solving for x, multiply both sides by 6:

5x = 30

Divide by 5:

x = 6

Same result. So that confirms it again. So whether I eliminate the denominators first or combine the fractions first, I end up with x = 6.

Another way to think about it: maybe using algebraic manipulation step by step. Let me try that. Starting with:

(x/2) + (x/3) = 5

I can factor out x from both terms:

x*(1/2 + 1/3) = 5

Calculating the sum inside the parentheses. 1/2 + 1/3. To add these, common denominator is 6:

(3/6 + 2/6) = 5/6

So the equation becomes:

x*(5/6) = 5

Which is the same as:

(5/6)x = 5

Then, solving for x, multiply both sides by the reciprocal of 5/6, which is 6/5:

x = 5*(6/5)

Simplifying, the 5s cancel out:

x = 6

Same answer again. So all methods lead to x = 6. Therefore, I can be confident that the solution is correct.

I think that's thorough enough. I considered different methods to solve the equation, checked my answer by substitution, and verified each step. All paths lead to the same solution, so I don't see any mistakes here.

**Final Answer**
The solution is \boxed{6}.

To solve the equation:

$$
\frac{x}{2} + \frac{x}{3} = 5
$$

we aim to isolate $ x $ by eliminating the fractions and simplifying the expression.

---

### Step 1: Eliminate the Denominators

The denominators are 2 and 3. The least common multiple (LCM) of 2 and 3 is 6. To eliminate the fractions, multiply both sides of the equation by 6:

$$
6 \left( \frac{x}{2} + \frac{x}{3} \right) = 6 \cdot 5
$$

Distribute the 6 across the terms inside the parentheses:

$$
6 \cdot \frac{x}{2} + 6 \cdot \frac{x}{3} = 30
$$

Simplify each term:

$$
3x + 2x = 30
$$

---

### Step 2: Combine Like Terms

$$
5x = 30
$$

---

### Step 3: Solve for $ x $

Divide both sides by 5:

$$
x = \frac{30}{5} = 6
$$

---

### Step 4: Verify the Solution

Substitute $ x = 6 $ back into the original equation:

$$
\frac{6}{2} + \frac{6}{3} = 3 + 2 = 5
$$

This matches the right-hand side of the equation, confirming the solution is correct.

---

### Final Answer

$$
\boxed{6}
$$

[Assistant]
Explanation:
Okay, so I need to solve the equation (x/2) + (x/3) = 5. Hmm, let me think about how to approach this. I remember that when you have fractions in an equation, it's often helpful to eliminate them by finding a common denominator or multiplying both sides by the least common multiple of the denominators. Let me try that.

First, let me identify the denominators here. They are 2 and 3. The least common multiple of 2 and 3 is 6, right? So if I multiply both sides of the equation by 6, that should eliminate the fractions. Let me write that down:

6 * [(x/2) + (x/3)] = 6 * 5

Now, distributing the 6 to each term inside the brackets:

6*(x/2) + 6*(x/3) = 30

Calculating each term:

For the first term, 6 divided by 2 is 3, so 3x. For the second term, 6 divided by 3 is 2, so 2x. So the equation becomes:

3x + 2x = 30

Combining like terms:

3x + 2x is 5x, so:

5x = 30

Now, to solve for x, I just need to divide both sides by 5:

x = 30 / 5

Which simplifies to:

x = 6

Wait, let me check if that makes sense. Let me substitute x = 6 back into the original equation to verify.

Original equation: (6/2) + (6/3) = 5

Calculating each term:

6 divided by 2 is 3, and 6 divided by 3 is 2. Adding them together: 3 + 2 = 5. Which matches the right side of the equation. So that checks out.

But just to make sure I didn't make any mistakes in my steps, let me go through them again. Starting with multiplying both sides by 6. That's a valid step because 6 is not zero, so it's allowed. Then distributing the 6 correctly: 6*(x/2) is indeed 3x and 6*(x/3) is 2x. Adding those gives 5x, which equals 30. Dividing by 5 gives x = 6. Yep, that all seems correct.

Alternatively, maybe I can solve it another way to confirm. Let me try combining the fractions first before multiplying. The original equation is (x/2) + (x/3) = 5. To combine the fractions, I need a common denominator, which is 6 again. So converting each term:

x/2 is equivalent to 3x/6, and x/3 is equivalent to 2x/6. Adding them together:

3x/6 + 2x/6 = (3x + 2x)/6 = 5x/6

So the equation becomes:

5x/6 = 5

Then, solving for x, multiply both sides by 6:

5x = 30

Divide by 5:

x = 6

Same result. So that confirms it again. So whether I eliminate the denominators first or combine the fractions first, I end up with x = 6.

Another way to think about it: maybe using algebraic manipulation step by step. Let me try that. Starting with:

(x/2) + (x/3) = 5

I can factor out x from both terms:

x*(1/2 + 1/3) = 5

Calculating the sum inside the parentheses. 1/2 + 1/3. To add these, common denominator is 6:

(3/6 + 2/6) = 5/6

So the equation becomes:

x*(5/6) = 5

Which is the same as:

(5/6)x = 5

Then, solving for x, multiply both sides by the reciprocal of 5/6, which is 6/5:

x = 5*(6/5)

Simplifying, the 5s cancel out:

x = 6

Same answer again. So all methods lead to x = 6. Therefore, I can be confident that the solution is correct.

I think that's thorough enough. I considered different methods to solve the equation, checked my answer by substitution, and verified each step. All paths lead to the same solution, so I don't see any mistakes here.

**Final Answer**
The solution is \boxed{6}.

To solve the equation:

$$
\frac{x}{2} + \frac{x}{3} = 5
$$

we aim to isolate $ x $ by eliminating the fractions and simplifying the expression.

---

### Step 1: Eliminate the Denominators

The denominators are 2 and 3. The least common multiple (LCM) of 2 and 3 is 6. To eliminate the fractions, multiply both sides of the equation by 6:

$$
6 \left( \frac{x}{2} + \frac{x}{3} \right) = 6 \cdot 5
$$

Distribute the 6 across the terms inside the parentheses:

$$
6 \cdot \frac{x}{2} + 6 \cdot \frac{x}{3} = 30
$$

Simplify each term:

$$
3x + 2x = 30
$$

---

### Step 2: Combine Like Terms

$$
5x = 30
$$

---

### Step 3: Solve for $ x $

Divide both sides by 5:

$$
x = \frac{30}{5} = 6
$$

---

### Step 4: Verify the Solution

Substitute $ x = 6 $ back into the original equation:

$$
\frac{6}{2} + \frac{6}{3} = 3 + 2 = 5
$$

This matches the right-hand side of the equation, confirming the solution is correct.

---

### Final Answer

$$
\boxed{6}
$$

Solution: Writing the left-hand side with a common denominator, we have$$\dfrac{x}{2} + \dfrac{x}{3} = \dfrac{3x}{6} + \dfrac{2x}{6} = \dfrac{5x}{6},$$ so our equation is $$\dfrac{5x}{6} = 5.$$Multiplying both sides by $\dfrac{6}{5}$ gives $$x = 5\cdot \dfrac{6}{5} = \boxed{6}.$$