Explanation:
Okay, so I need to solve the equation $3^{2x} + 19 = 10^x$. Let me think about how to approach this. Hmm, first, maybe I can simplify the equation or rewrite it in a way that makes it easier to solve. Let me see...

The left side has $3^{2x}$, which I know is the same as $(3^2)^x$, so that simplifies to $9^x$. So the equation becomes $9^x + 19 = 10^x$. That seems a bit simpler. Now, I have an equation with two exponential terms: $9^x$ and $10^x$.

I remember that equations with different bases can sometimes be tricky. Maybe I can take logarithms? But since there are two exponential terms, taking logarithms directly might not be straightforward. Let me think... If I subtract $9^x$ from both sides, I get $19 = 10^x - 9^x$. Not sure if that helps yet.

Alternatively, maybe I can let $y = x$ and try to find a substitution? Wait, maybe if I let $t = x$, but that doesn't help. Wait, perhaps I can write both sides in terms of the same base? But 9 and 10 don't share a common base. 9 is 3 squared, but 10 is 2 times 5. Not sure if that helps.

Another thought: maybe I can consider this as a transcendental equation, which might not have an algebraic solution, so I might need to use numerical methods or graphing. But since this is a problem likely expecting an exact answer, maybe there's a clever substitution or integer solution?

Let me check if there are integer solutions. Let me try plugging in some integer values for x.

Let me try x=1: Left side is 9^1 + 19 = 9 + 19 = 28. Right side is 10^1 = 10. Not equal.

x=2: Left side is 9^2 + 19 = 81 + 19 = 100. Right side is 10^2 = 100. Oh! They are equal! So x=2 is a solution.

Wait, so x=2 works. Let me check x=3: Left side is 9^3 + 19 = 729 + 19 = 748. Right side is 10^3 = 1000. Not equal.

x=0: Left side is 9^0 + 19 = 1 + 19 = 20. Right side is 10^0 = 1. Not equal.

x= -1: Left side is 9^{-1} + 19 = 1/9 + 19 ≈ 19.11. Right side is 10^{-1} = 0.1. Not equal.

So x=2 is a solution. But is there another solution? Let me check between x=1 and x=2. At x=1, left side is 28, right side is 10. At x=2, both are 100. So the left side increases from 28 to 100, and the right side increases from 10 to 100. So maybe they cross at x=2? Wait, but maybe there's another crossing? Let me check for x between 0 and 1. At x=0, left is 20, right is 1. At x=1, left is 28, right is 10. So left side is always above right side here. For x>2, let's check x=3: left is 748 vs right 1000. So left is less than right. So maybe there's another solution between x=2 and x=3? Wait, but at x=2, they are equal. Let me check the behavior of the functions.

Let me define f(x) = 10^x - 9^x - 19. We found that f(2) = 0. Let me check f(1) = 10 - 9 -19 = -18. f(2)=0. f(3)=1000 - 729 -19= 252. So f(x) increases from x=1 to x=2, crosses zero at x=2, then continues increasing? Wait, but 10^x grows faster than 9^x, so the difference 10^x - 9^x would increase as x increases. Therefore, f(x) is increasing for all x? Wait, is that true? Let me check the derivative.

If I take derivative of f(x) = 10^x - 9^x -19, then f’(x) = ln(10)*10^x - ln(9)*9^x. Since ln(10) > ln(9), and 10^x grows faster than 9^x, so f’(x) is positive for all x? Let me check for x=0: f’(0) = ln(10) - ln(9) ≈ 2.302 - 2.197 ≈ 0.105 >0. For x approaching negative infinity, 10^x and 9^x both approach zero, but which term dominates? Since 10^x approaches zero faster than 9^x? Wait, no. For negative x, say x = -a where a>0, then 10^{-a} = 1/(10^a), 9^{-a} = 1/(9^a). Since 10^a > 9^a, so 1/(10^a) < 1/(9^a). Therefore, 10^x < 9^x for x negative. Therefore, f’(x) = ln(10)*10^x - ln(9)*9^x. For x negative, since 10^x < 9^x, and ln(10) > ln(9), so ln(10)*10^x - ln(9)*9^x. Let me see, if I factor out 10^x, it's 10^x [ln(10) - ln(9)*(9/10)^x]. Since (9/10)^x for x negative is (10/9)^{|x|}, which is greater than 1. Therefore, ln(10) - ln(9)*(something greater than 1). Since ln(10) ≈ 2.302, ln(9) ≈ 2.197. Let me compute ln(10) - ln(9)*(something). For example, if x approaches negative infinity, (9/10)^x approaches infinity, so the term becomes negative infinity. Therefore, f’(x) approaches negative infinity? Wait, but that contradicts with earlier thought. Wait, maybe my analysis is wrong.

Alternatively, maybe f’(x) is positive for all x? Wait, but if x is very negative, then 10^x is much smaller than 9^x, so ln(10)*10^x - ln(9)*9^x would be negative? Let me take x = -1. Then f’(-1) = ln(10)/10 - ln(9)/9 ≈ 2.302/10 - 2.197/9 ≈ 0.2302 - 0.244 ≈ -0.0138. So negative. Therefore, f’(-1) is negative. Therefore, the function f(x) is decreasing at x=-1. Therefore, the derivative is not always positive. Therefore, there might be some critical points.

Therefore, maybe there are two solutions? Wait, but we found x=2 is a solution. Let me check if there is another solution. Let me check for x < 2. For example, when x approaches negative infinity, f(x) = 10^x - 9^x -19 approaches -19, since both 10^x and 9^x go to zero. At x=0, f(0) = 1 - 1 -19 = -19. At x=1, f(1)=10 -9 -19 = -18. At x=2, f(2)=0. So from x approaching -infty to x=2, f(x) goes from -19 to 0. But since the derivative is negative at x=-1, maybe there is a minimum somewhere? Let me check.

If f’(x) = ln(10)*10^x - ln(9)*9^x. Let me set derivative to zero: ln(10)*10^x = ln(9)*9^x. Let me divide both sides by 9^x: ln(10)*(10/9)^x = ln(9). Therefore, (10/9)^x = ln(9)/ln(10). Taking natural logarithm on both sides: x*ln(10/9) = ln(ln(9)/ln(10)). Therefore, x = [ln(ln(9)/ln(10))]/ln(10/9). Let me compute that.

First compute ln(9) ≈ 2.1972, ln(10) ≈ 2.3026. Therefore, ln(9)/ln(10) ≈ 2.1972 / 2.3026 ≈ 0.9542. Then ln(0.9542) ≈ -0.047. Then ln(10/9) ≈ ln(1.1111) ≈ 0.1054. Therefore, x ≈ -0.047 / 0.1054 ≈ -0.446. So there is a critical point at x ≈ -0.446.

Therefore, the function f(x) has a critical point at x ≈ -0.446. Let me check the value of f at that critical point. Let me approximate. Let me take x ≈ -0.446. Let me compute 10^x - 9^x -19.

First, 10^{-0.446} ≈ e^{-0.446 * ln10} ≈ e^{-0.446 * 2.3026} ≈ e^{-1.027} ≈ 0.359. Similarly, 9^{-0.446} ≈ e^{-0.446 * ln9} ≈ e^{-0.446 * 2.1972} ≈ e^{-0.979} ≈ 0.376. Therefore, f(x) ≈ 0.359 - 0.376 -19 ≈ -0.017 -19 ≈ -19.017. So the minimum value of f(x) is approximately -19.017, which is slightly less than -19. Therefore, the function f(x) approaches -19 as x approaches -infty, reaches a minimum of about -19.017 at x≈-0.446, then increases to f(2)=0. Therefore, since f(x) is increasing from x≈-0.446 onwards? Wait, the derivative was zero at x≈-0.446, and since for x > -0.446, the derivative is positive? Let me check at x=0, derivative was ≈0.105>0. So yes, after the critical point, the function is increasing. Therefore, since f(x) approaches -19 as x approaches -infty, reaches a minimum of about -19.017, then increases to f(2)=0, and then continues increasing to infinity. Therefore, the equation f(x)=0 has only one solution at x=2? Because before x≈-0.446, the function is decreasing from -19 to -19.017, then increasing back to 0 at x=2. Therefore, since it only crosses zero once at x=2.

Wait, but if the minimum is -19.017, which is less than -19, but as x approaches -infty, f(x) approaches -19. So the function approaches -19 from below? Wait, no. Wait, when x approaches -infty, 10^x and 9^x both approach zero, so f(x) approaches -19. But the minimum is at x≈-0.446, where f(x)≈-19.017. Therefore, the function f(x) is decreasing from x=-infty to x≈-0.446, reaching a minimum of -19.017, then increasing from there to infinity. Therefore, since f(x) approaches -19 as x approaches -infty, and reaches -19.017, then increases to 0 at x=2. Therefore, the equation f(x)=0 has only one solution at x=2.

Therefore, the only real solution is x=2.

But let me check if there's another solution for x>2. For example, as x increases beyond 2, f(x) = 10^x -9^x -19. At x=2, it's 0. At x=3, it's 1000 - 729 -19 = 252. At x=4, 10^4 -9^4 -19 = 10000 - 6561 -19 = 3420. So it's increasing. Therefore, since f(x) is increasing for x > -0.446, and since at x=2 it's zero, and increasing afterwards, there are no other solutions.

Therefore, the only real solution is x=2.

But let me check if there's a possibility of another solution between x=-infty and x=-0.446? Since f(x) approaches -19 as x approaches -infty, and reaches -19.017 at x≈-0.446. So it's always less than -19. Therefore, f(x) = 0 is not achieved there. Therefore, only x=2 is the solution.

Therefore, the answer is x=2.

But let me verify once again. Let me plug x=2 into original equation: 3^{2*2} +19 = 3^4 +19 = 81 +19 = 100. And 10^2 = 100. Yes, that works.

Is there a way to solve this algebraically? Maybe by substitution? Let me think.

Original equation: 9^x +19 = 10^x. Let me write it as 10^x - 9^x = 19.

Let me consider substituting t = (10/9)^x. Let me see. Let me divide both sides by 9^x: (10/9)^x - 1 = 19 / 9^x. Hmm, not sure. Alternatively, let me set t = (10/9)^x. Then 10^x = t * 9^x. Therefore, original equation: 9^x +19 = t * 9^x. Therefore, 19 = t * 9^x - 9^x = 9^x (t -1). Therefore, 9^x = 19 / (t -1). But t = (10/9)^x. Therefore, 9^x = 19 / [ (10/9)^x -1 ].

But 9^x is also equal to (9^x). Let me write 9^x = 19 / [ (10/9)^x -1 ].

Let me denote y = (10/9)^x. Then 9^x = (9^x). Wait, but y = (10/9)^x, so taking natural logarithm: ln y = x ln(10/9). Therefore, x = ln y / ln(10/9). Also, 9^x = e^{x ln 9} = e^{ [ln y / ln(10/9)] * ln 9 }.

This seems complicated. Maybe not helpful. Alternatively, since y = (10/9)^x, then 9^x = 10^x / y. Therefore, original equation: 10^x / y + 19 = 10^x. Therefore, 19 = 10^x - 10^x / y = 10^x (1 - 1/y). Therefore, 10^x = 19 / (1 - 1/y) = 19y / (y -1). But y = (10/9)^x, so 10^x = 19y / (y -1). Also, since y = (10/9)^x, then 10^x = y * 9^x. Therefore, substituting into previous equation: y * 9^x = 19y / (y -1). Assuming y ≠ 0, we can divide both sides by y: 9^x = 19 / (y -1). But y = (10/9)^x, so 9^x = 19 / [ (10/9)^x -1 ].

This seems to bring us back to where we were before. Maybe not helpful.

Alternatively, let me consider that 10^x - 9^x = 19. Let me think about possible integer solutions. We saw x=2 works. Maybe there's another integer? For x=1, 10 -9=1≠19. x=0: 1 -1=0≠19. x=3: 1000 - 729=271≠19. So no.

Alternatively, maybe x is a fraction? Let me try x= log_{10}(something). But this seems vague.

Alternatively, since we found x=2 works, and the function is increasing for x > -0.446, and since at x=2 it's zero, and for x>2 it's increasing, so only one solution. Therefore, x=2 is the only solution.

Therefore, the answer is \boxed{2}.

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

To solve the equation:

$$
3^{2x} + 19 = 10^x
$$

---

### Step 1: Simplify the Equation

We begin by simplifying the left-hand side:

$$
3^{2x} = (3^2)^x = 9^x
$$

So the equation becomes:

$$
9^x + 19 = 10^x
$$

---

### Step 2: Rearranging the Equation

Subtract $9^x$ from both sides:

$$
10^x - 9^x = 19
$$

Let’s define a function:

$$
f(x) = 10^x - 9^x - 19
$$

We are looking for the value(s) of $x$ such that $f(x) = 0$.

---

### Step 3: Try Integer Values

Let’s test some integer values of $x$:

- $x = 1$: $10^1 - 9^1 - 19 = 10 - 9 - 19 = -18$
- $x = 2$: $10^2 - 9^2 - 19 = 100 - 81 - 19 = 0$

So, $x = 2$ is a solution.

---

### Step 4: Analyze the Behavior of the Function

We define:

$$
f(x) = 10^x - 9^x - 19
$$

We analyze the derivative to understand the behavior of the function:

$$
f'(x) = \ln(10) \cdot 10^x - \ln(9) \cdot 9^x
$$

Since $\ln(10) > \ln(9)$ and $10^x$ grows faster than $9^x$, the derivative is positive for all $x > 0$, indicating that $f(x)$ is increasing for $x > 0$.

We also check the behavior as $x \to -\infty$:

- $10^x \to 0$, $9^x \to 0$, so $f(x) \to -19$

At $x = 2$, $f(x) = 0$, and since the function is increasing for $x > 0$, there are no other solutions for $x > 2$.

For $x < 2$, the function is decreasing (as the derivative is negative for $x < 0$), and since $f(x) \to -19$ as $x \to -\infty$, the function never reaches zero again for $x < 2$.

---

### Step 5: Conclusion

The function $f(x) = 10^x - 9^x - 19$ is increasing for $x > 0$, and it only crosses zero once at $x = 2$. Therefore, the only real solution to the equation is:

$$
\boxed{2}
$$

[Assistant]
Explanation:
Okay, so I need to solve the equation $3^{2x} + 19 = 10^x$. Let me think about how to approach this. Hmm, first, maybe I can simplify the equation or rewrite it in a way that makes it easier to solve. Let me see...

The left side has $3^{2x}$, which I know is the same as $(3^2)^x$, so that simplifies to $9^x$. So the equation becomes $9^x + 19 = 10^x$. That seems a bit simpler. Now, I have an equation with two exponential terms: $9^x$ and $10^x$.

I remember that equations with different bases can sometimes be tricky. Maybe I can take logarithms? But since there are two exponential terms, taking logarithms directly might not be straightforward. Let me think... If I subtract $9^x$ from both sides, I get $19 = 10^x - 9^x$. Not sure if that helps yet.

Alternatively, maybe I can let $y = x$ and try to find a substitution? Wait, maybe if I let $t = x$, but that doesn't help. Wait, perhaps I can write both sides in terms of the same base? But 9 and 10 don't share a common base. 9 is 3 squared, but 10 is 2 times 5. Not sure if that helps.

Another thought: maybe I can consider this as a transcendental equation, which might not have an algebraic solution, so I might need to use numerical methods or graphing. But since this is a problem likely expecting an exact answer, maybe there's a clever substitution or integer solution?

Let me check if there are integer solutions. Let me try plugging in some integer values for x.

Let me try x=1: Left side is 9^1 + 19 = 9 + 19 = 28. Right side is 10^1 = 10. Not equal.

x=2: Left side is 9^2 + 19 = 81 + 19 = 100. Right side is 10^2 = 100. Oh! They are equal! So x=2 is a solution.

Wait, so x=2 works. Let me check x=3: Left side is 9^3 + 19 = 729 + 19 = 748. Right side is 10^3 = 1000. Not equal.

x=0: Left side is 9^0 + 19 = 1 + 19 = 20. Right side is 10^0 = 1. Not equal.

x= -1: Left side is 9^{-1} + 19 = 1/9 + 19 ≈ 19.11. Right side is 10^{-1} = 0.1. Not equal.

So x=2 is a solution. But is there another solution? Let me check between x=1 and x=2. At x=1, left side is 28, right side is 10. At x=2, both are 100. So the left side increases from 28 to 100, and the right side increases from 10 to 100. So maybe they cross at x=2? Wait, but maybe there's another crossing? Let me check for x between 0 and 1. At x=0, left is 20, right is 1. At x=1, left is 28, right is 10. So left side is always above right side here. For x>2, let's check x=3: left is 748 vs right 1000. So left is less than right. So maybe there's another solution between x=2 and x=3? Wait, but at x=2, they are equal. Let me check the behavior of the functions.

Let me define f(x) = 10^x - 9^x - 19. We found that f(2) = 0. Let me check f(1) = 10 - 9 -19 = -18. f(2)=0. f(3)=1000 - 729 -19= 252. So f(x) increases from x=1 to x=2, crosses zero at x=2, then continues increasing? Wait, but 10^x grows faster than 9^x, so the difference 10^x - 9^x would increase as x increases. Therefore, f(x) is increasing for all x? Wait, is that true? Let me check the derivative.

If I take derivative of f(x) = 10^x - 9^x -19, then f’(x) = ln(10)*10^x - ln(9)*9^x. Since ln(10) > ln(9), and 10^x grows faster than 9^x, so f’(x) is positive for all x? Let me check for x=0: f’(0) = ln(10) - ln(9) ≈ 2.302 - 2.197 ≈ 0.105 >0. For x approaching negative infinity, 10^x and 9^x both approach zero, but which term dominates? Since 10^x approaches zero faster than 9^x? Wait, no. For negative x, say x = -a where a>0, then 10^{-a} = 1/(10^a), 9^{-a} = 1/(9^a). Since 10^a > 9^a, so 1/(10^a) < 1/(9^a). Therefore, 10^x < 9^x for x negative. Therefore, f’(x) = ln(10)*10^x - ln(9)*9^x. For x negative, since 10^x < 9^x, and ln(10) > ln(9), so ln(10)*10^x - ln(9)*9^x. Let me see, if I factor out 10^x, it's 10^x [ln(10) - ln(9)*(9/10)^x]. Since (9/10)^x for x negative is (10/9)^{|x|}, which is greater than 1. Therefore, ln(10) - ln(9)*(something greater than 1). Since ln(10) ≈ 2.302, ln(9) ≈ 2.197. Let me compute ln(10) - ln(9)*(something). For example, if x approaches negative infinity, (9/10)^x approaches infinity, so the term becomes negative infinity. Therefore, f’(x) approaches negative infinity? Wait, but that contradicts with earlier thought. Wait, maybe my analysis is wrong.

Alternatively, maybe f’(x) is positive for all x? Wait, but if x is very negative, then 10^x is much smaller than 9^x, so ln(10)*10^x - ln(9)*9^x would be negative? Let me take x = -1. Then f’(-1) = ln(10)/10 - ln(9)/9 ≈ 2.302/10 - 2.197/9 ≈ 0.2302 - 0.244 ≈ -0.0138. So negative. Therefore, f’(-1) is negative. Therefore, the function f(x) is decreasing at x=-1. Therefore, the derivative is not always positive. Therefore, there might be some critical points.

Therefore, maybe there are two solutions? Wait, but we found x=2 is a solution. Let me check if there is another solution. Let me check for x < 2. For example, when x approaches negative infinity, f(x) = 10^x - 9^x -19 approaches -19, since both 10^x and 9^x go to zero. At x=0, f(0) = 1 - 1 -19 = -19. At x=1, f(1)=10 -9 -19 = -18. At x=2, f(2)=0. So from x approaching -infty to x=2, f(x) goes from -19 to 0. But since the derivative is negative at x=-1, maybe there is a minimum somewhere? Let me check.

If f’(x) = ln(10)*10^x - ln(9)*9^x. Let me set derivative to zero: ln(10)*10^x = ln(9)*9^x. Let me divide both sides by 9^x: ln(10)*(10/9)^x = ln(9). Therefore, (10/9)^x = ln(9)/ln(10). Taking natural logarithm on both sides: x*ln(10/9) = ln(ln(9)/ln(10)). Therefore, x = [ln(ln(9)/ln(10))]/ln(10/9). Let me compute that.

First compute ln(9) ≈ 2.1972, ln(10) ≈ 2.3026. Therefore, ln(9)/ln(10) ≈ 2.1972 / 2.3026 ≈ 0.9542. Then ln(0.9542) ≈ -0.047. Then ln(10/9) ≈ ln(1.1111) ≈ 0.1054. Therefore, x ≈ -0.047 / 0.1054 ≈ -0.446. So there is a critical point at x ≈ -0.446.

Therefore, the function f(x) has a critical point at x ≈ -0.446. Let me check the value of f at that critical point. Let me approximate. Let me take x ≈ -0.446. Let me compute 10^x - 9^x -19.

First, 10^{-0.446} ≈ e^{-0.446 * ln10} ≈ e^{-0.446 * 2.3026} ≈ e^{-1.027} ≈ 0.359. Similarly, 9^{-0.446} ≈ e^{-0.446 * ln9} ≈ e^{-0.446 * 2.1972} ≈ e^{-0.979} ≈ 0.376. Therefore, f(x) ≈ 0.359 - 0.376 -19 ≈ -0.017 -19 ≈ -19.017. So the minimum value of f(x) is approximately -19.017, which is slightly less than -19. Therefore, the function f(x) approaches -19 as x approaches -infty, reaches a minimum of about -19.017 at x≈-0.446, then increases to f(2)=0. Therefore, since f(x) is increasing from x≈-0.446 onwards? Wait, the derivative was zero at x≈-0.446, and since for x > -0.446, the derivative is positive? Let me check at x=0, derivative was ≈0.105>0. So yes, after the critical point, the function is increasing. Therefore, since f(x) approaches -19 as x approaches -infty, reaches a minimum of about -19.017, then increases to f(2)=0, and then continues increasing to infinity. Therefore, the equation f(x)=0 has only one solution at x=2? Because before x≈-0.446, the function is decreasing from -19 to -19.017, then increasing back to 0 at x=2. Therefore, since it only crosses zero once at x=2.

Wait, but if the minimum is -19.017, which is less than -19, but as x approaches -infty, f(x) approaches -19. So the function approaches -19 from below? Wait, no. Wait, when x approaches -infty, 10^x and 9^x both approach zero, so f(x) approaches -19. But the minimum is at x≈-0.446, where f(x)≈-19.017. Therefore, the function f(x) is decreasing from x=-infty to x≈-0.446, reaching a minimum of -19.017, then increasing from there to infinity. Therefore, since f(x) approaches -19 as x approaches -infty, and reaches -19.017, then increases to 0 at x=2. Therefore, the equation f(x)=0 has only one solution at x=2.

Therefore, the only real solution is x=2.

But let me check if there's another solution for x>2. For example, as x increases beyond 2, f(x) = 10^x -9^x -19. At x=2, it's 0. At x=3, it's 1000 - 729 -19 = 252. At x=4, 10^4 -9^4 -19 = 10000 - 6561 -19 = 3420. So it's increasing. Therefore, since f(x) is increasing for x > -0.446, and since at x=2 it's zero, and increasing afterwards, there are no other solutions.

Therefore, the only real solution is x=2.

But let me check if there's a possibility of another solution between x=-infty and x=-0.446? Since f(x) approaches -19 as x approaches -infty, and reaches -19.017 at x≈-0.446. So it's always less than -19. Therefore, f(x) = 0 is not achieved there. Therefore, only x=2 is the solution.

Therefore, the answer is x=2.

But let me verify once again. Let me plug x=2 into original equation: 3^{2*2} +19 = 3^4 +19 = 81 +19 = 100. And 10^2 = 100. Yes, that works.

Is there a way to solve this algebraically? Maybe by substitution? Let me think.

Original equation: 9^x +19 = 10^x. Let me write it as 10^x - 9^x = 19.

Let me consider substituting t = (10/9)^x. Let me see. Let me divide both sides by 9^x: (10/9)^x - 1 = 19 / 9^x. Hmm, not sure. Alternatively, let me set t = (10/9)^x. Then 10^x = t * 9^x. Therefore, original equation: 9^x +19 = t * 9^x. Therefore, 19 = t * 9^x - 9^x = 9^x (t -1). Therefore, 9^x = 19 / (t -1). But t = (10/9)^x. Therefore, 9^x = 19 / [ (10/9)^x -1 ].

But 9^x is also equal to (9^x). Let me write 9^x = 19 / [ (10/9)^x -1 ].

Let me denote y = (10/9)^x. Then 9^x = (9^x). Wait, but y = (10/9)^x, so taking natural logarithm: ln y = x ln(10/9). Therefore, x = ln y / ln(10/9). Also, 9^x = e^{x ln 9} = e^{ [ln y / ln(10/9)] * ln 9 }.

This seems complicated. Maybe not helpful. Alternatively, since y = (10/9)^x, then 9^x = 10^x / y. Therefore, original equation: 10^x / y + 19 = 10^x. Therefore, 19 = 10^x - 10^x / y = 10^x (1 - 1/y). Therefore, 10^x = 19 / (1 - 1/y) = 19y / (y -1). But y = (10/9)^x, so 10^x = 19y / (y -1). Also, since y = (10/9)^x, then 10^x = y * 9^x. Therefore, substituting into previous equation: y * 9^x = 19y / (y -1). Assuming y ≠ 0, we can divide both sides by y: 9^x = 19 / (y -1). But y = (10/9)^x, so 9^x = 19 / [ (10/9)^x -1 ].

This seems to bring us back to where we were before. Maybe not helpful.

Alternatively, let me consider that 10^x - 9^x = 19. Let me think about possible integer solutions. We saw x=2 works. Maybe there's another integer? For x=1, 10 -9=1≠19. x=0: 1 -1=0≠19. x=3: 1000 - 729=271≠19. So no.

Alternatively, maybe x is a fraction? Let me try x= log_{10}(something). But this seems vague.

Alternatively, since we found x=2 works, and the function is increasing for x > -0.446, and since at x=2 it's zero, and for x>2 it's increasing, so only one solution. Therefore, x=2 is the only solution.

Therefore, the answer is \boxed{2}.

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

To solve the equation:

$$
3^{2x} + 19 = 10^x
$$

---

### Step 1: Simplify the Equation

We begin by simplifying the left-hand side:

$$
3^{2x} = (3^2)^x = 9^x
$$

So the equation becomes:

$$
9^x + 19 = 10^x
$$

---

### Step 2: Rearranging the Equation

Subtract $9^x$ from both sides:

$$
10^x - 9^x = 19
$$

Let’s define a function:

$$
f(x) = 10^x - 9^x - 19
$$

We are looking for the value(s) of $x$ such that $f(x) = 0$.

---

### Step 3: Try Integer Values

Let’s test some integer values of $x$:

- $x = 1$: $10^1 - 9^1 - 19 = 10 - 9 - 19 = -18$
- $x = 2$: $10^2 - 9^2 - 19 = 100 - 81 - 19 = 0$

So, $x = 2$ is a solution.

---

### Step 4: Analyze the Behavior of the Function

We define:

$$
f(x) = 10^x - 9^x - 19
$$

We analyze the derivative to understand the behavior of the function:

$$
f'(x) = \ln(10) \cdot 10^x - \ln(9) \cdot 9^x
$$

Since $\ln(10) > \ln(9)$ and $10^x$ grows faster than $9^x$, the derivative is positive for all $x > 0$, indicating that $f(x)$ is increasing for $x > 0$.

We also check the behavior as $x \to -\infty$:

- $10^x \to 0$, $9^x \to 0$, so $f(x) \to -19$

At $x = 2$, $f(x) = 0$, and since the function is increasing for $x > 0$, there are no other solutions for $x > 2$.

For $x < 2$, the function is decreasing (as the derivative is negative for $x < 0$), and since $f(x) \to -19$ as $x \to -\infty$, the function never reaches zero again for $x < 2$.

---

### Step 5: Conclusion

The function $f(x) = 10^x - 9^x - 19$ is increasing for $x > 0$, and it only crosses zero once at $x = 2$. Therefore, the only real solution to the equation is:

$$
\boxed{2}
$$

Solution:
Rewrite $3^{2x}$ as $(3^2)^x=9^x$, and subtract $9^x$ from both sides to obtain $19=10^x-9^x$.  There are no solutions of this equation for $x\leq 0$, because neither $10^x$ nor $9^x$ is greater than 1 if $x\leq 0$.  Trying $x=1$, $x=2$, and $x=3$, we see that $10^x-9^x$ is increasing for $x>0$, and it equals 19 when $x=\boxed{2}$.

Note: Using calculus, we could prove that $10^x-9^x$ is monotonically increasing for $x>0$, which would prove that the solution we found is unique.


Explanation:
Okay, so I need to find the value of y such that the projection of the vector [2, y, -5] onto the vector [1, -2, 1] is equal to (5/6)*[1, -2, 1]. Hmm, projections... Let me recall how that works.

First, the formula for the projection of a vector **a** onto another vector **b** is given by:

proj_b **a** = ( (**a** · **b**) / ||**b**||² ) * **b**

So, in this case, **a** is [2, y, -5] and **b** is [1, -2, 1]. The projection is given as (5/6)*[1, -2, 1]. Therefore, I can set up the equation using the projection formula and solve for y.

Let me compute the dot product **a** · **b** first. That would be:

(2)(1) + (y)(-2) + (-5)(1) = 2 - 2y - 5 = (-3 - 2y)

Then, the norm squared of **b** is ||**b**||² = (1)² + (-2)² + (1)² = 1 + 4 + 1 = 6.

So plugging into the projection formula:

proj_b **a** = ( (-3 - 2y)/6 ) * [1, -2, 1]

But according to the problem, this projection is equal to (5/6)*[1, -2, 1]. Therefore, we can set the scalar multiples equal to each other because the vectors are scalar multiples of [1, -2, 1].

So:

(-3 - 2y)/6 = 5/6

Since the denominators are the same, we can equate the numerators:

-3 - 2y = 5

Solving for y:

-2y = 5 + 3

-2y = 8

y = 8 / (-2) = -4

Wait, so y is -4? Let me check my steps again to make sure I didn't make a mistake.

First, the dot product: 2*1 is 2, y*(-2) is -2y, -5*1 is -5. So total is 2 - 2y -5 = -3 - 2y. That seems right.

Then ||**b**||² is 1 + 4 + 1 = 6. Correct.

So the scalar multiplier is (-3 - 2y)/6. The projection is that scalar times **b**, which is equal to (5/6)**b**. Therefore, the scalar must be 5/6. Therefore:

(-3 - 2y)/6 = 5/6 => -3 - 2y = 5 => -2y = 8 => y = -4. Seems correct.

But let me verify by plugging y = -4 back into the original vector and computing the projection.

Original vector **a** is [2, -4, -5]. Let's compute the projection.

First, **a** · **b** = 2*1 + (-4)*(-2) + (-5)*1 = 2 + 8 - 5 = 5. Wait, that's 5. Then divide by ||**b**||² which is 6. So the scalar is 5/6. Therefore, the projection is (5/6)*[1, -2, 1], which matches the given projection. So that checks out. Therefore, y = -4 is correct.

Wait, but hold on, when I first calculated the dot product with y, I had -3 - 2y. If y = -4, then that would be -3 -2*(-4) = -3 + 8 = 5. Which is correct. So that's exactly what I just did. So yes, that works. Therefore, the answer is y = -4.

But let me think again if there's another way this could be approached, just to confirm.

Alternatively, since the projection is given, maybe we can think about the fact that the projection vector is a scalar multiple of **b**, so the vector [2, y, -5] minus its projection should be orthogonal to **b**. That is, (**a** - proj_b **a**) · **b** = 0.

Let me try that approach as a check.

Let me denote proj_b **a** as (5/6)**b**. Therefore, **a** - (5/6)**b** should be orthogonal to **b**.

Compute **a** - (5/6)**b**:

**a** is [2, y, -5], (5/6)**b** is [5/6, -10/6, 5/6] = [5/6, -5/3, 5/6].

Subtracting:

[2 - 5/6, y - (-5/3), -5 - 5/6] = [ (12/6 - 5/6), y + 5/3, (-30/6 - 5/6) ] = [7/6, y + 5/3, -35/6]

Now take the dot product with **b** = [1, -2, 1]:

(7/6)(1) + (y + 5/3)(-2) + (-35/6)(1) = 7/6 - 2y - 10/3 - 35/6

Convert all terms to sixths:

7/6 - 2y - 20/6 - 35/6 = (7 - 20 - 35)/6 - 2y = (-48/6) - 2y = -8 - 2y

Set this equal to zero:

-8 - 2y = 0 => -2y = 8 => y = -4. Same result. So that confirms it again.

Therefore, regardless of the method, y is -4. So I think that's the correct answer.

But just to make sure, let me do the entire projection calculation again with y = -4.

Original vector: [2, -4, -5]

**b** = [1, -2, 1]

Dot product: 2*1 + (-4)*(-2) + (-5)*1 = 2 + 8 -5 = 5. Correct.

||**b**||² = 6. So scalar is 5/6. Therefore, projection is 5/6*[1, -2, 1], which is exactly what was given. Perfect.

Therefore, I can be confident that y = -4 is the correct answer.

**Final Answer**
The value of $ y $ is \boxed{-4}.

To find the value of $ y $ such that the projection of the vector
$$
\begin{pmatrix} 2 \\ y \\ -5 \end{pmatrix}
$$
onto the vector
$$
\begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}
$$
is equal to
$$
\frac{5}{6} \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix},
$$
we use the standard formula for the projection of a vector $ \mathbf{a} $ onto another vector $ \mathbf{b} $:

$$
\text{proj}_{\mathbf{b}} \mathbf{a} = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{b}\|^2} \mathbf{b}
$$

---

### Step 1: Compute the dot product $ \mathbf{a} \cdot \mathbf{b} $

Let
$$
\mathbf{a} = \begin{pmatrix} 2 \\ y \\ -5 \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}
$$

Then
$$
\mathbf{a} \cdot \mathbf{b} = 2 \cdot 1 + y \cdot (-2) + (-5) \cdot 1 = 2 - 2y - 5 = -3 - 2y
$$

---

### Step 2: Compute the squared norm of $ \mathbf{b} $

$$
\|\mathbf{b}\|^2 = 1^2 + (-2)^2 + 1^2 = 1 + 4 + 1 = 6
$$

---

### Step 3: Apply the projection formula

$$
\text{proj}_{\mathbf{b}} \mathbf{a} = \frac{-3 - 2y}{6} \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}
$$

We are told that this projection equals
$$
\frac{5}{6} \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}
$$

So we equate the scalar multipliers:

$$
\frac{-3 - 2y}{6} = \frac{5}{6}
$$

Multiply both sides by 6:

$$
-3 - 2y = 5
$$

Solve for $ y $:

$$
-2y = 5 + 3 = 8 \quad \Rightarrow \quad y = \frac{8}{-2} = -4
$$

---

### Step 4: Verification

Let’s verify by plugging $ y = -4 $ back into the original vector:

$$
\mathbf{a} = \begin{pmatrix} 2 \\ -4 \\ -5 \end{pmatrix}
$$

Compute the dot product:

$$
\mathbf{a} \cdot \mathbf{b} = 2 \cdot 1 + (-4) \cdot (-2) + (-5) \cdot 1 = 2 + 8 - 5 = 5
$$

Then the projection is:

$$
\frac{5}{6} \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}
$$

Which matches the given projection. So the solution is correct.

---

### Final Answer

$$
\boxed{-4}
$$

[Assistant]
Explanation:
Okay, so I need to find the value of y such that the projection of the vector [2, y, -5] onto the vector [1, -2, 1] is equal to (5/6)*[1, -2, 1]. Hmm, projections... Let me recall how that works.

First, the formula for the projection of a vector **a** onto another vector **b** is given by:

proj_b **a** = ( (**a** · **b**) / ||**b**||² ) * **b**

So, in this case, **a** is [2, y, -5] and **b** is [1, -2, 1]. The projection is given as (5/6)*[1, -2, 1]. Therefore, I can set up the equation using the projection formula and solve for y.

Let me compute the dot product **a** · **b** first. That would be:

(2)(1) + (y)(-2) + (-5)(1) = 2 - 2y - 5 = (-3 - 2y)

Then, the norm squared of **b** is ||**b**||² = (1)² + (-2)² + (1)² = 1 + 4 + 1 = 6.

So plugging into the projection formula:

proj_b **a** = ( (-3 - 2y)/6 ) * [1, -2, 1]

But according to the problem, this projection is equal to (5/6)*[1, -2, 1]. Therefore, we can set the scalar multiples equal to each other because the vectors are scalar multiples of [1, -2, 1].

So:

(-3 - 2y)/6 = 5/6

Since the denominators are the same, we can equate the numerators:

-3 - 2y = 5

Solving for y:

-2y = 5 + 3

-2y = 8

y = 8 / (-2) = -4

Wait, so y is -4? Let me check my steps again to make sure I didn't make a mistake.

First, the dot product: 2*1 is 2, y*(-2) is -2y, -5*1 is -5. So total is 2 - 2y -5 = -3 - 2y. That seems right.

Then ||**b**||² is 1 + 4 + 1 = 6. Correct.

So the scalar multiplier is (-3 - 2y)/6. The projection is that scalar times **b**, which is equal to (5/6)**b**. Therefore, the scalar must be 5/6. Therefore:

(-3 - 2y)/6 = 5/6 => -3 - 2y = 5 => -2y = 8 => y = -4. Seems correct.

But let me verify by plugging y = -4 back into the original vector and computing the projection.

Original vector **a** is [2, -4, -5]. Let's compute the projection.

First, **a** · **b** = 2*1 + (-4)*(-2) + (-5)*1 = 2 + 8 - 5 = 5. Wait, that's 5. Then divide by ||**b**||² which is 6. So the scalar is 5/6. Therefore, the projection is (5/6)*[1, -2, 1], which matches the given projection. So that checks out. Therefore, y = -4 is correct.

Wait, but hold on, when I first calculated the dot product with y, I had -3 - 2y. If y = -4, then that would be -3 -2*(-4) = -3 + 8 = 5. Which is correct. So that's exactly what I just did. So yes, that works. Therefore, the answer is y = -4.

But let me think again if there's another way this could be approached, just to confirm.

Alternatively, since the projection is given, maybe we can think about the fact that the projection vector is a scalar multiple of **b**, so the vector [2, y, -5] minus its projection should be orthogonal to **b**. That is, (**a** - proj_b **a**) · **b** = 0.

Let me try that approach as a check.

Let me denote proj_b **a** as (5/6)**b**. Therefore, **a** - (5/6)**b** should be orthogonal to **b**.

Compute **a** - (5/6)**b**:

**a** is [2, y, -5], (5/6)**b** is [5/6, -10/6, 5/6] = [5/6, -5/3, 5/6].

Subtracting:

[2 - 5/6, y - (-5/3), -5 - 5/6] = [ (12/6 - 5/6), y + 5/3, (-30/6 - 5/6) ] = [7/6, y + 5/3, -35/6]

Now take the dot product with **b** = [1, -2, 1]:

(7/6)(1) + (y + 5/3)(-2) + (-35/6)(1) = 7/6 - 2y - 10/3 - 35/6

Convert all terms to sixths:

7/6 - 2y - 20/6 - 35/6 = (7 - 20 - 35)/6 - 2y = (-48/6) - 2y = -8 - 2y

Set this equal to zero:

-8 - 2y = 0 => -2y = 8 => y = -4. Same result. So that confirms it again.

Therefore, regardless of the method, y is -4. So I think that's the correct answer.

But just to make sure, let me do the entire projection calculation again with y = -4.

Original vector: [2, -4, -5]

**b** = [1, -2, 1]

Dot product: 2*1 + (-4)*(-2) + (-5)*1 = 2 + 8 -5 = 5. Correct.

||**b**||² = 6. So scalar is 5/6. Therefore, projection is 5/6*[1, -2, 1], which is exactly what was given. Perfect.

Therefore, I can be confident that y = -4 is the correct answer.

**Final Answer**
The value of $ y $ is \boxed{-4}.

To find the value of $ y $ such that the projection of the vector
$$
\begin{pmatrix} 2 \\ y \\ -5 \end{pmatrix}
$$
onto the vector
$$
\begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}
$$
is equal to
$$
\frac{5}{6} \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix},
$$
we use the standard formula for the projection of a vector $ \mathbf{a} $ onto another vector $ \mathbf{b} $:

$$
\text{proj}_{\mathbf{b}} \mathbf{a} = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{b}\|^2} \mathbf{b}
$$

---

### Step 1: Compute the dot product $ \mathbf{a} \cdot \mathbf{b} $

Let
$$
\mathbf{a} = \begin{pmatrix} 2 \\ y \\ -5 \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}
$$

Then
$$
\mathbf{a} \cdot \mathbf{b} = 2 \cdot 1 + y \cdot (-2) + (-5) \cdot 1 = 2 - 2y - 5 = -3 - 2y
$$

---

### Step 2: Compute the squared norm of $ \mathbf{b} $

$$
\|\mathbf{b}\|^2 = 1^2 + (-2)^2 + 1^2 = 1 + 4 + 1 = 6
$$

---

### Step 3: Apply the projection formula

$$
\text{proj}_{\mathbf{b}} \mathbf{a} = \frac{-3 - 2y}{6} \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}
$$

We are told that this projection equals
$$
\frac{5}{6} \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}
$$

So we equate the scalar multipliers:

$$
\frac{-3 - 2y}{6} = \frac{5}{6}
$$

Multiply both sides by 6:

$$
-3 - 2y = 5
$$

Solve for $ y $:

$$
-2y = 5 + 3 = 8 \quad \Rightarrow \quad y = \frac{8}{-2} = -4
$$

---

### Step 4: Verification

Let’s verify by plugging $ y = -4 $ back into the original vector:

$$
\mathbf{a} = \begin{pmatrix} 2 \\ -4 \\ -5 \end{pmatrix}
$$

Compute the dot product:

$$
\mathbf{a} \cdot \mathbf{b} = 2 \cdot 1 + (-4) \cdot (-2) + (-5) \cdot 1 = 2 + 8 - 5 = 5
$$

Then the projection is:

$$
\frac{5}{6} \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}
$$

Which matches the given projection. So the solution is correct.

---

### Final Answer

$$
\boxed{-4}
$$

Solution:
The projection of $\begin{pmatrix} 2 \\ y \\ -5 \end{pmatrix}$ onto $\begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}$ is
\[\frac{\begin{pmatrix} 2 \\ y \\ -5 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}}{\begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}} \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} = \frac{-2y - 3}{6} \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}.\]Then $-2y - 3 = 5,$ so $y = \boxed{-4}.$


Explanation:
Okay, so I need to simplify this expression: (-k + 4) + (-2 + 3k). Let me think about how to approach this. Alright, first, I remember that when you have parentheses in an expression, you usually need to distribute any signs or coefficients, but in this case, there's just addition between the two parentheses. So maybe I can just remove the parentheses and combine like terms?

Let me write it out step by step. The original expression is (-k + 4) + (-2 + 3k). If I take away the parentheses, it should be -k + 4 - 2 + 3k. Wait, is that right? Let me check. The first part is (-k + 4), so when you add the second part, which is (-2 + 3k), you just add each term. So yes, it's like adding -2 and 3k to the first expression. So that would be -k + 4 - 2 + 3k.

Now, I need to combine like terms. The like terms here are the terms with k and the constant terms. Let me separate them. The terms with k are -k and 3k. The constant terms are 4 and -2.

Starting with the k terms: -k + 3k. Hmm, combining those. Let me think. If I have -1k + 3k, that's like (-1 + 3)k, which is 2k. Okay, so that simplifies to 2k.

Now the constants: 4 - 2. That's straightforward. 4 minus 2 is 2. So putting those together, the simplified expression should be 2k + 2.

Wait, let me double-check my steps. Original expression: (-k + 4) + (-2 + 3k). Remove parentheses: -k + 4 - 2 + 3k. Combine like terms: (-k + 3k) + (4 - 2). Which is 2k + 2. Yeah, that seems right.

But let me try another way to verify. Maybe rearrange the terms. Let's see. If I group the k terms first: (-k + 3k) and then the constants (4 - 2). So again, that's 2k + 2.

Alternatively, maybe I can factor something out? Wait, 2k + 2 can be factored as 2(k + 1), but the question just says to simplify, so maybe 2k + 2 is the simplest form. Unless there's a different way to present it, but I think that's it.

Let me check if I made any sign errors. The original expression has -k and then + (-2). So when you remove the parentheses, it's -k + 4 - 2 + 3k. Yes, because adding a negative 2 is the same as subtracting 2. So that's correct.

Another way to check is to plug in a value for k and see if both the original expression and the simplified one give the same result. Let me pick k = 1.

Original expression: (-1 + 4) + (-2 + 3*1) = (3) + (-2 + 3) = 3 + 1 = 4.

Simplified expression: 2*1 + 2 = 2 + 2 = 4. Okay, same result.

Let me try another value, maybe k = 0.

Original expression: (-0 + 4) + (-2 + 0) = (4) + (-2) = 2.

Simplified expression: 2*0 + 2 = 0 + 2 = 2. Also same.

How about k = 2?

Original: (-2 + 4) + (-2 + 6) = (2) + (4) = 6.

Simplified: 2*2 + 2 = 4 + 2 = 6. Perfect, same result.

So that seems to confirm that the simplification is correct.

Therefore, the simplified form of (-k + 4) + (-2 + 3k) is 2k + 2.

**Final Answer**
The simplified expression is \boxed{2k + 2}.

To simplify the expression $(-k + 4) + (-2 + 3k)$, we follow a systematic approach by first removing the parentheses and then combining like terms.

---

### Step 1: Remove Parentheses

We start by distributing the addition across the parentheses:

$$
(-k + 4) + (-2 + 3k) = -k + 4 - 2 + 3k
$$

---

### Step 2: Combine Like Terms

Now, we group and combine the terms with the variable $k$ and the constant terms separately.

- **Terms with $k$:**
  $-k + 3k = ( -1 + 3 )k = 2k$

- **Constant terms:**
  $4 - 2 = 2$

---

### Step 3: Write the Simplified Expression

Putting the simplified terms together:

$$
2k + 2
$$

---

### Final Answer

$$
\boxed{2k + 2}
$$

[Assistant]
Explanation:
Okay, so I need to simplify this expression: (-k + 4) + (-2 + 3k). Let me think about how to approach this. Alright, first, I remember that when you have parentheses in an expression, you usually need to distribute any signs or coefficients, but in this case, there's just addition between the two parentheses. So maybe I can just remove the parentheses and combine like terms?

Let me write it out step by step. The original expression is (-k + 4) + (-2 + 3k). If I take away the parentheses, it should be -k + 4 - 2 + 3k. Wait, is that right? Let me check. The first part is (-k + 4), so when you add the second part, which is (-2 + 3k), you just add each term. So yes, it's like adding -2 and 3k to the first expression. So that would be -k + 4 - 2 + 3k.

Now, I need to combine like terms. The like terms here are the terms with k and the constant terms. Let me separate them. The terms with k are -k and 3k. The constant terms are 4 and -2.

Starting with the k terms: -k + 3k. Hmm, combining those. Let me think. If I have -1k + 3k, that's like (-1 + 3)k, which is 2k. Okay, so that simplifies to 2k.

Now the constants: 4 - 2. That's straightforward. 4 minus 2 is 2. So putting those together, the simplified expression should be 2k + 2.

Wait, let me double-check my steps. Original expression: (-k + 4) + (-2 + 3k). Remove parentheses: -k + 4 - 2 + 3k. Combine like terms: (-k + 3k) + (4 - 2). Which is 2k + 2. Yeah, that seems right.

But let me try another way to verify. Maybe rearrange the terms. Let's see. If I group the k terms first: (-k + 3k) and then the constants (4 - 2). So again, that's 2k + 2.

Alternatively, maybe I can factor something out? Wait, 2k + 2 can be factored as 2(k + 1), but the question just says to simplify, so maybe 2k + 2 is the simplest form. Unless there's a different way to present it, but I think that's it.

Let me check if I made any sign errors. The original expression has -k and then + (-2). So when you remove the parentheses, it's -k + 4 - 2 + 3k. Yes, because adding a negative 2 is the same as subtracting 2. So that's correct.

Another way to check is to plug in a value for k and see if both the original expression and the simplified one give the same result. Let me pick k = 1.

Original expression: (-1 + 4) + (-2 + 3*1) = (3) + (-2 + 3) = 3 + 1 = 4.

Simplified expression: 2*1 + 2 = 2 + 2 = 4. Okay, same result.

Let me try another value, maybe k = 0.

Original expression: (-0 + 4) + (-2 + 0) = (4) + (-2) = 2.

Simplified expression: 2*0 + 2 = 0 + 2 = 2. Also same.

How about k = 2?

Original: (-2 + 4) + (-2 + 6) = (2) + (4) = 6.

Simplified: 2*2 + 2 = 4 + 2 = 6. Perfect, same result.

So that seems to confirm that the simplification is correct.

Therefore, the simplified form of (-k + 4) + (-2 + 3k) is 2k + 2.

**Final Answer**
The simplified expression is \boxed{2k + 2}.

To simplify the expression $(-k + 4) + (-2 + 3k)$, we follow a systematic approach by first removing the parentheses and then combining like terms.

---

### Step 1: Remove Parentheses

We start by distributing the addition across the parentheses:

$$
(-k + 4) + (-2 + 3k) = -k + 4 - 2 + 3k
$$

---

### Step 2: Combine Like Terms

Now, we group and combine the terms with the variable $k$ and the constant terms separately.

- **Terms with $k$:**
  $-k + 3k = ( -1 + 3 )k = 2k$

- **Constant terms:**
  $4 - 2 = 2$

---

### Step 3: Write the Simplified Expression

Putting the simplified terms together:

$$
2k + 2
$$

---

### Final Answer

$$
\boxed{2k + 2}
$$

Solution: We have $(-k+4) + (-2+3k) = -k + 4 -2 + 3k = \boxed{2k+2}$.


Explanation:
Okay, so I need to find the horizontal asymptote of the function y = (4x³ + 2x - 4) / (3x³ - 2x² + 5x - 1). Hmm, horizontal asymptotes... I remember they have something to do with the behavior of the function as x approaches positive or negative infinity. Let me think.

First, I recall that for rational functions, which are ratios of polynomials, the horizontal asymptote depends on the degrees of the numerator and the denominator. If the degrees are the same, the horizontal asymptote is the ratio of the leading coefficients. If the degree of the numerator is less than the denominator, the horizontal asymptote is y = 0. If the degree of the numerator is higher, there's no horizontal asymptote, but maybe an oblique asymptote? Wait, but the question specifically asks for a horizontal asymptote, so maybe we can ignore the cases where there isn't one?

Let me check the degrees here. The numerator is 4x³ + 2x - 4. The highest power is x³, so degree 3. The denominator is 3x³ - 2x² + 5x - 1. Also degree 3. So since the degrees are equal, the horizontal asymptote should be the ratio of the leading coefficients. The leading term of the numerator is 4x³, and the leading term of the denominator is 3x³. So the ratio is 4/3. Therefore, the horizontal asymptote is y = 4/3.

Wait, but maybe I should verify this by actually dividing the numerator and denominator by x³ to see what happens as x approaches infinity? Let me try that.

Dividing numerator and denominator by x³:

Numerator becomes (4x³ + 2x - 4)/x³ = 4 + 2/x² - 4/x³

Denominator becomes (3x³ - 2x² + 5x - 1)/x³ = 3 - 2/x + 5/x² - 1/x³

So the function simplifies to [4 + 2/x² - 4/x³] / [3 - 2/x + 5/x² - 1/x³]

Now, as x approaches infinity, the terms with 1/x², 1/x³, etc., all go to zero. Therefore, the expression approaches 4 / 3. So that confirms the horizontal asymptote is y = 4/3.

But wait, the question says "at what value of y is there a horizontal asymptote". So maybe they just want the y-value, which is 4/3? But let me make sure there isn't another horizontal asymptote as x approaches negative infinity? Wait, for horizontal asymptotes, we check both positive and negative infinity, but if the degrees are the same, the horizontal asymptote is the same in both cases. So even as x approaches negative infinity, the terms with 1/x², etc., still go to zero, so the limit is still 4/3. Therefore, the horizontal asymptote is y = 4/3.

But let me check if there's any possibility that the function might cross the horizontal asymptote or something? But the question isn't asking about that. It's just asking for the value of y where the horizontal asymptote exists. So regardless of whether the function crosses it or not, the horizontal asymptote is y = 4/3.

Alternatively, maybe I should check if there's a case where the horizontal asymptote isn't just the ratio of leading coefficients? For example, if there are some other terms? But no, when the degrees are equal, the horizontal asymptote is always the ratio of the leading coefficients.

Wait, let me think again. Suppose the numerator and denominator are both degree 3. Let me write them as:

Numerator: 4x³ + ...

Denominator: 3x³ + ...

So when x is very large, the lower degree terms become negligible. So the function behaves like 4x³ / 3x³ = 4/3. Therefore, the horizontal asymptote is y = 4/3.

Alternatively, maybe I can use L’Hospital’s Rule? Wait, but L’Hospital’s Rule is for indeterminate forms like infinity/infinity or 0/0. If I take the limit as x approaches infinity of (4x³ + ...)/(3x³ + ...), that is infinity/infinity, so maybe applying L’Hospital’s Rule? Let me try that.

Take derivative of numerator: 12x² + 2

Derivative of denominator: 9x² - 4x + 5

So the limit becomes (12x² + 2)/(9x² - 4x + 5). Still infinity/infinity. Apply L’Hospital again:

Derivative of numerator: 24x

Derivative of denominator: 18x - 4

Limit becomes 24x / (18x - 4). Still infinity/infinity. Apply L’Hospital again:

Derivative of numerator: 24

Derivative of denominator: 18

So the limit is 24/18 = 4/3. So same result. Therefore, the horizontal asymptote is y = 4/3.

Therefore, the answer should be 4/3.

But let me check if there's any mistake here. For example, maybe the denominator has a factor that cancels with the numerator? But the question is about horizontal asymptote, which is about the behavior at infinity, not about holes or anything. Even if there were common factors, the horizontal asymptote would still be determined by the leading coefficients. Unless the degrees are different.

Wait, but if there were common factors, then the function would simplify, but since the degrees are the same, even if there is a common factor, the horizontal asymptote would still be the ratio of the leading coefficients. For example, if both numerator and denominator had a common factor of x - a, then after canceling, the degrees would be reduced by 1. But in this case, the original degrees are both 3, so if they had a common factor, the simplified function would have degrees 2 and 2, so horizontal asymptote would be ratio of leading coefficients. But in this case, the original function is given, and we don't know if there are common factors.

But maybe I should check if the denominator can be factored? Let me try to factor the denominator: 3x³ - 2x² + 5x - 1.

Trying rational roots. The possible rational roots are ±1, ±1/3. Let me test x=1: 3(1)^3 - 2(1)^2 + 5(1) -1 = 3 - 2 + 5 -1 = 5 ≠ 0. x=1/3: 3*(1/27) - 2*(1/9) + 5*(1/3) -1 = 1/9 - 2/9 + 5/3 -1 = (-1/9) + 5/3 -1 = (-1/9) + (15/9 - 9/9) = (-1/9 + 6/9) = 5/9 ≠ 0. x=-1: 3(-1)^3 - 2(-1)^2 +5(-1) -1 = -3 -2 -5 -1 = -11 ≠ 0. x= -1/3: 3*(-1/3)^3 - 2*(-1/3)^2 +5*(-1/3) -1 = 3*(-1/27) - 2*(1/9) -5/3 -1 = -1/9 - 2/9 -5/3 -1 = (-3/9) -5/3 -1 = (-1/3) -5/3 -1 = (-6/3) -1 = -2 -1 = -3 ≠ 0. So no rational roots. Therefore, the denominator doesn't factor nicely with rational roots. Therefore, there are no common factors with the numerator? Let me check the numerator: 4x³ + 2x -4. Let me see if it has any common factors with denominator. Let me check if they share any roots. Suppose there is a common root, say α. Then 4α³ + 2α -4 = 0 and 3α³ - 2α² +5α -1 = 0. Let me try to solve these equations.

From the first equation: 4α³ = -2α +4 => α³ = (-2α +4)/4. Substitute into the second equation:

3*(-2α +4)/4 - 2α² +5α -1 = 0. Multiply through by 4 to eliminate denominator:

3*(-2α +4) -8α² +20α -4 = 0 => -6α +12 -8α² +20α -4 = 0 => (-8α²) +14α +8 = 0 => Multiply by -1: 8α² -14α -8 = 0.

Solve quadratic equation: 8α² -14α -8 = 0. Discriminant D = 196 + 256 = 452. sqrt(452) is irrational. Therefore, α is irrational. Therefore, there are no common roots between numerator and denominator. Therefore, the function doesn't have any holes, and the horizontal asymptote is indeed y = 4/3.

Therefore, the answer is 4/3.

But just to make sure, let me plug in a very large x value and see. Let's take x = 1000.

Numerator: 4*(1000)^3 + 2*(1000) -4 ≈ 4*10^9 + 2000 -4 ≈ 4.000002*10^9

Denominator: 3*(1000)^3 - 2*(1000)^2 +5*(1000) -1 ≈ 3*10^9 - 2*10^6 + 5000 -1 ≈ 3*10^9 - 2,000,000 + 5000 ≈ 2.998005*10^9

So the ratio is approximately 4.000002 / 2.998005 ≈ 4 / 3 ≈ 1.3333... So that checks out.

Alternatively, if I take x approaching negative infinity, say x = -1000.

Numerator: 4*(-1000)^3 + 2*(-1000) -4 = -4*10^9 -2000 -4 ≈ -4.000004*10^9

Denominator: 3*(-1000)^3 - 2*(-1000)^2 +5*(-1000) -1 = -3*10^9 - 2*10^6 -5000 -1 ≈ -3.002001*10^9

So the ratio is (-4.000004*10^9)/(-3.002001*10^9) ≈ 4/3 ≈ 1.3333... So same result.

Therefore, regardless of the direction, the horizontal asymptote is y = 4/3.

So I think that's solid. Therefore, the answer is 4/3.

**Final Answer**
The horizontal asymptote occurs at \boxed{\dfrac{4}{3}}.

To determine the horizontal asymptote of the function
$$
y = \frac{4x^3 + 2x - 4}{3x^3 - 2x^2 + 5x - 1},
$$
we analyze the behavior of the function as $ x \to \pm\infty $.

---

### Step 1: Identify the Degrees of the Numerator and Denominator

- The **numerator** is a cubic polynomial: $ 4x^3 + 2x - 4 $, so its degree is **3**.
- The **denominator** is also a cubic polynomial: $ 3x^3 - 2x^2 + 5x - 1 $, so its degree is **3**.

Since the degrees of the numerator and denominator are equal, the function has a **horizontal asymptote**.

---

### Step 2: Use the Rule for Horizontal Asymptotes of Rational Functions

For a rational function of the form
$$
y = \frac{a_n x^n + \cdots}{b_m x^m + \cdots},
$$
- If $ n = m $, the horizontal asymptote is
$$
y = \frac{a_n}{b_m},
$$
where $ a_n $ and $ b_m $ are the leading coefficients of the numerator and denominator, respectively.

In our case:
- Leading coefficient of the numerator: $ 4 $
- Leading coefficient of the denominator: $ 3 $

So, the horizontal asymptote is
$$
y = \frac{4}{3}.
$$

---

### Step 3: Confirm with Limit Analysis

To verify, we can divide both the numerator and denominator by $ x^3 $, the highest power of $ x $:

$$
y = \frac{4 + \frac{2}{x^2} - \frac{4}{x^3}}{3 - \frac{2}{x} + \frac{5}{x^2} - \frac{1}{x^3}}.
$$

As $ x \to \pm\infty $, the terms with $ \frac{1}{x} $, $ \frac{1}{x^2} $, and $ \frac{1}{x^3} $ all approach 0. Therefore, the expression simplifies to:

$$
y \to \frac{4}{3}.
$$

This confirms that the horizontal asymptote is indeed $ y = \frac{4}{3} $.

---

### Step 4: Consider the Behavior at Infinity

We also checked the behavior as $ x \to \infty $ and $ x \to -\infty $, and in both cases, the function approaches the same value $ \frac{4}{3} $. This is consistent with the rule for horizontal asymptotes when the degrees are equal.

---

### Final Answer

$$
\boxed{\dfrac{4}{3}}
$$

[Assistant]
Explanation:
Okay, so I need to find the horizontal asymptote of the function y = (4x³ + 2x - 4) / (3x³ - 2x² + 5x - 1). Hmm, horizontal asymptotes... I remember they have something to do with the behavior of the function as x approaches positive or negative infinity. Let me think.

First, I recall that for rational functions, which are ratios of polynomials, the horizontal asymptote depends on the degrees of the numerator and the denominator. If the degrees are the same, the horizontal asymptote is the ratio of the leading coefficients. If the degree of the numerator is less than the denominator, the horizontal asymptote is y = 0. If the degree of the numerator is higher, there's no horizontal asymptote, but maybe an oblique asymptote? Wait, but the question specifically asks for a horizontal asymptote, so maybe we can ignore the cases where there isn't one?

Let me check the degrees here. The numerator is 4x³ + 2x - 4. The highest power is x³, so degree 3. The denominator is 3x³ - 2x² + 5x - 1. Also degree 3. So since the degrees are equal, the horizontal asymptote should be the ratio of the leading coefficients. The leading term of the numerator is 4x³, and the leading term of the denominator is 3x³. So the ratio is 4/3. Therefore, the horizontal asymptote is y = 4/3.

Wait, but maybe I should verify this by actually dividing the numerator and denominator by x³ to see what happens as x approaches infinity? Let me try that.

Dividing numerator and denominator by x³:

Numerator becomes (4x³ + 2x - 4)/x³ = 4 + 2/x² - 4/x³

Denominator becomes (3x³ - 2x² + 5x - 1)/x³ = 3 - 2/x + 5/x² - 1/x³

So the function simplifies to [4 + 2/x² - 4/x³] / [3 - 2/x + 5/x² - 1/x³]

Now, as x approaches infinity, the terms with 1/x², 1/x³, etc., all go to zero. Therefore, the expression approaches 4 / 3. So that confirms the horizontal asymptote is y = 4/3.

But wait, the question says "at what value of y is there a horizontal asymptote". So maybe they just want the y-value, which is 4/3? But let me make sure there isn't another horizontal asymptote as x approaches negative infinity? Wait, for horizontal asymptotes, we check both positive and negative infinity, but if the degrees are the same, the horizontal asymptote is the same in both cases. So even as x approaches negative infinity, the terms with 1/x², etc., still go to zero, so the limit is still 4/3. Therefore, the horizontal asymptote is y = 4/3.

But let me check if there's any possibility that the function might cross the horizontal asymptote or something? But the question isn't asking about that. It's just asking for the value of y where the horizontal asymptote exists. So regardless of whether the function crosses it or not, the horizontal asymptote is y = 4/3.

Alternatively, maybe I should check if there's a case where the horizontal asymptote isn't just the ratio of leading coefficients? For example, if there are some other terms? But no, when the degrees are equal, the horizontal asymptote is always the ratio of the leading coefficients.

Wait, let me think again. Suppose the numerator and denominator are both degree 3. Let me write them as:

Numerator: 4x³ + ...

Denominator: 3x³ + ...

So when x is very large, the lower degree terms become negligible. So the function behaves like 4x³ / 3x³ = 4/3. Therefore, the horizontal asymptote is y = 4/3.

Alternatively, maybe I can use L’Hospital’s Rule? Wait, but L’Hospital’s Rule is for indeterminate forms like infinity/infinity or 0/0. If I take the limit as x approaches infinity of (4x³ + ...)/(3x³ + ...), that is infinity/infinity, so maybe applying L’Hospital’s Rule? Let me try that.

Take derivative of numerator: 12x² + 2

Derivative of denominator: 9x² - 4x + 5

So the limit becomes (12x² + 2)/(9x² - 4x + 5). Still infinity/infinity. Apply L’Hospital again:

Derivative of numerator: 24x

Derivative of denominator: 18x - 4

Limit becomes 24x / (18x - 4). Still infinity/infinity. Apply L’Hospital again:

Derivative of numerator: 24

Derivative of denominator: 18

So the limit is 24/18 = 4/3. So same result. Therefore, the horizontal asymptote is y = 4/3.

Therefore, the answer should be 4/3.

But let me check if there's any mistake here. For example, maybe the denominator has a factor that cancels with the numerator? But the question is about horizontal asymptote, which is about the behavior at infinity, not about holes or anything. Even if there were common factors, the horizontal asymptote would still be determined by the leading coefficients. Unless the degrees are different.

Wait, but if there were common factors, then the function would simplify, but since the degrees are the same, even if there is a common factor, the horizontal asymptote would still be the ratio of the leading coefficients. For example, if both numerator and denominator had a common factor of x - a, then after canceling, the degrees would be reduced by 1. But in this case, the original degrees are both 3, so if they had a common factor, the simplified function would have degrees 2 and 2, so horizontal asymptote would be ratio of leading coefficients. But in this case, the original function is given, and we don't know if there are common factors.

But maybe I should check if the denominator can be factored? Let me try to factor the denominator: 3x³ - 2x² + 5x - 1.

Trying rational roots. The possible rational roots are ±1, ±1/3. Let me test x=1: 3(1)^3 - 2(1)^2 + 5(1) -1 = 3 - 2 + 5 -1 = 5 ≠ 0. x=1/3: 3*(1/27) - 2*(1/9) + 5*(1/3) -1 = 1/9 - 2/9 + 5/3 -1 = (-1/9) + 5/3 -1 = (-1/9) + (15/9 - 9/9) = (-1/9 + 6/9) = 5/9 ≠ 0. x=-1: 3(-1)^3 - 2(-1)^2 +5(-1) -1 = -3 -2 -5 -1 = -11 ≠ 0. x= -1/3: 3*(-1/3)^3 - 2*(-1/3)^2 +5*(-1/3) -1 = 3*(-1/27) - 2*(1/9) -5/3 -1 = -1/9 - 2/9 -5/3 -1 = (-3/9) -5/3 -1 = (-1/3) -5/3 -1 = (-6/3) -1 = -2 -1 = -3 ≠ 0. So no rational roots. Therefore, the denominator doesn't factor nicely with rational roots. Therefore, there are no common factors with the numerator? Let me check the numerator: 4x³ + 2x -4. Let me see if it has any common factors with denominator. Let me check if they share any roots. Suppose there is a common root, say α. Then 4α³ + 2α -4 = 0 and 3α³ - 2α² +5α -1 = 0. Let me try to solve these equations.

From the first equation: 4α³ = -2α +4 => α³ = (-2α +4)/4. Substitute into the second equation:

3*(-2α +4)/4 - 2α² +5α -1 = 0. Multiply through by 4 to eliminate denominator:

3*(-2α +4) -8α² +20α -4 = 0 => -6α +12 -8α² +20α -4 = 0 => (-8α²) +14α +8 = 0 => Multiply by -1: 8α² -14α -8 = 0.

Solve quadratic equation: 8α² -14α -8 = 0. Discriminant D = 196 + 256 = 452. sqrt(452) is irrational. Therefore, α is irrational. Therefore, there are no common roots between numerator and denominator. Therefore, the function doesn't have any holes, and the horizontal asymptote is indeed y = 4/3.

Therefore, the answer is 4/3.

But just to make sure, let me plug in a very large x value and see. Let's take x = 1000.

Numerator: 4*(1000)^3 + 2*(1000) -4 ≈ 4*10^9 + 2000 -4 ≈ 4.000002*10^9

Denominator: 3*(1000)^3 - 2*(1000)^2 +5*(1000) -1 ≈ 3*10^9 - 2*10^6 + 5000 -1 ≈ 3*10^9 - 2,000,000 + 5000 ≈ 2.998005*10^9

So the ratio is approximately 4.000002 / 2.998005 ≈ 4 / 3 ≈ 1.3333... So that checks out.

Alternatively, if I take x approaching negative infinity, say x = -1000.

Numerator: 4*(-1000)^3 + 2*(-1000) -4 = -4*10^9 -2000 -4 ≈ -4.000004*10^9

Denominator: 3*(-1000)^3 - 2*(-1000)^2 +5*(-1000) -1 = -3*10^9 - 2*10^6 -5000 -1 ≈ -3.002001*10^9

So the ratio is (-4.000004*10^9)/(-3.002001*10^9) ≈ 4/3 ≈ 1.3333... So same result.

Therefore, regardless of the direction, the horizontal asymptote is y = 4/3.

So I think that's solid. Therefore, the answer is 4/3.

**Final Answer**
The horizontal asymptote occurs at \boxed{\dfrac{4}{3}}.

To determine the horizontal asymptote of the function
$$
y = \frac{4x^3 + 2x - 4}{3x^3 - 2x^2 + 5x - 1},
$$
we analyze the behavior of the function as $ x \to \pm\infty $.

---

### Step 1: Identify the Degrees of the Numerator and Denominator

- The **numerator** is a cubic polynomial: $ 4x^3 + 2x - 4 $, so its degree is **3**.
- The **denominator** is also a cubic polynomial: $ 3x^3 - 2x^2 + 5x - 1 $, so its degree is **3**.

Since the degrees of the numerator and denominator are equal, the function has a **horizontal asymptote**.

---

### Step 2: Use the Rule for Horizontal Asymptotes of Rational Functions

For a rational function of the form
$$
y = \frac{a_n x^n + \cdots}{b_m x^m + \cdots},
$$
- If $ n = m $, the horizontal asymptote is
$$
y = \frac{a_n}{b_m},
$$
where $ a_n $ and $ b_m $ are the leading coefficients of the numerator and denominator, respectively.

In our case:
- Leading coefficient of the numerator: $ 4 $
- Leading coefficient of the denominator: $ 3 $

So, the horizontal asymptote is
$$
y = \frac{4}{3}.
$$

---

### Step 3: Confirm with Limit Analysis

To verify, we can divide both the numerator and denominator by $ x^3 $, the highest power of $ x $:

$$
y = \frac{4 + \frac{2}{x^2} - \frac{4}{x^3}}{3 - \frac{2}{x} + \frac{5}{x^2} - \frac{1}{x^3}}.
$$

As $ x \to \pm\infty $, the terms with $ \frac{1}{x} $, $ \frac{1}{x^2} $, and $ \frac{1}{x^3} $ all approach 0. Therefore, the expression simplifies to:

$$
y \to \frac{4}{3}.
$$

This confirms that the horizontal asymptote is indeed $ y = \frac{4}{3} $.

---

### Step 4: Consider the Behavior at Infinity

We also checked the behavior as $ x \to \infty $ and $ x \to -\infty $, and in both cases, the function approaches the same value $ \frac{4}{3} $. This is consistent with the rule for horizontal asymptotes when the degrees are equal.

---

### Final Answer

$$
\boxed{\dfrac{4}{3}}
$$

Solution:
When the degrees of the numerator and denominator are the same in a rational function, the horizontal asymptote is the coefficient of the highest degree in the numerator divided by the coefficient of the highest degree in the denominator. To see this, divide the numerator and denominator by $x^3$ to write the expression as  \[
\frac{4+\frac{2}{x^2}-\frac{4}{x^3}}{3-\frac{2}{x}+\frac{5}{x^2}-\frac{1}{x^3}}
\]As $x\to\infty$ or $x\to-\infty$, the terms involving $x$ approach 0, which means that the whole expression approaches 4/3. Therefore, there is only one horizontal asymptote, and it is at $y=\boxed{\frac43}$.


Explanation:
Okay, so I need to figure out how many elements are in the intersection of two sets: the set of all prime numbers less than 30 and the set of all odd numbers greater than zero. Let me start by recalling what each of these sets contains and then find their common elements.

First, let me define the two sets clearly.

The first set is all prime numbers less than 30. Prime numbers are numbers greater than 1 that have no positive divisors other than 1 and themselves. So I need to list all primes less than 30. Let me remember the primes: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29. Wait, let me check if I missed any. Starting from 2:

- 2 is prime
- 3 is prime
- 4 is not prime (divisible by 2)
- 5 is prime
- 6 is not
- 7 is prime
- 8 not
- 9 not (divisible by 3)
- 10 not
- 11 prime
- 12 not
- 13 prime
- 14 not
- 15 not
- 16 not
- 17 prime
- 18 not
- 19 prime
- 20 not
- 21 not
- 22 not
- 23 prime
- 24 not
- 25 not (5 squared)
- 26 not
- 27 not
- 28 not
- 29 prime
- 30 is not, since we're looking for less than 30.

So the primes less than 30 are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29. Let me count them. That's 10 primes. Wait, let me count again:

1. 2
2. 3
3. 5
4. 7
5. 11
6. 13
7. 17
8. 19
9. 23
10. 29

Yes, 10 primes. Okay, so the first set has 10 elements.

Now the second set is all odd numbers greater than zero. So odd numbers are numbers that are not divisible by 2. The set of all odd numbers greater than zero would be 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, etc. But since we are looking for the intersection with the primes less than 30, we don't need to consider numbers beyond 29. However, the second set is infinite, but since we are intersecting with the first set (which is finite), the intersection will be the primes less than 30 that are also odd numbers greater than zero.

But wait, the second set is all odd numbers greater than zero. So, the intersection would be all numbers that are both prime less than 30 and odd numbers greater than zero. However, note that the only even prime number is 2. All other primes are odd. Therefore, if we take the set of primes less than 30, and remove the even prime (which is 2), then the rest are odd. Therefore, the intersection of the two sets would be all primes less than 30 except 2. But let me verify this.

Alternatively, since the second set includes all odd numbers greater than zero, which includes all odd primes, but also includes composite odd numbers. However, the first set is only primes. Therefore, the intersection would be the primes that are odd numbers greater than zero. Since primes except 2 are odd, so the intersection is all primes less than 30 except 2. Therefore, the number of elements in the intersection would be the number of primes less than 30 minus 1 (for the prime 2). Since there are 10 primes less than 30, subtracting 1 gives 9. Therefore, the answer is 9?

Wait, but let me check again. Let me list the primes less than 30 and see which ones are odd. The primes less than 30 are:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29.

Now, which of these are odd? Well, 2 is even, so the rest are odd. So 3, 5, 7, 11, 13, 17, 19, 23, 29. That's 9 numbers. Therefore, the intersection is these 9 numbers. Therefore, the answer is 9.

But let me make sure that the second set is indeed all odd numbers greater than zero. So, the second set includes numbers like 1, 3, 5, 7,... So, 1 is in the second set. Is 1 considered here? The problem says "the set of all the odd numbers greater than zero." So, greater than zero, so starting from 1. But 1 is not a prime number. Therefore, the intersection would be the primes less than 30 that are also in the set of odd numbers greater than zero. Since primes are greater than 1, so all primes are greater than 1, so they are in the set of odd numbers greater than zero, except for 2. Therefore, the intersection is all primes less than 30 except 2. Therefore, 10 - 1 = 9.

But let me check if 1 is included in the second set. The problem says "the set of all the odd numbers greater than zero." So, numbers greater than zero, so starting at 1. So 1 is included. However, 1 is not prime. Therefore, the intersection would be primes less than 30 that are odd. Since primes except 2 are odd, so the intersection is primes less than 30 except 2. Therefore, 9 elements.

Alternatively, maybe I can think of the intersection as the set of numbers that are both prime less than 30 and odd numbers greater than zero. So, numbers that are primes less than 30 and also odd and greater than zero. Since all primes except 2 are odd, and all primes are greater than zero. Therefore, the intersection is primes less than 30 except 2. Therefore, 9 elements.

But let me check if there's any other number that is in both sets. For example, is there a prime number less than 30 that is not odd? Only 2. So, the only prime that is even is 2. Therefore, the rest are odd. Therefore, the intersection is all primes less than 30 except 2. Therefore, 10 - 1 = 9.

Alternatively, maybe I can list the elements of both sets and see their intersection.

First set (primes less than 30): {2, 3, 5, 7, 11, 13, 17, 19, 23, 29}

Second set (odd numbers greater than zero): {1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, ...}

So the intersection would be numbers that are in both sets. So, looking at the first set, which numbers are also in the second set? Well, all numbers in the first set except 2 are odd, so they are in the second set. However, 2 is not in the second set because it's even. Therefore, the intersection is {3, 5, 7, 11, 13, 17, 19, 23, 29}, which has 9 elements.

Therefore, the answer is 9.

But let me check if there's any other number in the first set that is not odd. Well, the only even prime is 2, so no others. Therefore, the intersection is indeed 9 elements.

Alternatively, maybe I can think of the second set as all odd numbers greater than zero. So, the intersection is the set of primes less than 30 that are odd. Since primes are either 2 or odd, so the intersection is primes less than 30 minus {2}. Therefore, 10 - 1 = 9.

Therefore, I think the answer is 9.

But just to be thorough, let me list the numbers in the intersection:

From the first set: 3, 5, 7, 11, 13, 17, 19, 23, 29. Let me count them:

1. 3
2. 5
3. 7
4. 11
5. 13
6. 17
7. 19
8. 23
9. 29

Yes, that's 9 numbers. Therefore, the intersection has 9 elements.

Therefore, the answer is 9.

But let me check once more if I made a mistake in counting the primes less than 30. Let me list them again:

Primes less than 30:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29.

Yes, that's 10 primes. So removing 2, we have 9 primes left. Therefore, the intersection is 9 elements.

Alternatively, if I think of the set of odd numbers greater than zero, which includes 1, 3, 5,... So, the primes in that set would be all primes except 2. Therefore, the intersection is primes less than 30 except 2. Therefore, 9 elements.

Therefore, I think the answer is 9.

But just to make sure, let me check if there's any other number that might be in both sets. For example, is 1 in the first set? No, because 1 is not a prime. So, the intersection is only the primes that are odd. So, yes, 9 elements.

Therefore, I think the answer is 9.

**Final Answer**
The number of elements in the intersection is \boxed{9}.

To determine how many elements are in the intersection of the set of all prime numbers less than 30 and the set of all odd numbers greater than zero, we proceed step by step.

---

### Step 1: Identify the First Set – Prime Numbers Less Than 30

A **prime number** is a natural number greater than 1 that has no positive divisors other than 1 and itself. Listing all such numbers less than 30:

$$
\{2, 3, 5, 7, 11, 13, 17, 19, 23, 29\}
$$

This set contains **10 elements**.

---

### Step 2: Identify the Second Set – Odd Numbers Greater Than Zero

An **odd number** is any integer that is not divisible by 2. The set of all odd numbers greater than zero includes:

$$
\{1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, \dots\}
$$

This is an **infinite** set, but since we are only interested in the intersection with the first set (which is finite), we only need to consider the odd numbers that are also in the first set.

---

### Step 3: Find the Intersection

We are looking for numbers that are **both**:

- Prime numbers less than 30, and
- Odd numbers greater than zero.

From the first set, the only even prime is **2**. All other primes are odd. Therefore, the intersection will consist of all the primes less than 30 **except** 2.

So, the intersection is:

$$
\{3, 5, 7, 11, 13, 17, 19, 23, 29\}
$$

This set contains **9 elements**.

---

### Final Answer

$$
\boxed{9}
$$

[Assistant]
Explanation:
Okay, so I need to figure out how many elements are in the intersection of two sets: the set of all prime numbers less than 30 and the set of all odd numbers greater than zero. Let me start by recalling what each of these sets contains and then find their common elements.

First, let me define the two sets clearly.

The first set is all prime numbers less than 30. Prime numbers are numbers greater than 1 that have no positive divisors other than 1 and themselves. So I need to list all primes less than 30. Let me remember the primes: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29. Wait, let me check if I missed any. Starting from 2:

- 2 is prime
- 3 is prime
- 4 is not prime (divisible by 2)
- 5 is prime
- 6 is not
- 7 is prime
- 8 not
- 9 not (divisible by 3)
- 10 not
- 11 prime
- 12 not
- 13 prime
- 14 not
- 15 not
- 16 not
- 17 prime
- 18 not
- 19 prime
- 20 not
- 21 not
- 22 not
- 23 prime
- 24 not
- 25 not (5 squared)
- 26 not
- 27 not
- 28 not
- 29 prime
- 30 is not, since we're looking for less than 30.

So the primes less than 30 are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29. Let me count them. That's 10 primes. Wait, let me count again:

1. 2
2. 3
3. 5
4. 7
5. 11
6. 13
7. 17
8. 19
9. 23
10. 29

Yes, 10 primes. Okay, so the first set has 10 elements.

Now the second set is all odd numbers greater than zero. So odd numbers are numbers that are not divisible by 2. The set of all odd numbers greater than zero would be 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, etc. But since we are looking for the intersection with the primes less than 30, we don't need to consider numbers beyond 29. However, the second set is infinite, but since we are intersecting with the first set (which is finite), the intersection will be the primes less than 30 that are also odd numbers greater than zero.

But wait, the second set is all odd numbers greater than zero. So, the intersection would be all numbers that are both prime less than 30 and odd numbers greater than zero. However, note that the only even prime number is 2. All other primes are odd. Therefore, if we take the set of primes less than 30, and remove the even prime (which is 2), then the rest are odd. Therefore, the intersection of the two sets would be all primes less than 30 except 2. But let me verify this.

Alternatively, since the second set includes all odd numbers greater than zero, which includes all odd primes, but also includes composite odd numbers. However, the first set is only primes. Therefore, the intersection would be the primes that are odd numbers greater than zero. Since primes except 2 are odd, so the intersection is all primes less than 30 except 2. Therefore, the number of elements in the intersection would be the number of primes less than 30 minus 1 (for the prime 2). Since there are 10 primes less than 30, subtracting 1 gives 9. Therefore, the answer is 9?

Wait, but let me check again. Let me list the primes less than 30 and see which ones are odd. The primes less than 30 are:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29.

Now, which of these are odd? Well, 2 is even, so the rest are odd. So 3, 5, 7, 11, 13, 17, 19, 23, 29. That's 9 numbers. Therefore, the intersection is these 9 numbers. Therefore, the answer is 9.

But let me make sure that the second set is indeed all odd numbers greater than zero. So, the second set includes numbers like 1, 3, 5, 7,... So, 1 is in the second set. Is 1 considered here? The problem says "the set of all the odd numbers greater than zero." So, greater than zero, so starting from 1. But 1 is not a prime number. Therefore, the intersection would be the primes less than 30 that are also in the set of odd numbers greater than zero. Since primes are greater than 1, so all primes are greater than 1, so they are in the set of odd numbers greater than zero, except for 2. Therefore, the intersection is all primes less than 30 except 2. Therefore, 10 - 1 = 9.

But let me check if 1 is included in the second set. The problem says "the set of all the odd numbers greater than zero." So, numbers greater than zero, so starting at 1. So 1 is included. However, 1 is not prime. Therefore, the intersection would be primes less than 30 that are odd. Since primes except 2 are odd, so the intersection is primes less than 30 except 2. Therefore, 9 elements.

Alternatively, maybe I can think of the intersection as the set of numbers that are both prime less than 30 and odd numbers greater than zero. So, numbers that are primes less than 30 and also odd and greater than zero. Since all primes except 2 are odd, and all primes are greater than zero. Therefore, the intersection is primes less than 30 except 2. Therefore, 9 elements.

But let me check if there's any other number that is in both sets. For example, is there a prime number less than 30 that is not odd? Only 2. So, the only prime that is even is 2. Therefore, the rest are odd. Therefore, the intersection is all primes less than 30 except 2. Therefore, 10 - 1 = 9.

Alternatively, maybe I can list the elements of both sets and see their intersection.

First set (primes less than 30): {2, 3, 5, 7, 11, 13, 17, 19, 23, 29}

Second set (odd numbers greater than zero): {1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, ...}

So the intersection would be numbers that are in both sets. So, looking at the first set, which numbers are also in the second set? Well, all numbers in the first set except 2 are odd, so they are in the second set. However, 2 is not in the second set because it's even. Therefore, the intersection is {3, 5, 7, 11, 13, 17, 19, 23, 29}, which has 9 elements.

Therefore, the answer is 9.

But let me check if there's any other number in the first set that is not odd. Well, the only even prime is 2, so no others. Therefore, the intersection is indeed 9 elements.

Alternatively, maybe I can think of the second set as all odd numbers greater than zero. So, the intersection is the set of primes less than 30 that are odd. Since primes are either 2 or odd, so the intersection is primes less than 30 minus {2}. Therefore, 10 - 1 = 9.

Therefore, I think the answer is 9.

But just to be thorough, let me list the numbers in the intersection:

From the first set: 3, 5, 7, 11, 13, 17, 19, 23, 29. Let me count them:

1. 3
2. 5
3. 7
4. 11
5. 13
6. 17
7. 19
8. 23
9. 29

Yes, that's 9 numbers. Therefore, the intersection has 9 elements.

Therefore, the answer is 9.

But let me check once more if I made a mistake in counting the primes less than 30. Let me list them again:

Primes less than 30:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29.

Yes, that's 10 primes. So removing 2, we have 9 primes left. Therefore, the intersection is 9 elements.

Alternatively, if I think of the set of odd numbers greater than zero, which includes 1, 3, 5,... So, the primes in that set would be all primes except 2. Therefore, the intersection is primes less than 30 except 2. Therefore, 9 elements.

Therefore, I think the answer is 9.

But just to make sure, let me check if there's any other number that might be in both sets. For example, is 1 in the first set? No, because 1 is not a prime. So, the intersection is only the primes that are odd. So, yes, 9 elements.

Therefore, I think the answer is 9.

**Final Answer**
The number of elements in the intersection is \boxed{9}.

To determine how many elements are in the intersection of the set of all prime numbers less than 30 and the set of all odd numbers greater than zero, we proceed step by step.

---

### Step 1: Identify the First Set – Prime Numbers Less Than 30

A **prime number** is a natural number greater than 1 that has no positive divisors other than 1 and itself. Listing all such numbers less than 30:

$$
\{2, 3, 5, 7, 11, 13, 17, 19, 23, 29\}
$$

This set contains **10 elements**.

---

### Step 2: Identify the Second Set – Odd Numbers Greater Than Zero

An **odd number** is any integer that is not divisible by 2. The set of all odd numbers greater than zero includes:

$$
\{1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, \dots\}
$$

This is an **infinite** set, but since we are only interested in the intersection with the first set (which is finite), we only need to consider the odd numbers that are also in the first set.

---

### Step 3: Find the Intersection

We are looking for numbers that are **both**:

- Prime numbers less than 30, and
- Odd numbers greater than zero.

From the first set, the only even prime is **2**. All other primes are odd. Therefore, the intersection will consist of all the primes less than 30 **except** 2.

So, the intersection is:

$$
\{3, 5, 7, 11, 13, 17, 19, 23, 29\}
$$

This set contains **9 elements**.

---

### Final Answer

$$
\boxed{9}
$$

Solution: In other words, we're looking for the number of positive odd prime numbers less than 30. We go through all odd numbers less than 30 and note how many of them are prime. We get that 3, 5, 7, 11, 13, 17, 19, 23, and 29 are all of the positive odd prime numbers less than 30, a total of $\boxed{9}$ elements in the intersection.


Explanation:
Okay, so I need to find the value of this expression: (26² - 24² - 10)² - 10². Hmm, let me break this down step by step. First, maybe I should compute the parts inside the parentheses first. Let me recall some algebraic identities that might help simplify things.

Wait, there's a difference of squares formula: a² - b² = (a - b)(a + b). Maybe I can use that here. Let me check if that applies to 26² - 24². Yes, that's exactly a difference of squares where a is 26 and b is 24. So, applying the formula, 26² - 24² would be (26 - 24)(26 + 24). Let me compute that.

26 minus 24 is 2, and 26 plus 24 is 50. So multiplying those together, 2 * 50 equals 100. So, 26² - 24² is 100. Then, the original expression inside the first parentheses is 100 minus 10. So that would be 100 - 10 = 90.

So now the entire expression simplifies to (90)² - 10². Wait, that's another difference of squares! So, (90)² - (10)² can be written as (90 - 10)(90 + 10). Let me compute that.

90 minus 10 is 80, and 90 plus 10 is 100. Multiplying those together, 80 * 100. That's straightforward: 80 * 100 = 8000.

Wait, so is the answer 8000? Let me double-check my steps to make sure I didn't make a mistake.

First step: 26² - 24². Using the difference of squares formula: (26 - 24)(26 + 24) = 2 * 50 = 100. Then subtract 10: 100 - 10 = 90. Then square that: 90². Then subtract 10². So, 90² - 10². Applying difference of squares again: (90 - 10)(90 + 10) = 80 * 100 = 8000.

Hmm, seems correct. Let me verify by calculating each part numerically without using the formula to see if I get the same result.

Calculating 26²: 26*26. Let me compute that. 25² is 625, so 26² is 625 + 25 + 26 = 676? Wait, no. Wait, 25² is 625, then 26² is (25 + 1)² = 25² + 2*25*1 + 1² = 625 + 50 + 1 = 676. Yes, that's right.

Then 24²: 24*24. Let me compute that. 20² is 400, 4² is 16, and then cross terms: 2*20*4 = 160. So (20 + 4)² = 400 + 160 + 16 = 576. So 24² is 576.

So 26² - 24² is 676 - 576 = 100. Then subtract 10: 100 - 10 = 90. Then square that: 90² is 8100. Then subtract 10² which is 100. So 8100 - 100 = 8000.

Yes, same result. So that seems to confirm it.

Alternatively, maybe I can compute the entire original expression step by step without factoring. Let me try that approach as another check.

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

First compute 26²: 676. Then 24²: 576. Then subtract them: 676 - 576 = 100. Then subtract 10: 100 - 10 = 90. Then square that: 90² = 8100. Then subtract 10²: 8100 - 100 = 8000. Yep, same answer.

So both methods give me 8000. Therefore, I think that's the correct answer.

But just to be thorough, let me check if there's another way someone might approach this problem and maybe make a mistake. For example, someone might incorrectly compute 26² - 24² - 10 as (26² - 24²) - 10, which is what I did, but maybe someone might think to subtract 10 first before squaring? Wait, no, the parentheses are around 26² - 24² - 10, so the order of operations is correct.

Alternatively, someone might miscalculate 26² or 24². Let me check those again.

26 squared: 26*26. Let me do 26*20 + 26*6 = 520 + 156 = 676. Correct.

24 squared: 24*24. 20*24 + 4*24 = 480 + 96 = 576. Correct.

So 676 - 576 is indeed 100. Then 100 - 10 is 90. Squared is 8100. Minus 10 squared is 8100 - 100 = 8000.

Another possible mistake could be in the difference of squares step. For example, if someone thought that (a² - b² - c²) is something else, but in this case, the expression is (a² - b² - c)² - c², which is different.

Alternatively, if someone tried to expand (90)² - 10² without recognizing the difference of squares, they might compute 90² as 8100 and 10² as 100, then subtract them. Which is exactly what I did.

Alternatively, someone might think that (a² - b² - c)² - c² can be factored in some other way, but I think the way I did it is the most straightforward.

Alternatively, maybe someone could make an arithmetic mistake when multiplying 80 * 100. But 80*100 is clearly 8000.

So, all steps check out. Therefore, I think the answer is 8000.

**Final Answer**
\boxed{8000}

To evaluate the expression:

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

we proceed step by step, simplifying the inner parts first and applying algebraic identities where possible.

---

### Step 1: Simplify the inner expression

We start with the expression inside the first set of parentheses:

$$
26^2 - 24^2 - 10
$$

We recognize that $26^2 - 24^2$ is a **difference of squares**, which can be factored as:

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

Now subtract 10:

$$
100 - 10 = 90
$$

So the expression becomes:

$$
(90)^2 - 10^2
$$

---

### Step 2: Apply the difference of squares again

Now we have:

$$
(90)^2 - (10)^2
$$

This is another **difference of squares**, which can be factored as:

$$
(90 - 10)(90 + 10) = 80 \times 100 = 8000
$$

---

### Final Answer

$$
\boxed{8000}
$$

[Assistant]
Explanation:
Okay, so I need to find the value of this expression: (26² - 24² - 10)² - 10². Hmm, let me break this down step by step. First, maybe I should compute the parts inside the parentheses first. Let me recall some algebraic identities that might help simplify things.

Wait, there's a difference of squares formula: a² - b² = (a - b)(a + b). Maybe I can use that here. Let me check if that applies to 26² - 24². Yes, that's exactly a difference of squares where a is 26 and b is 24. So, applying the formula, 26² - 24² would be (26 - 24)(26 + 24). Let me compute that.

26 minus 24 is 2, and 26 plus 24 is 50. So multiplying those together, 2 * 50 equals 100. So, 26² - 24² is 100. Then, the original expression inside the first parentheses is 100 minus 10. So that would be 100 - 10 = 90.

So now the entire expression simplifies to (90)² - 10². Wait, that's another difference of squares! So, (90)² - (10)² can be written as (90 - 10)(90 + 10). Let me compute that.

90 minus 10 is 80, and 90 plus 10 is 100. Multiplying those together, 80 * 100. That's straightforward: 80 * 100 = 8000.

Wait, so is the answer 8000? Let me double-check my steps to make sure I didn't make a mistake.

First step: 26² - 24². Using the difference of squares formula: (26 - 24)(26 + 24) = 2 * 50 = 100. Then subtract 10: 100 - 10 = 90. Then square that: 90². Then subtract 10². So, 90² - 10². Applying difference of squares again: (90 - 10)(90 + 10) = 80 * 100 = 8000.

Hmm, seems correct. Let me verify by calculating each part numerically without using the formula to see if I get the same result.

Calculating 26²: 26*26. Let me compute that. 25² is 625, so 26² is 625 + 25 + 26 = 676? Wait, no. Wait, 25² is 625, then 26² is (25 + 1)² = 25² + 2*25*1 + 1² = 625 + 50 + 1 = 676. Yes, that's right.

Then 24²: 24*24. Let me compute that. 20² is 400, 4² is 16, and then cross terms: 2*20*4 = 160. So (20 + 4)² = 400 + 160 + 16 = 576. So 24² is 576.

So 26² - 24² is 676 - 576 = 100. Then subtract 10: 100 - 10 = 90. Then square that: 90² is 8100. Then subtract 10² which is 100. So 8100 - 100 = 8000.

Yes, same result. So that seems to confirm it.

Alternatively, maybe I can compute the entire original expression step by step without factoring. Let me try that approach as another check.

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

First compute 26²: 676. Then 24²: 576. Then subtract them: 676 - 576 = 100. Then subtract 10: 100 - 10 = 90. Then square that: 90² = 8100. Then subtract 10²: 8100 - 100 = 8000. Yep, same answer.

So both methods give me 8000. Therefore, I think that's the correct answer.

But just to be thorough, let me check if there's another way someone might approach this problem and maybe make a mistake. For example, someone might incorrectly compute 26² - 24² - 10 as (26² - 24²) - 10, which is what I did, but maybe someone might think to subtract 10 first before squaring? Wait, no, the parentheses are around 26² - 24² - 10, so the order of operations is correct.

Alternatively, someone might miscalculate 26² or 24². Let me check those again.

26 squared: 26*26. Let me do 26*20 + 26*6 = 520 + 156 = 676. Correct.

24 squared: 24*24. 20*24 + 4*24 = 480 + 96 = 576. Correct.

So 676 - 576 is indeed 100. Then 100 - 10 is 90. Squared is 8100. Minus 10 squared is 8100 - 100 = 8000.

Another possible mistake could be in the difference of squares step. For example, if someone thought that (a² - b² - c²) is something else, but in this case, the expression is (a² - b² - c)² - c², which is different.

Alternatively, if someone tried to expand (90)² - 10² without recognizing the difference of squares, they might compute 90² as 8100 and 10² as 100, then subtract them. Which is exactly what I did.

Alternatively, someone might think that (a² - b² - c)² - c² can be factored in some other way, but I think the way I did it is the most straightforward.

Alternatively, maybe someone could make an arithmetic mistake when multiplying 80 * 100. But 80*100 is clearly 8000.

So, all steps check out. Therefore, I think the answer is 8000.

**Final Answer**
\boxed{8000}

To evaluate the expression:

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

we proceed step by step, simplifying the inner parts first and applying algebraic identities where possible.

---

### Step 1: Simplify the inner expression

We start with the expression inside the first set of parentheses:

$$
26^2 - 24^2 - 10
$$

We recognize that $26^2 - 24^2$ is a **difference of squares**, which can be factored as:

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

Now subtract 10:

$$
100 - 10 = 90
$$

So the expression becomes:

$$
(90)^2 - 10^2
$$

---

### Step 2: Apply the difference of squares again

Now we have:

$$
(90)^2 - (10)^2
$$

This is another **difference of squares**, which can be factored as:

$$
(90 - 10)(90 + 10) = 80 \times 100 = 8000
$$

---

### Final Answer

$$
\boxed{8000}
$$

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