Explanation:
Okay, let's try to solve this equation: 3^{2x} + 19 = 10^x. Hmm, so it's an exponential equation with different bases, 3 and 10. I need to find the value of x that makes this equation true. Let me think about how to approach this.

First, maybe I can simplify the equation a bit. The term 3^{2x} can be rewritten using exponent rules. Since 3^{2x} is the same as (3^2)^x, which is 9^x. So the equation becomes 9^x + 19 = 10^x. That seems a bit simpler. Now I have 9^x + 19 = 10^x.

Hmm, so now I have two exponential terms with different bases. I remember that equations with different bases can sometimes be solved by taking logarithms, but I need to see if that's feasible here. Alternatively, maybe I can rearrange the equation to get all terms on one side and see if there's a way to factor or find a common base. Let me try rearranging.

Subtract 9^x from both sides: 19 = 10^x - 9^x. So the equation is 10^x - 9^x = 19. Hmm, not sure if that helps directly. Maybe I can consider substituting variables here. Let me think. Let me denote y = x, but that's not helpful. Maybe set t = something else.

Wait, since both 10^x and 9^x are exponential functions, maybe I can take logarithms of both sides? But the problem is that there are two exponential terms on the left. If I take log of both sides, I would get log(9^x + 19) = log(10^x), which might not help because of the addition inside the logarithm. That seems complicated.

Alternatively, maybe I can think of this as a function and try to find its roots. Let me define a function f(x) = 10^x - 9^x - 19. Then, I need to find x such that f(x) = 0. To solve this, maybe I can try plugging in some integer values for x to see if I can find a solution. Let's try that.

Let me test x=2: 10^2 - 9^2 -19 = 100 - 81 -19 = 0. Wait, that's zero! So x=2 is a solution? Let me check again. 10^2 is 100, 9^2 is 81, so 100 -81 is 19, minus 19 is 0. Yes! So x=2 is a solution.

Wait, so is there another solution? Because sometimes exponential equations can have more than one solution. Let me check another value. For example, x=1: 10^1 -9^1 -19 = 10 -9 -19 = -18. So negative. x=3: 10^3 -9^3 -19 = 1000 - 729 -19 = 252. So positive. So between x=1 and x=2, f(x) goes from -18 to 0, so crosses zero at x=2. Then from x=2 to x=3, it goes from 0 to 252. What about for x less than 2? Let me try x=0: 10^0 -9^0 -19 = 1 -1 -19 = -19. Still negative. How about x= -1? 10^{-1} -9^{-1} -19 = 1/10 - 1/9 -19 ≈ 0.1 - 0.111 -19 ≈ -19.011, still negative. So maybe x=2 is the only solution?

But let me check if there's another solution somewhere else. Let me think about the behavior of the function f(x) = 10^x -9^x -19. For very large x, since 10^x grows faster than 9^x, the function will go to infinity as x approaches infinity. For x approaching negative infinity, 10^x and 9^x both approach zero, so f(x) approaches -19. Therefore, since f(x) crosses zero at x=2, and since it's increasing? Wait, let's check the derivative to see if the function is increasing or decreasing.

Let me compute f'(x) = derivative of 10^x -9^x -19. The derivative of a^x is ln(a)*a^x, so f'(x) = ln(10)*10^x - ln(9)*9^x.

So for x=2: ln(10)*10^2 - ln(9)*9^2 ≈ 2.3026*100 - 2.1972*81 ≈ 230.26 - 177.97 ≈ 52.29, which is positive. So at x=2, the function is increasing. What about for x approaching negative infinity? The derivative would be dominated by the term with the smaller base, since 9^x is smaller than 10^x for x negative. Wait, actually for negative x, 9^x = (1/9)^{|x|}, and 10^x = (1/10)^{|x|}, so 10^x is smaller than 9^x when x is negative. Therefore, ln(10)*10^x - ln(9)*9^x would be negative because ln(10)*something smaller minus ln(9)*something larger. So the derivative is negative for x negative. So the function f(x) is decreasing for x negative, reaches some minimum, and then increases for x positive. So maybe there's another solution where the function crosses zero again?

Wait, but earlier when I checked x=1: f(1) = -18, x=2: 0, x=3: positive. So if the function is increasing for x beyond some point, maybe there's only one crossing? But let me check if there's another solution.

Wait, maybe when x is very large, the function is increasing, and since it's always increasing after a certain point? Wait, but the derivative f'(x) = ln(10)*10^x - ln(9)*9^x. Let me see when f'(x) is positive. Since 10^x grows faster than 9^x, there must be some x where f'(x) becomes positive. Wait, for x=0: ln(10)*1 - ln(9)*1 ≈ 2.3026 - 2.1972 ≈ 0.1054, which is positive. So derivative is positive at x=0, which means the function is increasing at x=0. But earlier, for x negative, derivative was negative. Therefore, the function has a minimum somewhere between x negative and x=0? Wait, but how come at x=0, derivative is positive? If the derivative is negative for x negative, and positive at x=0, there must be a critical point where derivative is zero. Let me find where f'(x)=0.

Set ln(10)*10^x - ln(9)*9^x = 0. So ln(10)/ln(9) = (9/10)^x. Let me compute ln(10)/ln(9). Let me calculate that. ln(10) ≈ 2.302585093, ln(9) ≈ 2.197224577. So ln(10)/ln(9) ≈ 2.302585093 / 2.197224577 ≈ 1.048. So 1.048 = (9/10)^x. Taking natural logarithm on both sides: ln(1.048) = x * ln(9/10). Therefore, x = ln(1.048)/ln(9/10). Compute ln(1.048) ≈ 0.0468. ln(9/10) ≈ ln(0.9) ≈ -0.10536. Therefore, x ≈ 0.0468 / (-0.10536) ≈ -0.444. So there is a critical point at x ≈ -0.444. So the function f(x) decreases until x ≈ -0.444, then increases afterward. So the minimum of f(x) is around x ≈ -0.444. Let me check the value of f(x) at that minimum.

But since f(x) approaches -19 as x approaches negative infinity, and reaches some minimum at x≈-0.444, then increases to f(0)=1 -1 -19 = -19? Wait, wait, f(0)=10^0 -9^0 -19 = 1 -1 -19 = -19. So at x=0, f(x) is -19. Wait, but if the function decreases until x≈-0.444, then increases afterwards, and at x=0 it's -19. So the minimum at x≈-0.444 must be less than or equal to -19? Wait, but f(x) approaches -19 as x approaches -infty. So maybe the minimum is -19? Wait, that can't be. Wait, if the function is decreasing from x=-infty to x≈-0.444, then increasing from x≈-0.444 to +infty, and at x approaching -infty it approaches -19, while at x=0, it's -19. Therefore, the minimum must be somewhere between x=-infty and x=0? Wait, but if at x approaching -infty, f(x) approaches -19, and at x=0, it's -19. So if the function decreases then increases, then the minimum must be lower than -19? But how?

Wait, maybe my calculation is wrong. Let me re-examine. Let me compute f(-0.444). Let me approximate x ≈ -0.444. Let me compute 10^x -9^x -19. Let me compute 10^{-0.444} ≈ 1 / 10^{0.444}. Let me calculate 10^{0.444}. Natural log: ln(10^{0.444}) = 0.444 * ln(10) ≈ 0.444 * 2.3026 ≈ 1.023. So exponentiate: e^{1.023} ≈ 2.78. Therefore, 10^{-0.444} ≈ 1 / 2.78 ≈ 0.359. Similarly, 9^{-0.444} = 1 / 9^{0.444}. Let me compute 9^{0.444}. ln(9^{0.444}) = 0.444 * ln(9) ≈ 0.444 * 2.1972 ≈ 0.975. Exponentiate: e^{0.975} ≈ 2.65. Therefore, 9^{-0.444} ≈ 1 / 2.65 ≈ 0.377. Therefore, f(-0.444) ≈ 0.359 - 0.377 -19 ≈ -0.018 -19 ≈ -19.018. So the minimum of f(x) is approximately -19.018, which is slightly less than -19. So the function f(x) reaches a minimum of about -19.018, then increases back to -19 at x=0, and then keeps increasing. Therefore, since the minimum is below -19, then the function crosses zero only once at x=2? Wait, but how?

Wait, because for x approaching infinity, f(x) approaches infinity, and since f(x) is increasing after x≈-0.444, then if at x=2, f(x)=0, and after that it keeps increasing, so there is only one solution at x=2. But wait, maybe there's another solution between x approaching negative infinity and x=-0.444? But since f(x) approaches -19 as x approaches negative infinity, and at the minimum point it is -19.018, which is less than -19, so the function is decreasing from -19 to -19.018 as x goes from -infty to -0.444? Wait, no. Wait, if as x approaches -infty, f(x) approaches -19. Then, the function decreases to the minimum at x≈-0.444, which is -19.018, and then increases back to -19 at x=0. Therefore, the function never crosses zero in that region. Because the minimum is -19.018, so it's always less than zero in that range. Then from x=0 onwards, it increases from -19, passes through f(1)=-18, f(2)=0, then increases further. So only crosses zero once at x=2. Therefore, the only real solution is x=2.

But wait, just to be thorough, maybe there is another solution for some x > 2? Wait, but f(x) is increasing after x≈-0.444. So for x>2, f(x) continues increasing, going from 0 up to infinity. So no other zeros there. For x between -0.444 and 0, f(x) increases from -19.018 to -19, still negative. For x between 0 and 2, f(x) increases from -19 to 0. So only crosses zero at x=2. Therefore, the equation has only one real solution, which is x=2.

Wait, but let me check x=2 again. Original equation: 3^{2x} + 19 = 10^x. Plugging x=2: 3^4 + 19 = 81 + 19 = 100, and 10^2=100. Yes, that works. So x=2 is definitely a solution.

Is there another solution? Let me think. For example, if I consider the original equation again: 9^x +19 =10^x. Suppose I think of this as 10^x -9^x =19. If there is another x where 10^x -9^x equals 19, but since we saw that f(x)=10^x -9^x -19 crosses zero only once at x=2, then there is only one solution.

Alternatively, maybe there is a way to solve this equation algebraically? Let me think. Maybe taking logarithms? But since there are two exponentials, perhaps not straightforward. Let me try to make substitution. Let me set y = 10^x. Then, since 9^x can be written in terms of y? Wait, but 9^x = (9/10)^x * 10^x = (9/10)^x * y. So 9^x = y * (9/10)^x. But that seems to complicate things because of the (9/10)^x term. Alternatively, maybe express both terms in terms of base 10 or base 3?

Alternatively, take logarithms of both sides. But again, because of addition, it's tricky. Wait, but if I rearrange the equation: 10^x - 9^x =19. Let me think. For x=2, it's 100 -81=19. Maybe that's the only integer solution, and since the function is strictly increasing after x≈-0.444, so only one solution. Therefore, perhaps x=2 is the only solution.

Alternatively, maybe we can use the Lambert W function? But that might be overcomplicating. Let me think. The equation is 10^x -9^x=19. Let me try to write this as 10^x = 9^x +19. Let me divide both sides by 9^x: (10/9)^x = 1 + 19 * (9^{-x}). Let me set t = (10/9)^x. Then, note that (9^{-x}) = (9^x)^{-1} = ( (10/9)^x * 9^x )^{-1}? Wait, maybe not helpful. Alternatively, since t = (10/9)^x, then ln(t) = x ln(10/9). Also, 9^{-x} = (9^x)^{-1} = [ (10^x)/(10/9)^x ]^{-1} ??? Maybe not helpful. Let me think again.

Let me set t = 10^x. Then, 9^x = (9/10)^x * t. So equation becomes t - (9/10)^x * t = 19. Then, t [ 1 - (9/10)^x ] =19. But (9/10)^x = (9/10)^{log_{10} t / log_{10} 10} }? Wait, since t =10^x, so x = log_{10} t. Therefore, (9/10)^x = (9/10)^{log_{10} t} = t^{log_{10}(9/10)}. Because a^{log_b c} = c^{log_b a}. Let me verify that. Let me recall that a^{log_b c} = e^{ln a * log_b c} = e^{ln a * (ln c / ln b)} = e^{(ln c / ln b) * ln a} = e^{ln c * log_b a} = c^{log_b a}. Yes, so that identity holds. Therefore, (9/10)^x = t^{log_{10}(9/10)}. Since log_{10}(9/10) is log_{10}9 - 1 ≈ 0.9542 -1 ≈ -0.0458. Therefore, the equation becomes t[1 - t^{-0.0458}] = 19. Which is t - t^{1 - 0.0458} = 19, so t - t^{0.9542} = 19.

Hmm, this seems more complicated. So, t - t^{0.9542} =19. Not sure if that helps. Maybe not. Maybe this approach isn't helpful.

Alternatively, maybe consider that for x=2 is the solution, and since it's the only solution, we can just state that. But the question is to solve for x, so maybe there is a way to show that x=2 is the only solution.

Alternatively, think about the function g(x) = 10^x - 9^x. Its behavior: as x increases, g(x) increases because the difference between two exponentials with different bases. Since 10^x grows faster than 9^x. Also, since for x=2, g(x)=19, and since g(x) is increasing, then x=2 is the unique solution.

But how do we know that g(x) is increasing? Earlier, we saw that the derivative is positive for x> critical point, but for x < critical point, derivative is negative. However, if the function g(x) reaches a minimum at x≈-0.444, but since we saw that at x=2, g(x)=19 and it's increasing after that, then if we can show that g(x) is increasing for x >= some value, then since g(2)=19 and g(x) increases beyond that, x=2 is the only solution.

Alternatively, maybe consider that for x >= 1, 10^x -9^x is increasing? Let me check the derivative for x >=1. At x=1, derivative f'(1)= ln(10)*10 - ln(9)*9 ≈ 2.3026*10 - 2.1972*9 ≈ 23.026 - 19.7748 ≈ 3.25, which is positive. So for x >=1, derivative is positive, so function is increasing, and since at x=2, f(x)=0, then for x>2, f(x) is positive. Therefore, since f(x) is increasing for x >=1, and f(2)=0, then x=2 is the only solution for x >=1.

For x <1, since the function f(x) is decreasing from x=-infty to x≈-0.444, then increasing from x≈-0.444 to +infty. But since at x= -infty, f(x) approaches -19, and at x≈-0.444, it reaches -19.018, then increases back to -19 at x=0, then increases further. Therefore, for x <2, the function f(x) is less than zero except at x=2 where it's zero. Therefore, x=2 is the only solution.

Therefore, the solution is x=2.

But just to be thorough, let me check another value near x=2. For example, x=1.5: 10^1.5 - 9^1.5 -19. Let me compute 10^1.5 = sqrt(10^3) = sqrt(1000) ≈ 31.6227766. 9^1.5 = sqrt(9^3) = sqrt(729) = 27. So 31.6227766 -27 -19 ≈ 31.6227766 -46 ≈ -14.377. So still negative. At x=2, it's zero, and at x=2.5, let's see: 10^2.5 = sqrt(10^5) = sqrt(100000) ≈ 316.227766. 9^2.5 = sqrt(9^5) = sqrt(59049) ≈ 243. So 316.227766 -243 -19 ≈ 316.227766 -262 ≈ 54.227. Positive. So the function crosses zero between 1.5 and 2.5. Wait, but we already know that at x=2 it is zero. So that's consistent. So x=2 is the only crossing.

Therefore, after all this analysis, I can conclude that x=2 is the only real solution to the equation.

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

To solve the equation:

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

we begin by simplifying the left-hand side using exponent rules. Note that $3^{2x} = (3^2)^x = 9^x$. So the equation becomes:

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

Rewriting this:

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

This is now an equation involving two exponential functions with different bases, which can be challenging to solve explicitly using elementary algebra. Let's define a function:

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

We are looking for values of $x$ where $f(x) = 0$.

---

### Step 1: Try Integer Values

Let’s test integer values of $x$ to find any potential solutions.

- For $x = 1$: $10^1 - 9^1 - 19 = 10 - 9 - 19 = -18$ → Not a solution.
- For $x = 2$: $10^2 - 9^2 - 19 = 100 - 81 - 19 = 0$ → **Solution found!**
- For $x = 3$: $10^3 - 9^3 - 19 = 1000 - 729 - 19 = 252$ → Not zero.

Thus, we have at least one solution: $x = 2$.

---

### Step 2: Analyze the Function Behavior

To ensure this is the **only** solution, consider the function:

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

We analyze its behavior:

- As $x \to \infty$, $10^x$ grows much faster than $9^x$, so $f(x) \to \infty$.
- As $x \to -\infty$, both $10^x$ and $9^x$ approach 0, so $f(x) \to -19$.

Now, compute the derivative:

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

Since $\ln(10) > \ln(9)$ and $10^x > 9^x$ for $x > 0$, the derivative is **positive** for all $x > 0$, which means $f(x)$ is **strictly increasing** in that region.

Furthermore, we find:

- $f(2) = 0$
- $f(1) = -18$
- $f(3) = 252$

So the function **crosses zero exactly once** (at $x = 2$), moving from negative to positive as $x$ increases.

---

### Step 3: Confirm Uniqueness

Since:

- $f(x)$ is continuous,
- $f(x) < 0$ for all $x < 2$,
- $f(x) > 0$ for all $x > 2$,
- $f(x)$ increases for $x > 0$,

there is **no other value** of $x$ such that $f(x) = 0$. Hence, $x = 2$ is the **only** real solution.

---

### Final Answer

$$
\boxed{2}
$$

[Assistant]
Explanation:
Okay, let's try to solve this equation: 3^{2x} + 19 = 10^x. Hmm, so it's an exponential equation with different bases, 3 and 10. I need to find the value of x that makes this equation true. Let me think about how to approach this.

First, maybe I can simplify the equation a bit. The term 3^{2x} can be rewritten using exponent rules. Since 3^{2x} is the same as (3^2)^x, which is 9^x. So the equation becomes 9^x + 19 = 10^x. That seems a bit simpler. Now I have 9^x + 19 = 10^x.

Hmm, so now I have two exponential terms with different bases. I remember that equations with different bases can sometimes be solved by taking logarithms, but I need to see if that's feasible here. Alternatively, maybe I can rearrange the equation to get all terms on one side and see if there's a way to factor or find a common base. Let me try rearranging.

Subtract 9^x from both sides: 19 = 10^x - 9^x. So the equation is 10^x - 9^x = 19. Hmm, not sure if that helps directly. Maybe I can consider substituting variables here. Let me think. Let me denote y = x, but that's not helpful. Maybe set t = something else.

Wait, since both 10^x and 9^x are exponential functions, maybe I can take logarithms of both sides? But the problem is that there are two exponential terms on the left. If I take log of both sides, I would get log(9^x + 19) = log(10^x), which might not help because of the addition inside the logarithm. That seems complicated.

Alternatively, maybe I can think of this as a function and try to find its roots. Let me define a function f(x) = 10^x - 9^x - 19. Then, I need to find x such that f(x) = 0. To solve this, maybe I can try plugging in some integer values for x to see if I can find a solution. Let's try that.

Let me test x=2: 10^2 - 9^2 -19 = 100 - 81 -19 = 0. Wait, that's zero! So x=2 is a solution? Let me check again. 10^2 is 100, 9^2 is 81, so 100 -81 is 19, minus 19 is 0. Yes! So x=2 is a solution.

Wait, so is there another solution? Because sometimes exponential equations can have more than one solution. Let me check another value. For example, x=1: 10^1 -9^1 -19 = 10 -9 -19 = -18. So negative. x=3: 10^3 -9^3 -19 = 1000 - 729 -19 = 252. So positive. So between x=1 and x=2, f(x) goes from -18 to 0, so crosses zero at x=2. Then from x=2 to x=3, it goes from 0 to 252. What about for x less than 2? Let me try x=0: 10^0 -9^0 -19 = 1 -1 -19 = -19. Still negative. How about x= -1? 10^{-1} -9^{-1} -19 = 1/10 - 1/9 -19 ≈ 0.1 - 0.111 -19 ≈ -19.011, still negative. So maybe x=2 is the only solution?

But let me check if there's another solution somewhere else. Let me think about the behavior of the function f(x) = 10^x -9^x -19. For very large x, since 10^x grows faster than 9^x, the function will go to infinity as x approaches infinity. For x approaching negative infinity, 10^x and 9^x both approach zero, so f(x) approaches -19. Therefore, since f(x) crosses zero at x=2, and since it's increasing? Wait, let's check the derivative to see if the function is increasing or decreasing.

Let me compute f'(x) = derivative of 10^x -9^x -19. The derivative of a^x is ln(a)*a^x, so f'(x) = ln(10)*10^x - ln(9)*9^x.

So for x=2: ln(10)*10^2 - ln(9)*9^2 ≈ 2.3026*100 - 2.1972*81 ≈ 230.26 - 177.97 ≈ 52.29, which is positive. So at x=2, the function is increasing. What about for x approaching negative infinity? The derivative would be dominated by the term with the smaller base, since 9^x is smaller than 10^x for x negative. Wait, actually for negative x, 9^x = (1/9)^{|x|}, and 10^x = (1/10)^{|x|}, so 10^x is smaller than 9^x when x is negative. Therefore, ln(10)*10^x - ln(9)*9^x would be negative because ln(10)*something smaller minus ln(9)*something larger. So the derivative is negative for x negative. So the function f(x) is decreasing for x negative, reaches some minimum, and then increases for x positive. So maybe there's another solution where the function crosses zero again?

Wait, but earlier when I checked x=1: f(1) = -18, x=2: 0, x=3: positive. So if the function is increasing for x beyond some point, maybe there's only one crossing? But let me check if there's another solution.

Wait, maybe when x is very large, the function is increasing, and since it's always increasing after a certain point? Wait, but the derivative f'(x) = ln(10)*10^x - ln(9)*9^x. Let me see when f'(x) is positive. Since 10^x grows faster than 9^x, there must be some x where f'(x) becomes positive. Wait, for x=0: ln(10)*1 - ln(9)*1 ≈ 2.3026 - 2.1972 ≈ 0.1054, which is positive. So derivative is positive at x=0, which means the function is increasing at x=0. But earlier, for x negative, derivative was negative. Therefore, the function has a minimum somewhere between x negative and x=0? Wait, but how come at x=0, derivative is positive? If the derivative is negative for x negative, and positive at x=0, there must be a critical point where derivative is zero. Let me find where f'(x)=0.

Set ln(10)*10^x - ln(9)*9^x = 0. So ln(10)/ln(9) = (9/10)^x. Let me compute ln(10)/ln(9). Let me calculate that. ln(10) ≈ 2.302585093, ln(9) ≈ 2.197224577. So ln(10)/ln(9) ≈ 2.302585093 / 2.197224577 ≈ 1.048. So 1.048 = (9/10)^x. Taking natural logarithm on both sides: ln(1.048) = x * ln(9/10). Therefore, x = ln(1.048)/ln(9/10). Compute ln(1.048) ≈ 0.0468. ln(9/10) ≈ ln(0.9) ≈ -0.10536. Therefore, x ≈ 0.0468 / (-0.10536) ≈ -0.444. So there is a critical point at x ≈ -0.444. So the function f(x) decreases until x ≈ -0.444, then increases afterward. So the minimum of f(x) is around x ≈ -0.444. Let me check the value of f(x) at that minimum.

But since f(x) approaches -19 as x approaches negative infinity, and reaches some minimum at x≈-0.444, then increases to f(0)=1 -1 -19 = -19? Wait, wait, f(0)=10^0 -9^0 -19 = 1 -1 -19 = -19. So at x=0, f(x) is -19. Wait, but if the function decreases until x≈-0.444, then increases afterwards, and at x=0 it's -19. So the minimum at x≈-0.444 must be less than or equal to -19? Wait, but f(x) approaches -19 as x approaches -infty. So maybe the minimum is -19? Wait, that can't be. Wait, if the function is decreasing from x=-infty to x≈-0.444, then increasing from x≈-0.444 to +infty, and at x approaching -infty it approaches -19, while at x=0, it's -19. Therefore, the minimum must be somewhere between x=-infty and x=0? Wait, but if at x approaching -infty, f(x) approaches -19, and at x=0, it's -19. So if the function decreases then increases, then the minimum must be lower than -19? But how?

Wait, maybe my calculation is wrong. Let me re-examine. Let me compute f(-0.444). Let me approximate x ≈ -0.444. Let me compute 10^x -9^x -19. Let me compute 10^{-0.444} ≈ 1 / 10^{0.444}. Let me calculate 10^{0.444}. Natural log: ln(10^{0.444}) = 0.444 * ln(10) ≈ 0.444 * 2.3026 ≈ 1.023. So exponentiate: e^{1.023} ≈ 2.78. Therefore, 10^{-0.444} ≈ 1 / 2.78 ≈ 0.359. Similarly, 9^{-0.444} = 1 / 9^{0.444}. Let me compute 9^{0.444}. ln(9^{0.444}) = 0.444 * ln(9) ≈ 0.444 * 2.1972 ≈ 0.975. Exponentiate: e^{0.975} ≈ 2.65. Therefore, 9^{-0.444} ≈ 1 / 2.65 ≈ 0.377. Therefore, f(-0.444) ≈ 0.359 - 0.377 -19 ≈ -0.018 -19 ≈ -19.018. So the minimum of f(x) is approximately -19.018, which is slightly less than -19. So the function f(x) reaches a minimum of about -19.018, then increases back to -19 at x=0, and then keeps increasing. Therefore, since the minimum is below -19, then the function crosses zero only once at x=2? Wait, but how?

Wait, because for x approaching infinity, f(x) approaches infinity, and since f(x) is increasing after x≈-0.444, then if at x=2, f(x)=0, and after that it keeps increasing, so there is only one solution at x=2. But wait, maybe there's another solution between x approaching negative infinity and x=-0.444? But since f(x) approaches -19 as x approaches negative infinity, and at the minimum point it is -19.018, which is less than -19, so the function is decreasing from -19 to -19.018 as x goes from -infty to -0.444? Wait, no. Wait, if as x approaches -infty, f(x) approaches -19. Then, the function decreases to the minimum at x≈-0.444, which is -19.018, and then increases back to -19 at x=0. Therefore, the function never crosses zero in that region. Because the minimum is -19.018, so it's always less than zero in that range. Then from x=0 onwards, it increases from -19, passes through f(1)=-18, f(2)=0, then increases further. So only crosses zero once at x=2. Therefore, the only real solution is x=2.

But wait, just to be thorough, maybe there is another solution for some x > 2? Wait, but f(x) is increasing after x≈-0.444. So for x>2, f(x) continues increasing, going from 0 up to infinity. So no other zeros there. For x between -0.444 and 0, f(x) increases from -19.018 to -19, still negative. For x between 0 and 2, f(x) increases from -19 to 0. So only crosses zero at x=2. Therefore, the equation has only one real solution, which is x=2.

Wait, but let me check x=2 again. Original equation: 3^{2x} + 19 = 10^x. Plugging x=2: 3^4 + 19 = 81 + 19 = 100, and 10^2=100. Yes, that works. So x=2 is definitely a solution.

Is there another solution? Let me think. For example, if I consider the original equation again: 9^x +19 =10^x. Suppose I think of this as 10^x -9^x =19. If there is another x where 10^x -9^x equals 19, but since we saw that f(x)=10^x -9^x -19 crosses zero only once at x=2, then there is only one solution.

Alternatively, maybe there is a way to solve this equation algebraically? Let me think. Maybe taking logarithms? But since there are two exponentials, perhaps not straightforward. Let me try to make substitution. Let me set y = 10^x. Then, since 9^x can be written in terms of y? Wait, but 9^x = (9/10)^x * 10^x = (9/10)^x * y. So 9^x = y * (9/10)^x. But that seems to complicate things because of the (9/10)^x term. Alternatively, maybe express both terms in terms of base 10 or base 3?

Alternatively, take logarithms of both sides. But again, because of addition, it's tricky. Wait, but if I rearrange the equation: 10^x - 9^x =19. Let me think. For x=2, it's 100 -81=19. Maybe that's the only integer solution, and since the function is strictly increasing after x≈-0.444, so only one solution. Therefore, perhaps x=2 is the only solution.

Alternatively, maybe we can use the Lambert W function? But that might be overcomplicating. Let me think. The equation is 10^x -9^x=19. Let me try to write this as 10^x = 9^x +19. Let me divide both sides by 9^x: (10/9)^x = 1 + 19 * (9^{-x}). Let me set t = (10/9)^x. Then, note that (9^{-x}) = (9^x)^{-1} = ( (10/9)^x * 9^x )^{-1}? Wait, maybe not helpful. Alternatively, since t = (10/9)^x, then ln(t) = x ln(10/9). Also, 9^{-x} = (9^x)^{-1} = [ (10^x)/(10/9)^x ]^{-1} ??? Maybe not helpful. Let me think again.

Let me set t = 10^x. Then, 9^x = (9/10)^x * t. So equation becomes t - (9/10)^x * t = 19. Then, t [ 1 - (9/10)^x ] =19. But (9/10)^x = (9/10)^{log_{10} t / log_{10} 10} }? Wait, since t =10^x, so x = log_{10} t. Therefore, (9/10)^x = (9/10)^{log_{10} t} = t^{log_{10}(9/10)}. Because a^{log_b c} = c^{log_b a}. Let me verify that. Let me recall that a^{log_b c} = e^{ln a * log_b c} = e^{ln a * (ln c / ln b)} = e^{(ln c / ln b) * ln a} = e^{ln c * log_b a} = c^{log_b a}. Yes, so that identity holds. Therefore, (9/10)^x = t^{log_{10}(9/10)}. Since log_{10}(9/10) is log_{10}9 - 1 ≈ 0.9542 -1 ≈ -0.0458. Therefore, the equation becomes t[1 - t^{-0.0458}] = 19. Which is t - t^{1 - 0.0458} = 19, so t - t^{0.9542} = 19.

Hmm, this seems more complicated. So, t - t^{0.9542} =19. Not sure if that helps. Maybe not. Maybe this approach isn't helpful.

Alternatively, maybe consider that for x=2 is the solution, and since it's the only solution, we can just state that. But the question is to solve for x, so maybe there is a way to show that x=2 is the only solution.

Alternatively, think about the function g(x) = 10^x - 9^x. Its behavior: as x increases, g(x) increases because the difference between two exponentials with different bases. Since 10^x grows faster than 9^x. Also, since for x=2, g(x)=19, and since g(x) is increasing, then x=2 is the unique solution.

But how do we know that g(x) is increasing? Earlier, we saw that the derivative is positive for x> critical point, but for x < critical point, derivative is negative. However, if the function g(x) reaches a minimum at x≈-0.444, but since we saw that at x=2, g(x)=19 and it's increasing after that, then if we can show that g(x) is increasing for x >= some value, then since g(2)=19 and g(x) increases beyond that, x=2 is the only solution.

Alternatively, maybe consider that for x >= 1, 10^x -9^x is increasing? Let me check the derivative for x >=1. At x=1, derivative f'(1)= ln(10)*10 - ln(9)*9 ≈ 2.3026*10 - 2.1972*9 ≈ 23.026 - 19.7748 ≈ 3.25, which is positive. So for x >=1, derivative is positive, so function is increasing, and since at x=2, f(x)=0, then for x>2, f(x) is positive. Therefore, since f(x) is increasing for x >=1, and f(2)=0, then x=2 is the only solution for x >=1.

For x <1, since the function f(x) is decreasing from x=-infty to x≈-0.444, then increasing from x≈-0.444 to +infty. But since at x= -infty, f(x) approaches -19, and at x≈-0.444, it reaches -19.018, then increases back to -19 at x=0, then increases further. Therefore, for x <2, the function f(x) is less than zero except at x=2 where it's zero. Therefore, x=2 is the only solution.

Therefore, the solution is x=2.

But just to be thorough, let me check another value near x=2. For example, x=1.5: 10^1.5 - 9^1.5 -19. Let me compute 10^1.5 = sqrt(10^3) = sqrt(1000) ≈ 31.6227766. 9^1.5 = sqrt(9^3) = sqrt(729) = 27. So 31.6227766 -27 -19 ≈ 31.6227766 -46 ≈ -14.377. So still negative. At x=2, it's zero, and at x=2.5, let's see: 10^2.5 = sqrt(10^5) = sqrt(100000) ≈ 316.227766. 9^2.5 = sqrt(9^5) = sqrt(59049) ≈ 243. So 316.227766 -243 -19 ≈ 316.227766 -262 ≈ 54.227. Positive. So the function crosses zero between 1.5 and 2.5. Wait, but we already know that at x=2 it is zero. So that's consistent. So x=2 is the only crossing.

Therefore, after all this analysis, I can conclude that x=2 is the only real solution to the equation.

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

To solve the equation:

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

we begin by simplifying the left-hand side using exponent rules. Note that $3^{2x} = (3^2)^x = 9^x$. So the equation becomes:

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

Rewriting this:

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

This is now an equation involving two exponential functions with different bases, which can be challenging to solve explicitly using elementary algebra. Let's define a function:

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

We are looking for values of $x$ where $f(x) = 0$.

---

### Step 1: Try Integer Values

Let’s test integer values of $x$ to find any potential solutions.

- For $x = 1$: $10^1 - 9^1 - 19 = 10 - 9 - 19 = -18$ → Not a solution.
- For $x = 2$: $10^2 - 9^2 - 19 = 100 - 81 - 19 = 0$ → **Solution found!**
- For $x = 3$: $10^3 - 9^3 - 19 = 1000 - 729 - 19 = 252$ → Not zero.

Thus, we have at least one solution: $x = 2$.

---

### Step 2: Analyze the Function Behavior

To ensure this is the **only** solution, consider the function:

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

We analyze its behavior:

- As $x \to \infty$, $10^x$ grows much faster than $9^x$, so $f(x) \to \infty$.
- As $x \to -\infty$, both $10^x$ and $9^x$ approach 0, so $f(x) \to -19$.

Now, compute the derivative:

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

Since $\ln(10) > \ln(9)$ and $10^x > 9^x$ for $x > 0$, the derivative is **positive** for all $x > 0$, which means $f(x)$ is **strictly increasing** in that region.

Furthermore, we find:

- $f(2) = 0$
- $f(1) = -18$
- $f(3) = 252$

So the function **crosses zero exactly once** (at $x = 2$), moving from negative to positive as $x$ increases.

---

### Step 3: Confirm Uniqueness

Since:

- $f(x)$ is continuous,
- $f(x) < 0$ for all $x < 2$,
- $f(x) > 0$ for all $x > 2$,
- $f(x)$ increases for $x > 0$,

there is **no other value** of $x$ such that $f(x) = 0$. Hence, $x = 2$ is the **only** real solution.

---

### Final Answer

$$
\boxed{2}
$$

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) times [1, -2, 1]. Hmm, projections... Let me recall how that works.

From what I remember, the projection of a vector **a** onto another vector **b** is given by the formula:

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

So, the projection is a scalar multiple of **b**, right? And in this case, they say that the projection is (5/6)*[1, -2, 1]. So, if I can compute the projection using the formula and set it equal to that, I can solve for y.

First, let me write down the given vectors. The vector **a** is [2, y, -5], and the vector **b** is [1, -2, 1]. Let me compute the dot product **a** · **b** first.

Calculating the dot product:

**a** · **b** = (2)(1) + (y)(-2) + (-5)(1) = 2 - 2y - 5 = (2 - 5) - 2y = -3 - 2y.

Okay, so that's the dot product. Next, I need to compute ||**b**||², which is the squared magnitude of **b**.

Calculating ||**b**||²:

||**b**||² = (1)² + (-2)² + (1)² = 1 + 4 + 1 = 6.

So, the denominator is 6.

Therefore, the projection vector should be:

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

But they say the projection is (5/6)*[1, -2, 1]. Therefore, the scalar multiplier in front of **b** must be equal to 5/6.

So, setting the scalar equal:

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

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

-3 - 2y = 5

Now, solving for y:

-2y = 5 + 3 = 8

Therefore, y = 8 / (-2) = -4

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

First, the dot product: 2*1 + y*(-2) + (-5)*1. Yes, 2 - 2y -5 = -3 -2y. Correct. Then ||**b**||²: 1 + 4 + 1 = 6. Correct. Then setting (-3 - 2y)/6 = 5/6. Multiply both sides by 6: -3 - 2y = 5. Then adding 3 to both sides: -2y = 8, so y = -4. Seems right.

But let me verify by plugging back y = -4 into the original projection. Let me compute the projection again with y = -4.

So **a** becomes [2, -4, -5]. Then **a** · **b** = 2*1 + (-4)*(-2) + (-5)*1 = 2 + 8 -5 = 5.

Then, proj_b **a** is (5 / 6) * [1, -2, 1], which matches the given projection. Perfect! So that checks out.

Therefore, the value of y is indeed -4.

But let me think again if there's another way to approach this problem, maybe by components? For example, since the projection vector is given, maybe each component must match? Let me try that.

The projection vector is (5/6)[1, -2, 1] = [5/6, -10/6, 5/6] = [5/6, -5/3, 5/6].

On the other hand, using the projection formula, the projection vector is [(-3 - 2y)/6 * 1, (-3 - 2y)/6 * (-2), (-3 - 2y)/6 * 1]. Let me compute each component:

First component: (-3 - 2y)/6

Second component: (-3 - 2y)/6 * (-2) = [ ( -3 - 2y ) * (-2) ] / 6 = (6 + 4y)/6 = 1 + (4y)/6 = 1 + (2y)/3

Third component: same as first component: (-3 - 2y)/6

But according to the given projection, the first component is 5/6, so:

(-3 - 2y)/6 = 5/6 => -3 - 2y = 5 => same as before => y = -4

Alternatively, check the second component. If y = -4, then the second component would be:

(-3 - 2*(-4))/6 * (-2) = (-3 + 8)/6 * (-2) = (5/6)*(-2) = -10/6 = -5/3, which matches the given second component. Similarly, the third component is same as first, so 5/6. So that's consistent.

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

Is there any chance I messed up the sign somewhere? Let me check the dot product again with y=-4.

Original **a** is [2, -4, -5], **b** is [1, -2, 1].

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

Then, proj_b **a** is (5 / ||**b**||² ) * **b** = (5 / 6) * **b**, which is exactly the given projection.

Yes, that's correct. So the answer should be y = -4.

Alternatively, could there be a different interpretation of the question? Like, maybe the projection is onto a vector, but maybe they meant something else? But no, the projection of **a** onto **b** is standard.

Alternatively, could the question have meant orthogonal projection? But that's the same as what I used.

Alternatively, maybe the user made a typo? But no, the problem seems straightforward.

So, I think 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 $, we are given that the projection of the vector
$$
\begin{pmatrix} 2 \\ y \\ -5 \end{pmatrix}
$$
onto the vector
$$
\begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}
$$
is
$$
\frac{5}{6} \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}.
$$

---

### Step 1: Recall the projection formula

The projection of a vector $ \mathbf{a} $ onto a vector $ \mathbf{b} $ is given by:

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

---

### Step 2: Apply the formula

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

First, compute the dot product $ \mathbf{a} \cdot \mathbf{b} $:

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

Next, compute the squared norm of $ \mathbf{b} $:

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

Thus, the projection becomes:

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

---

### Step 3: Set projection equal to the given result

We are told that the projection is:

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

Since the projection must be a scalar multiple of $ \mathbf{b} $, the scalar in front of the vector must be equal:

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

Multiply both sides by 6:

$$
-3 - 2y = 5
$$

Solving for $ y $:

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

---

### Step 4: Verification

Plugging $ y = -4 $ into the original vector $ \mathbf{a} $ gives:

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

Compute the dot product:

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

So the projection is:

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

Which matches the given projection. Hence, the value of $ y $ is confirmed.

---

### 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) times [1, -2, 1]. Hmm, projections... Let me recall how that works.

From what I remember, the projection of a vector **a** onto another vector **b** is given by the formula:

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

So, the projection is a scalar multiple of **b**, right? And in this case, they say that the projection is (5/6)*[1, -2, 1]. So, if I can compute the projection using the formula and set it equal to that, I can solve for y.

First, let me write down the given vectors. The vector **a** is [2, y, -5], and the vector **b** is [1, -2, 1]. Let me compute the dot product **a** · **b** first.

Calculating the dot product:

**a** · **b** = (2)(1) + (y)(-2) + (-5)(1) = 2 - 2y - 5 = (2 - 5) - 2y = -3 - 2y.

Okay, so that's the dot product. Next, I need to compute ||**b**||², which is the squared magnitude of **b**.

Calculating ||**b**||²:

||**b**||² = (1)² + (-2)² + (1)² = 1 + 4 + 1 = 6.

So, the denominator is 6.

Therefore, the projection vector should be:

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

But they say the projection is (5/6)*[1, -2, 1]. Therefore, the scalar multiplier in front of **b** must be equal to 5/6.

So, setting the scalar equal:

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

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

-3 - 2y = 5

Now, solving for y:

-2y = 5 + 3 = 8

Therefore, y = 8 / (-2) = -4

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

First, the dot product: 2*1 + y*(-2) + (-5)*1. Yes, 2 - 2y -5 = -3 -2y. Correct. Then ||**b**||²: 1 + 4 + 1 = 6. Correct. Then setting (-3 - 2y)/6 = 5/6. Multiply both sides by 6: -3 - 2y = 5. Then adding 3 to both sides: -2y = 8, so y = -4. Seems right.

But let me verify by plugging back y = -4 into the original projection. Let me compute the projection again with y = -4.

So **a** becomes [2, -4, -5]. Then **a** · **b** = 2*1 + (-4)*(-2) + (-5)*1 = 2 + 8 -5 = 5.

Then, proj_b **a** is (5 / 6) * [1, -2, 1], which matches the given projection. Perfect! So that checks out.

Therefore, the value of y is indeed -4.

But let me think again if there's another way to approach this problem, maybe by components? For example, since the projection vector is given, maybe each component must match? Let me try that.

The projection vector is (5/6)[1, -2, 1] = [5/6, -10/6, 5/6] = [5/6, -5/3, 5/6].

On the other hand, using the projection formula, the projection vector is [(-3 - 2y)/6 * 1, (-3 - 2y)/6 * (-2), (-3 - 2y)/6 * 1]. Let me compute each component:

First component: (-3 - 2y)/6

Second component: (-3 - 2y)/6 * (-2) = [ ( -3 - 2y ) * (-2) ] / 6 = (6 + 4y)/6 = 1 + (4y)/6 = 1 + (2y)/3

Third component: same as first component: (-3 - 2y)/6

But according to the given projection, the first component is 5/6, so:

(-3 - 2y)/6 = 5/6 => -3 - 2y = 5 => same as before => y = -4

Alternatively, check the second component. If y = -4, then the second component would be:

(-3 - 2*(-4))/6 * (-2) = (-3 + 8)/6 * (-2) = (5/6)*(-2) = -10/6 = -5/3, which matches the given second component. Similarly, the third component is same as first, so 5/6. So that's consistent.

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

Is there any chance I messed up the sign somewhere? Let me check the dot product again with y=-4.

Original **a** is [2, -4, -5], **b** is [1, -2, 1].

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

Then, proj_b **a** is (5 / ||**b**||² ) * **b** = (5 / 6) * **b**, which is exactly the given projection.

Yes, that's correct. So the answer should be y = -4.

Alternatively, could there be a different interpretation of the question? Like, maybe the projection is onto a vector, but maybe they meant something else? But no, the projection of **a** onto **b** is standard.

Alternatively, could the question have meant orthogonal projection? But that's the same as what I used.

Alternatively, maybe the user made a typo? But no, the problem seems straightforward.

So, I think 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 $, we are given that the projection of the vector
$$
\begin{pmatrix} 2 \\ y \\ -5 \end{pmatrix}
$$
onto the vector
$$
\begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}
$$
is
$$
\frac{5}{6} \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}.
$$

---

### Step 1: Recall the projection formula

The projection of a vector $ \mathbf{a} $ onto a vector $ \mathbf{b} $ is given by:

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

---

### Step 2: Apply the formula

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

First, compute the dot product $ \mathbf{a} \cdot \mathbf{b} $:

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

Next, compute the squared norm of $ \mathbf{b} $:

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

Thus, the projection becomes:

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

---

### Step 3: Set projection equal to the given result

We are told that the projection is:

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

Since the projection must be a scalar multiple of $ \mathbf{b} $, the scalar in front of the vector must be equal:

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

Multiply both sides by 6:

$$
-3 - 2y = 5
$$

Solving for $ y $:

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

---

### Step 4: Verification

Plugging $ y = -4 $ into the original vector $ \mathbf{a} $ gives:

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

Compute the dot product:

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

So the projection is:

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

Which matches the given projection. Hence, the value of $ y $ is confirmed.

---

### 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). Hmm, let's see. Alright, first, I remember that simplifying expressions usually involves combining like terms. Like terms are terms that have the same variable part, right? So, variables with the same exponent. In this case, the variables are k and the constants.

Let me start by writing down the expression again to make sure I have it right: (-k + 4) + (-2 + 3k). Okay, so there are two sets of parentheses here. Maybe I should distribute the addition sign? Wait, actually, since it's just addition, I can remove the parentheses without changing the signs, right? Let me try that.

So, if I take away the parentheses, it becomes -k + 4 - 2 + 3k. Let me check that again. The first part is (-k + 4), and then adding (-2 + 3k). So when you add two terms, you just combine them. So yeah, removing the parentheses, the second part is -2 + 3k. So combining all the terms together: -k + 4 - 2 + 3k.

Now, let me identify the like terms. The terms with k are -k and 3k. The constant terms are 4 and -2. So I need to combine those. Let me handle the k terms first.

-k + 3k. Let's see, if I have -1k + 3k, that would be... Hmm, combining coefficients: -1 + 3 is 2, so 2k. Okay, that seems right.

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 check again step by step to make sure I didn't make a mistake.

Original expression: (-k + 4) + (-2 + 3k).

First, remove parentheses: -k + 4 - 2 + 3k.

Combining like terms: (-k + 3k) + (4 - 2).

Calculating each part: (-k + 3k) is 2k, and (4 - 2) is 2.

So yes, that gives 2k + 2.

Wait, but maybe I should write it as 2k + 2 or 2 + 2k? Either way is correct, but usually, we write the terms with variables first, so 2k + 2.

Alternatively, I can factor out a 2, but the question just says to simplify, so combining like terms is sufficient.

Let me verify by plugging in a value for k. Let's choose k = 1.

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

Simplified expression: 2*1 + 2 = 2 + 2 = 4.

Okay, that checks out. Let me try another value, like k = 0.

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

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

Same result. How about k = 2?

Original expression: (-2 + 4) + (-2 + 3*2) = (2) + (-2 + 6) = 2 + 4 = 6.

Simplified expression: 2*2 + 2 = 4 + 2 = 6.

Perfect, that's the same. So my simplified expression seems correct.

Alternatively, maybe I can think of it as combining the terms step by step. Let me try that approach again.

Starting with (-k + 4) + (-2 + 3k).

I can rearrange the terms using the commutative property of addition. That is, I can swap the order of addition. Let me do that to group the like terms.

So, -k + 3k + 4 - 2.

Which is the same as (-k + 3k) + (4 - 2), which we already did. So again, 2k + 2.

Another way: Maybe combine the constants first? Let me see.

Original expression: (-k + 4) + (-2 + 3k).

Let me group the constants and the variables separately.

Constants: 4 and -2.

Variables: -k and 3k.

So, constants: 4 - 2 = 2.

Variables: -k + 3k = 2k.

So altogether, 2k + 2.

Same result.

I think that's solid. Therefore, the simplified form of the expression is 2k + 2.

But wait, let me check if there's another way someone might approach this problem and maybe make a mistake. For example, if someone forgets to distribute the negative sign or something? Wait, in this case, there's no distribution needed because it's just addition.

But if the problem had something like (-k + 4) + (-2) + 3k, maybe. Wait, no, the original problem is (-k + 4) + (-2 + 3k).

So in that case, removing parentheses, it's -k + 4 - 2 + 3k.

Another possible mistake might be combining 4 - 2 as -6? But no, 4 - 2 is 2.

Or maybe combining -k + 3k as -4k? But that would be wrong. Because -k + 3k is like -1k + 3k = 2k.

So, as long as I remember that, it's okay.

Therefore, I think I'm confident that the simplified expression is 2k + 2.

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

To simplify the expression $(-k + 4) + (-2 + 3k)$, we proceed step by step, focusing on combining like terms.

---

### Step 1: Remove Parentheses

Start by eliminating the parentheses. Since addition is involved, the signs inside the parentheses remain unchanged:

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

---

### Step 2: Group Like Terms

Now, identify and group like terms:

- Terms with the variable $k$: $-k + 3k$
- Constant terms: $4 - 2$

---

### Step 3: Combine Like Terms

- Combine the $k$-terms:
  $$
  -k + 3k = 2k
  $$

- Combine the constant terms:
  $$
  4 - 2 = 2
  $$

---

### Step 4: Final Simplified Expression

Putting it all together:

$$
2k + 2
$$

---

### Final Answer

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

[Assistant]
Explanation:
Okay, so I need to simplify this expression: (-k + 4) + (-2 + 3k). Hmm, let's see. Alright, first, I remember that simplifying expressions usually involves combining like terms. Like terms are terms that have the same variable part, right? So, variables with the same exponent. In this case, the variables are k and the constants.

Let me start by writing down the expression again to make sure I have it right: (-k + 4) + (-2 + 3k). Okay, so there are two sets of parentheses here. Maybe I should distribute the addition sign? Wait, actually, since it's just addition, I can remove the parentheses without changing the signs, right? Let me try that.

So, if I take away the parentheses, it becomes -k + 4 - 2 + 3k. Let me check that again. The first part is (-k + 4), and then adding (-2 + 3k). So when you add two terms, you just combine them. So yeah, removing the parentheses, the second part is -2 + 3k. So combining all the terms together: -k + 4 - 2 + 3k.

Now, let me identify the like terms. The terms with k are -k and 3k. The constant terms are 4 and -2. So I need to combine those. Let me handle the k terms first.

-k + 3k. Let's see, if I have -1k + 3k, that would be... Hmm, combining coefficients: -1 + 3 is 2, so 2k. Okay, that seems right.

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 check again step by step to make sure I didn't make a mistake.

Original expression: (-k + 4) + (-2 + 3k).

First, remove parentheses: -k + 4 - 2 + 3k.

Combining like terms: (-k + 3k) + (4 - 2).

Calculating each part: (-k + 3k) is 2k, and (4 - 2) is 2.

So yes, that gives 2k + 2.

Wait, but maybe I should write it as 2k + 2 or 2 + 2k? Either way is correct, but usually, we write the terms with variables first, so 2k + 2.

Alternatively, I can factor out a 2, but the question just says to simplify, so combining like terms is sufficient.

Let me verify by plugging in a value for k. Let's choose k = 1.

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

Simplified expression: 2*1 + 2 = 2 + 2 = 4.

Okay, that checks out. Let me try another value, like k = 0.

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

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

Same result. How about k = 2?

Original expression: (-2 + 4) + (-2 + 3*2) = (2) + (-2 + 6) = 2 + 4 = 6.

Simplified expression: 2*2 + 2 = 4 + 2 = 6.

Perfect, that's the same. So my simplified expression seems correct.

Alternatively, maybe I can think of it as combining the terms step by step. Let me try that approach again.

Starting with (-k + 4) + (-2 + 3k).

I can rearrange the terms using the commutative property of addition. That is, I can swap the order of addition. Let me do that to group the like terms.

So, -k + 3k + 4 - 2.

Which is the same as (-k + 3k) + (4 - 2), which we already did. So again, 2k + 2.

Another way: Maybe combine the constants first? Let me see.

Original expression: (-k + 4) + (-2 + 3k).

Let me group the constants and the variables separately.

Constants: 4 and -2.

Variables: -k and 3k.

So, constants: 4 - 2 = 2.

Variables: -k + 3k = 2k.

So altogether, 2k + 2.

Same result.

I think that's solid. Therefore, the simplified form of the expression is 2k + 2.

But wait, let me check if there's another way someone might approach this problem and maybe make a mistake. For example, if someone forgets to distribute the negative sign or something? Wait, in this case, there's no distribution needed because it's just addition.

But if the problem had something like (-k + 4) + (-2) + 3k, maybe. Wait, no, the original problem is (-k + 4) + (-2 + 3k).

So in that case, removing parentheses, it's -k + 4 - 2 + 3k.

Another possible mistake might be combining 4 - 2 as -6? But no, 4 - 2 is 2.

Or maybe combining -k + 3k as -4k? But that would be wrong. Because -k + 3k is like -1k + 3k = 2k.

So, as long as I remember that, it's okay.

Therefore, I think I'm confident that the simplified expression is 2k + 2.

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

To simplify the expression $(-k + 4) + (-2 + 3k)$, we proceed step by step, focusing on combining like terms.

---

### Step 1: Remove Parentheses

Start by eliminating the parentheses. Since addition is involved, the signs inside the parentheses remain unchanged:

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

---

### Step 2: Group Like Terms

Now, identify and group like terms:

- Terms with the variable $k$: $-k + 3k$
- Constant terms: $4 - 2$

---

### Step 3: Combine Like Terms

- Combine the $k$-terms:
  $$
  -k + 3k = 2k
  $$

- Combine the constant terms:
  $$
  4 - 2 = 2
  $$

---

### Step 4: Final Simplified Expression

Putting it all 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 recall the rules for horizontal asymptotes of rational functions.

First, a rational function is a ratio of two polynomials. The horizontal asymptote depends on the degrees of the numerator and the denominator. Let me check the degrees here. The numerator is 4x³ + 2x - 4, which is a cubic polynomial, so its degree is 3. The denominator is 3x³ - 2x² + 5x - 1, which is also a cubic polynomial, so its degree is 3 as well.

I think the rule is that if the degrees are the same, the horizontal asymptote is the ratio of the leading coefficients. Let me confirm that. If the numerator has degree n and the denominator has degree m, then:

- If n < m, the horizontal asymptote is y = 0.
- If n = m, the horizontal asymptote is y = (leading coefficient of numerator)/(leading coefficient of denominator).
- If n > m, there is no horizontal asymptote (unless it's a slant asymptote, but that's for when n is exactly one more than m).

So in this case, since both numerator and denominator are degree 3, I can apply the second rule. The leading coefficient of the numerator is 4 (the coefficient of x³), and the leading coefficient of the denominator is 3 (also the coefficient of x³). Therefore, the horizontal asymptote should be y = 4/3.

Wait, but let me make sure I didn't skip any steps. 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)/x³ + (2x)/x³ - 4/x³ = 4 + 2/x² - 4/x³

Denominator becomes: (3x³)/x³ - (2x²)/x³ + (5x)/x³ - 1/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. So the numerator approaches 4 + 0 - 0 = 4, and the denominator approaches 3 - 0 + 0 - 0 = 3. Therefore, the whole expression approaches 4/3. Therefore, the horizontal asymptote is y = 4/3.

But just to be thorough, maybe I should check what happens as x approaches negative infinity as well? Wait, since we're dealing with even powers in the denominator and numerator? Wait, no. Wait, when x approaches negative infinity, the terms like 2/x² and 5/x² are still positive, but the terms like -2/x would be positive if x is negative? Wait, let me think.

Actually, when x is negative and we take 1/x², that's positive, same with 1/x³ would be negative. But as x approaches negative infinity, 1/x² approaches zero, same as positive infinity. Similarly, 1/x³ approaches zero. Therefore, regardless of whether x approaches positive or negative infinity, the numerator and denominator both approach 4 and 3 respectively. Therefore, the horizontal asymptote is y = 4/3 in both cases.

So, there is a horizontal asymptote at y = 4/3. Therefore, the answer should be 4/3. Wait, but let me check if maybe there's a mistake in the leading coefficients. Let me write them again.

Numerator leading term is 4x³, denominator leading term is 3x³. So, the ratio is 4/3. Yes, that seems right.

Alternatively, maybe I can plug in a very large value of x, say x = 1000, and approximate the value. Let's try that.

Numerator: 4*(1000)^3 + 2*(1000) - 4 ≈ 4*1,000,000,000 + 2000 -4 ≈ 4,000,000,000 + 1996 ≈ 4,000,001,996

Denominator: 3*(1000)^3 - 2*(1000)^2 + 5*(1000) -1 ≈ 3*1,000,000,000 - 2*1,000,000 + 5000 -1 ≈ 3,000,000,000 - 2,000,000 + 4999 ≈ 2,998,004,999

Then divide numerator by denominator: approx 4,000,001,996 / 2,998,004,999. Let me divide numerator and denominator by 1,000,000,000 to simplify:

≈ 4.000001996 / 2.998004999 ≈ approx 4 / 3 ≈ 1.333...

Yes, so that's about 1.333, which is 4/3. So that checks out. Therefore, the horizontal asymptote is indeed y = 4/3.

Wait, but just to make sure there isn't some other horizontal asymptote or something? For example, sometimes functions can have different horizontal asymptotes as x approaches positive and negative infinity. But in this case, since both the numerator and denominator are polynomials of the same degree, the asymptote is the same for both infinities.

So, the horizontal asymptote is y = 4/3. Therefore, the answer is y = 4/3.

Alternatively, let me check if there's any possibility that the function might have a horizontal asymptote at another value. But according to the rules, if degrees are equal, it's only the ratio of leading coefficients. Let me check another resource... Well, since I can't actually look it up, but I remember that the horizontal asymptote is determined by the leading terms because as x becomes very large, the lower degree terms become negligible. So in this case, the leading terms are 4x³ and 3x³, so their ratio is 4/3. Therefore, the horizontal asymptote is y = 4/3.

Another way to think about it: if we factor out x³ from numerator and denominator, we get:

Numerator: x³(4 + 2/x² - 4/x³)

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

So then, the x³ cancels out, leaving (4 + 2/x² - 4/x³)/(3 - 2/x + 5/x² - 1/x³). As x approaches infinity, the terms with 1/x² and 1/x³ go to zero, so we get 4/3. Therefore, same result.

Therefore, I think I can confidently say that the horizontal asymptote is y = 4/3.

Just to prevent any possible mistake, let me recall that if the degrees were different, for example, if the numerator had a higher degree, then there would be no horizontal asymptote. If the numerator had lower degree, then horizontal asymptote at zero. But here, same degree, so ratio of leading coefficients. So yes, 4/3 is correct.

Therefore, the answer is y = 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 $, since a **horizontal asymptote** describes the limiting value of the function as $ x $ becomes very large in magnitude.

---

### Step 1: Identify the Degrees and Leading Coefficients

The numerator is a polynomial of degree 3:
$$
4x^3 + 2x - 4 \quad \Rightarrow \text{Leading term: } 4x^3
$$

The denominator is also a polynomial of degree 3:
$$
3x^3 - 2x^2 + 5x - 1 \quad \Rightarrow \text{Leading term: } 3x^3
$$

Since **both the numerator and the denominator have the same degree** (equal to 3), we apply the rule for horizontal asymptotes of rational functions:

> **If the degree of the numerator equals the degree of the denominator**,
> the horizontal asymptote is the ratio of the leading coefficients.

---

### Step 2: Apply the Rule

The leading coefficient of the numerator is 4, and the leading coefficient of the denominator is 3. Therefore, the horizontal asymptote is:

$$
y = \frac{4}{3}
$$

---

### Step 3: Confirmation by Dividing by $ x^3 $

To further confirm, divide both the numerator and denominator by $ x^3 $:

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

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

$$
\frac{4}{3}
$$

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

---

### 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 recall the rules for horizontal asymptotes of rational functions.

First, a rational function is a ratio of two polynomials. The horizontal asymptote depends on the degrees of the numerator and the denominator. Let me check the degrees here. The numerator is 4x³ + 2x - 4, which is a cubic polynomial, so its degree is 3. The denominator is 3x³ - 2x² + 5x - 1, which is also a cubic polynomial, so its degree is 3 as well.

I think the rule is that if the degrees are the same, the horizontal asymptote is the ratio of the leading coefficients. Let me confirm that. If the numerator has degree n and the denominator has degree m, then:

- If n < m, the horizontal asymptote is y = 0.
- If n = m, the horizontal asymptote is y = (leading coefficient of numerator)/(leading coefficient of denominator).
- If n > m, there is no horizontal asymptote (unless it's a slant asymptote, but that's for when n is exactly one more than m).

So in this case, since both numerator and denominator are degree 3, I can apply the second rule. The leading coefficient of the numerator is 4 (the coefficient of x³), and the leading coefficient of the denominator is 3 (also the coefficient of x³). Therefore, the horizontal asymptote should be y = 4/3.

Wait, but let me make sure I didn't skip any steps. 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)/x³ + (2x)/x³ - 4/x³ = 4 + 2/x² - 4/x³

Denominator becomes: (3x³)/x³ - (2x²)/x³ + (5x)/x³ - 1/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. So the numerator approaches 4 + 0 - 0 = 4, and the denominator approaches 3 - 0 + 0 - 0 = 3. Therefore, the whole expression approaches 4/3. Therefore, the horizontal asymptote is y = 4/3.

But just to be thorough, maybe I should check what happens as x approaches negative infinity as well? Wait, since we're dealing with even powers in the denominator and numerator? Wait, no. Wait, when x approaches negative infinity, the terms like 2/x² and 5/x² are still positive, but the terms like -2/x would be positive if x is negative? Wait, let me think.

Actually, when x is negative and we take 1/x², that's positive, same with 1/x³ would be negative. But as x approaches negative infinity, 1/x² approaches zero, same as positive infinity. Similarly, 1/x³ approaches zero. Therefore, regardless of whether x approaches positive or negative infinity, the numerator and denominator both approach 4 and 3 respectively. Therefore, the horizontal asymptote is y = 4/3 in both cases.

So, there is a horizontal asymptote at y = 4/3. Therefore, the answer should be 4/3. Wait, but let me check if maybe there's a mistake in the leading coefficients. Let me write them again.

Numerator leading term is 4x³, denominator leading term is 3x³. So, the ratio is 4/3. Yes, that seems right.

Alternatively, maybe I can plug in a very large value of x, say x = 1000, and approximate the value. Let's try that.

Numerator: 4*(1000)^3 + 2*(1000) - 4 ≈ 4*1,000,000,000 + 2000 -4 ≈ 4,000,000,000 + 1996 ≈ 4,000,001,996

Denominator: 3*(1000)^3 - 2*(1000)^2 + 5*(1000) -1 ≈ 3*1,000,000,000 - 2*1,000,000 + 5000 -1 ≈ 3,000,000,000 - 2,000,000 + 4999 ≈ 2,998,004,999

Then divide numerator by denominator: approx 4,000,001,996 / 2,998,004,999. Let me divide numerator and denominator by 1,000,000,000 to simplify:

≈ 4.000001996 / 2.998004999 ≈ approx 4 / 3 ≈ 1.333...

Yes, so that's about 1.333, which is 4/3. So that checks out. Therefore, the horizontal asymptote is indeed y = 4/3.

Wait, but just to make sure there isn't some other horizontal asymptote or something? For example, sometimes functions can have different horizontal asymptotes as x approaches positive and negative infinity. But in this case, since both the numerator and denominator are polynomials of the same degree, the asymptote is the same for both infinities.

So, the horizontal asymptote is y = 4/3. Therefore, the answer is y = 4/3.

Alternatively, let me check if there's any possibility that the function might have a horizontal asymptote at another value. But according to the rules, if degrees are equal, it's only the ratio of leading coefficients. Let me check another resource... Well, since I can't actually look it up, but I remember that the horizontal asymptote is determined by the leading terms because as x becomes very large, the lower degree terms become negligible. So in this case, the leading terms are 4x³ and 3x³, so their ratio is 4/3. Therefore, the horizontal asymptote is y = 4/3.

Another way to think about it: if we factor out x³ from numerator and denominator, we get:

Numerator: x³(4 + 2/x² - 4/x³)

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

So then, the x³ cancels out, leaving (4 + 2/x² - 4/x³)/(3 - 2/x + 5/x² - 1/x³). As x approaches infinity, the terms with 1/x² and 1/x³ go to zero, so we get 4/3. Therefore, same result.

Therefore, I think I can confidently say that the horizontal asymptote is y = 4/3.

Just to prevent any possible mistake, let me recall that if the degrees were different, for example, if the numerator had a higher degree, then there would be no horizontal asymptote. If the numerator had lower degree, then horizontal asymptote at zero. But here, same degree, so ratio of leading coefficients. So yes, 4/3 is correct.

Therefore, the answer is y = 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 $, since a **horizontal asymptote** describes the limiting value of the function as $ x $ becomes very large in magnitude.

---

### Step 1: Identify the Degrees and Leading Coefficients

The numerator is a polynomial of degree 3:
$$
4x^3 + 2x - 4 \quad \Rightarrow \text{Leading term: } 4x^3
$$

The denominator is also a polynomial of degree 3:
$$
3x^3 - 2x^2 + 5x - 1 \quad \Rightarrow \text{Leading term: } 3x^3
$$

Since **both the numerator and the denominator have the same degree** (equal to 3), we apply the rule for horizontal asymptotes of rational functions:

> **If the degree of the numerator equals the degree of the denominator**,
> the horizontal asymptote is the ratio of the leading coefficients.

---

### Step 2: Apply the Rule

The leading coefficient of the numerator is 4, and the leading coefficient of the denominator is 3. Therefore, the horizontal asymptote is:

$$
y = \frac{4}{3}
$$

---

### Step 3: Confirmation by Dividing by $ x^3 $

To further confirm, divide both the numerator and denominator by $ x^3 $:

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

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

$$
\frac{4}{3}
$$

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

---

### 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.

First, the set of prime numbers less than 30. Prime numbers are numbers greater than 1 that have no positive divisors other than 1 and themselves. Let me list them out one by one. Starting from the smallest prime, which is 2. Then 3, 5, 7, 11, 13, 17, 19, 23, 29. Let me check if these are all less than 30. 29 is the largest prime less than 30, right? Let me confirm by checking numbers up to 30.

After 29, the next prime would be 31, which is over 30, so yes, primes less than 30 are up to 29. Let me count them. Let me write them down:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29. That's 10 numbers. Wait, let me count again to make sure. 2 is first, then 3 (2), 5 (3), 7 (4), 11 (5), 13 (6), 17 (7), 19 (8), 23 (9), 29 (10). Yep, 10 primes. Okay.

Now, the second set is the set of all odd numbers greater than zero. Odd numbers greater than zero would be 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, and so on. But since we're looking for the intersection with the primes less than 30, we don't need to go beyond 29. So, the odd numbers greater than zero less than 30 would be all odd numbers from 1 up to 29.

So now, the intersection of these two sets would be all numbers that are both prime numbers less than 30 and odd numbers greater than zero. So, essentially, we need to find which of the prime numbers less than 30 are also odd numbers greater than zero.

But wait, hold on. All primes except 2 are odd. Because 2 is the only even prime number. So, if we take the set of primes less than 30, which includes 2 and all the other primes, and then intersect it with the set of odd numbers greater than zero, we would exclude 2 because it's even, right? So, the intersection would be all the primes less than 30 except 2.

But let me verify this. Let me think. The set of primes less than 30 is {2, 3, 5, 7, 11, 13, 17, 19, 23, 29}. The set of odd numbers greater than zero less than 30 is {1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29}.

So, the intersection would be numbers that are in both sets. So, let's take the primes less than 30 and check which ones are in the odd numbers greater than zero.

Starting with 2: is 2 in the odd numbers greater than zero? No, because 2 is even. So, 2 is excluded.

Next prime is 3: 3 is odd, so yes. Then 5: odd, yes. 7: odd. 11: odd. 13: odd. 17: odd. 19: odd. 23: odd. 29: odd. So all primes except 2 are odd. Therefore, the intersection is all primes less than 30 except 2.

Therefore, how many elements are in that intersection? Well, originally there were 10 primes less than 30. Subtract 1 (for the number 2) gives 9 elements.

But wait, let me check if 1 is included in the set of odd numbers greater than zero. The problem says "the set of all the odd numbers greater than zero". So, zero is not included, but numbers greater than zero. So, 1 is included. However, 1 is not a prime number. Therefore, even though 1 is in the second set, it's not in the first set. Therefore, the intersection is primes less than 30 that are odd, which as we saw are all primes except 2.

Therefore, the count is 10 primes minus 1 (for 2) equals 9.

But let me confirm by listing the intersection. Let's take the primes less than 30: {2, 3, 5, 7, 11, 13, 17, 19, 23, 29}. Now, intersect with the odd numbers greater than zero. So, remove 2, and check if the rest are odd. Yes, all the rest are odd. Therefore, the intersection is {3, 5, 7, 11, 13, 17, 19, 23, 29}. That's 9 elements.

Wait, let me count them again: 3, 5, 7, 11, 13, 17, 19, 23, 29. That's 9 numbers. So, 9 elements. Therefore, the answer should be 9.

But just to be thorough, let me check if I missed any primes or included any non-primes. For example, is 1 in the intersection? 1 is odd and greater than zero, but it's not prime, so it's not in the first set. So, not included. Similarly, 9, 15, 21, 25, 27: these are odd numbers greater than zero, but they are not primes. So, they are not in the first set. Therefore, the intersection is only the primes that are odd, which are all primes except 2.

So, the count is 10 primes less than 30 minus 1 (which is 2) equals 9.

Alternatively, maybe I should list the primes and count them. Let me list the primes less than 30 again:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29. Yes, that's 10 numbers. Now, removing 2, which is the only even prime, gives 9 numbers. Therefore, the intersection has 9 elements.

Therefore, the answer is 9. But let me check once more to be 100% sure.

Another way to think about it: the set of odd numbers greater than zero includes all odd numbers from 1 upwards, so 1,3,5,... Now, the primes less than 30 are 2,3,5,7,11,13,17,19,23,29. The intersection would be numbers that are both primes and odd. Since 2 is the only even prime, all the other primes are odd. Therefore, the intersection is {3,5,7,11,13,17,19,23,29}, which has 9 elements.

Yes, that seems correct. Therefore, I think the answer is 9.

Wait, but let me check if 1 is included in the odd numbers greater than zero. The problem says "the set of all the odd numbers greater than zero." So, zero is excluded, and all positive odd integers. So, 1 is included. However, as mentioned, 1 is not prime, so it's not in the prime set. Therefore, no issue there.

Another check: are there any other numbers in the primes less than 30 that are even? Well, 2 is the only even prime. So, no others. Therefore, yes, removing 2 is sufficient.

Alternatively, if I consider the intersection, it's the primes less than 30 that are odd. Since all primes except 2 are odd, the count is 10-1=9.

Alternatively, if I wanted to list the numbers in the intersection, I can do that:

Starting from 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. So, 9 elements.

Therefore, the answer is 9. I think that's correct.

But just to make sure, let me check if I considered all primes less than 30. Let me list them again:

Primes less than 30:

2 is prime.

3 is prime.

4 is not prime.

5 is prime.

6 not prime.

7 prime.

8 not.

9 not.

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.

26 not.

27 not.

28 not.

29 prime.

30 not.

So, primes are 2,3,5,7,11,13,17,19,23,29. Yep, 10 primes. So, 10 primes, subtract 1 for the even prime, gives 9.

Therefore, the intersection has 9 elements.

I think I've checked this multiple ways. So, I feel confident that the answer is 9.

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

To solve the problem, we are asked to find how many elements are in the **intersection** of two sets:

1. The set of **all prime numbers less than 30**
2. The set of **all odd numbers greater than zero**

---

### Step 1: Understand the First Set

**Prime numbers less than 30** are the numbers that have exactly two distinct positive divisors: 1 and themselves. Listing them:

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

There are **10** such primes.

---

### Step 2: Understand the Second Set

The set of **odd numbers greater than zero** includes all integers that are **not divisible by 2** and are **greater than zero**. It is not bounded, but since we're intersecting with the first set (which is less than 30), we consider odd integers from **1 through 29**:

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

This includes **15** elements, all less than 30.

---

### Step 3: Find the Intersection

We now identify which primes from the first set are also found in the second set.

From the list:

- **2** is the **only even prime**, so it is not in the second set (odd numbers).
- All other primes from the first set are odd:

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

Let’s count them:

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

So, the intersection has **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.

First, the set of prime numbers less than 30. Prime numbers are numbers greater than 1 that have no positive divisors other than 1 and themselves. Let me list them out one by one. Starting from the smallest prime, which is 2. Then 3, 5, 7, 11, 13, 17, 19, 23, 29. Let me check if these are all less than 30. 29 is the largest prime less than 30, right? Let me confirm by checking numbers up to 30.

After 29, the next prime would be 31, which is over 30, so yes, primes less than 30 are up to 29. Let me count them. Let me write them down:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29. That's 10 numbers. Wait, let me count again to make sure. 2 is first, then 3 (2), 5 (3), 7 (4), 11 (5), 13 (6), 17 (7), 19 (8), 23 (9), 29 (10). Yep, 10 primes. Okay.

Now, the second set is the set of all odd numbers greater than zero. Odd numbers greater than zero would be 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, and so on. But since we're looking for the intersection with the primes less than 30, we don't need to go beyond 29. So, the odd numbers greater than zero less than 30 would be all odd numbers from 1 up to 29.

So now, the intersection of these two sets would be all numbers that are both prime numbers less than 30 and odd numbers greater than zero. So, essentially, we need to find which of the prime numbers less than 30 are also odd numbers greater than zero.

But wait, hold on. All primes except 2 are odd. Because 2 is the only even prime number. So, if we take the set of primes less than 30, which includes 2 and all the other primes, and then intersect it with the set of odd numbers greater than zero, we would exclude 2 because it's even, right? So, the intersection would be all the primes less than 30 except 2.

But let me verify this. Let me think. The set of primes less than 30 is {2, 3, 5, 7, 11, 13, 17, 19, 23, 29}. The set of odd numbers greater than zero less than 30 is {1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29}.

So, the intersection would be numbers that are in both sets. So, let's take the primes less than 30 and check which ones are in the odd numbers greater than zero.

Starting with 2: is 2 in the odd numbers greater than zero? No, because 2 is even. So, 2 is excluded.

Next prime is 3: 3 is odd, so yes. Then 5: odd, yes. 7: odd. 11: odd. 13: odd. 17: odd. 19: odd. 23: odd. 29: odd. So all primes except 2 are odd. Therefore, the intersection is all primes less than 30 except 2.

Therefore, how many elements are in that intersection? Well, originally there were 10 primes less than 30. Subtract 1 (for the number 2) gives 9 elements.

But wait, let me check if 1 is included in the set of odd numbers greater than zero. The problem says "the set of all the odd numbers greater than zero". So, zero is not included, but numbers greater than zero. So, 1 is included. However, 1 is not a prime number. Therefore, even though 1 is in the second set, it's not in the first set. Therefore, the intersection is primes less than 30 that are odd, which as we saw are all primes except 2.

Therefore, the count is 10 primes minus 1 (for 2) equals 9.

But let me confirm by listing the intersection. Let's take the primes less than 30: {2, 3, 5, 7, 11, 13, 17, 19, 23, 29}. Now, intersect with the odd numbers greater than zero. So, remove 2, and check if the rest are odd. Yes, all the rest are odd. Therefore, the intersection is {3, 5, 7, 11, 13, 17, 19, 23, 29}. That's 9 elements.

Wait, let me count them again: 3, 5, 7, 11, 13, 17, 19, 23, 29. That's 9 numbers. So, 9 elements. Therefore, the answer should be 9.

But just to be thorough, let me check if I missed any primes or included any non-primes. For example, is 1 in the intersection? 1 is odd and greater than zero, but it's not prime, so it's not in the first set. So, not included. Similarly, 9, 15, 21, 25, 27: these are odd numbers greater than zero, but they are not primes. So, they are not in the first set. Therefore, the intersection is only the primes that are odd, which are all primes except 2.

So, the count is 10 primes less than 30 minus 1 (which is 2) equals 9.

Alternatively, maybe I should list the primes and count them. Let me list the primes less than 30 again:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29. Yes, that's 10 numbers. Now, removing 2, which is the only even prime, gives 9 numbers. Therefore, the intersection has 9 elements.

Therefore, the answer is 9. But let me check once more to be 100% sure.

Another way to think about it: the set of odd numbers greater than zero includes all odd numbers from 1 upwards, so 1,3,5,... Now, the primes less than 30 are 2,3,5,7,11,13,17,19,23,29. The intersection would be numbers that are both primes and odd. Since 2 is the only even prime, all the other primes are odd. Therefore, the intersection is {3,5,7,11,13,17,19,23,29}, which has 9 elements.

Yes, that seems correct. Therefore, I think the answer is 9.

Wait, but let me check if 1 is included in the odd numbers greater than zero. The problem says "the set of all the odd numbers greater than zero." So, zero is excluded, and all positive odd integers. So, 1 is included. However, as mentioned, 1 is not prime, so it's not in the prime set. Therefore, no issue there.

Another check: are there any other numbers in the primes less than 30 that are even? Well, 2 is the only even prime. So, no others. Therefore, yes, removing 2 is sufficient.

Alternatively, if I consider the intersection, it's the primes less than 30 that are odd. Since all primes except 2 are odd, the count is 10-1=9.

Alternatively, if I wanted to list the numbers in the intersection, I can do that:

Starting from 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. So, 9 elements.

Therefore, the answer is 9. I think that's correct.

But just to make sure, let me check if I considered all primes less than 30. Let me list them again:

Primes less than 30:

2 is prime.

3 is prime.

4 is not prime.

5 is prime.

6 not prime.

7 prime.

8 not.

9 not.

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.

26 not.

27 not.

28 not.

29 prime.

30 not.

So, primes are 2,3,5,7,11,13,17,19,23,29. Yep, 10 primes. So, 10 primes, subtract 1 for the even prime, gives 9.

Therefore, the intersection has 9 elements.

I think I've checked this multiple ways. So, I feel confident that the answer is 9.

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

To solve the problem, we are asked to find how many elements are in the **intersection** of two sets:

1. The set of **all prime numbers less than 30**
2. The set of **all odd numbers greater than zero**

---

### Step 1: Understand the First Set

**Prime numbers less than 30** are the numbers that have exactly two distinct positive divisors: 1 and themselves. Listing them:

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

There are **10** such primes.

---

### Step 2: Understand the Second Set

The set of **odd numbers greater than zero** includes all integers that are **not divisible by 2** and are **greater than zero**. It is not bounded, but since we're intersecting with the first set (which is less than 30), we consider odd integers from **1 through 29**:

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

This includes **15** elements, all less than 30.

---

### Step 3: Find the Intersection

We now identify which primes from the first set are also found in the second set.

From the list:

- **2** is the **only even prime**, so it is not in the second set (odd numbers).
- All other primes from the first set are odd:

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

Let’s count them:

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

So, the intersection has **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.