Okay, so I need to solve the equation 3^{2x} + 19 = 10^x. Hmm, let me think about how to approach this. First, I remember that equations involving exponents with different bases can sometimes be tricky. Maybe I can rewrite 3^{2x} as (3^x)^2, right? Because 3^{2x} is the same as (3^x)^2. So the equation becomes (3^x)^2 + 19 = 10^x. Hmm, that might not immediately help, but perhaps I can set a substitution here. Let me try letting y = 3^x. Then the equation becomes y^2 + 19 = 10^x. But wait, 10^x is another exponential term with base 10. Since y is 3^x, maybe I can express 10^x in terms of y? Let me see. Since 10^x = (10/3)^x * 3^x = (10/3)^x * y. But that might complicate things more. Alternatively, maybe I can take logarithms on both sides? Let's try that. If I take log of both sides, but the equation is y^2 + 19 = 10^x. Taking log would lead to log(y^2 + 19) = x log(10). But log(y^2 + 19) is not straightforward. Maybe that's not the best approach.

Alternatively, maybe I can consider this equation as 3^{2x} = 10^x - 19. Then, taking natural log on both sides: ln(3^{2x}) = ln(10^x - 19). Which simplifies to 2x ln(3) = ln(10^x - 19). But this still leaves me with x on both sides, and inside a logarithm. It's not easy to solve algebraically. Maybe I can use numerical methods or graphing?

Wait, maybe I can check for integer solutions first. Let's try plugging in some integer values for x.

For x=0: 3^{0} + 19 = 1 + 19 = 20. 10^0 = 1. So 20 ≠ 1. Not a solution.

x=1: 3^{2} + 19 = 9 + 19 = 28. 10^1 = 10. Not equal.

x=2: 3^{4} + 19 = 81 + 19 = 100. 10^2 = 100. Oh! They are equal here. So x=2 is a solution.

x=3: 3^{6} + 19 = 729 + 19 = 748. 10^3 = 1000. Not equal.

x=4: 3^{8} + 19 = 6561 + 19 = 6580. 10^4 = 10000. Not equal.

What about negative x? Let's check x=-1: 3^{-2} +19 = 1/9 +19 ≈ 19.111. 10^{-1}=0.1. Not equal.

x=-2: 3^{-4} +19 ≈ 0.0123 +19 ≈19.0123. 10^{-2}=0.01. Not equal.

So x=2 is a solution. But wait, is there another solution? Let's see. The function 3^{2x} +19 is exponential growth, and 10^x is also exponential growth. But 10^x grows faster than 3^{2x} since 10 > 9. So after a certain point, 10^x will overtake 3^{2x} +19. But maybe they intersect again somewhere else?

Wait, let's analyze the behavior. For x approaching negative infinity, 3^{2x} approaches 0, so 3^{2x} +19 approaches 19. 10^x approaches 0. So near negative infinity, 19 vs 0. So 3^{2x} +19 is above 10^x there. At x=2, they are equal. For x>2, 10^x grows faster, so 10^x will surpass 3^{2x} +19. So maybe there's another solution where 3^{2x} +19 equals 10^x for some x between negative infinity and 2. Wait, but when x is negative, 10^x is positive but decreasing, and 3^{2x} is also positive and decreasing. Let's check x=1: 3^{2} +19=28 vs 10=10. So 28>10. At x=2, both are 100. For x between 1 and 2, 10^x increases from 10 to 100, while 3^{2x} +19 increases from 28 to 100. So they meet at x=2. But what about between x=0 and x=2? At x=0, 3^{0}+19=20 vs 1. At x=1, 28 vs10. So 3^{2x} +19 is always above 10^x from x=0 to x=2. Then at x=2, they meet. Then for x>2, 10^x overtakes. For x<0, 3^{2x}+19 is 19 plus a small positive number, while 10^x is a small positive number. So maybe there's another solution for x <0 where 10^x is close to 19. Let's check x= -0. Let's try x= -0. Let's see x= -1: 3^{-2} +19 ≈19.111 vs 10^{-1}=0.1. So 19.111>0.1. x=-2: 3^{-4}+19≈0.0123+19≈19.0123 vs 10^{-2}=0.01. Still 19.0123>0.01. x=-3: 3^{-6}+19≈0.00137+19≈19.00137 vs 10^{-3}=0.001. Still 19.00137>0.001. So even as x approaches negative infinity, 3^{2x} +19 approaches 19, which is greater than 10^x approaching 0. So maybe there's no solution for x<0. Let's check x= log_10(19)/2. Wait, is that possible? Wait, maybe set 3^{2x} +19 =10^x. Let's take natural log on both sides. Wait, but earlier that led to 2x ln3 = ln(10^x -19). Not helpful. Maybe try to approximate the solution. Let's suppose that x is between 2 and 3. Wait, but at x=2, both sides are 100. At x=3, left side is 3^6 +19=729+19=748, right side is 1000. So between x=2 and x=3, 10^x increases from 100 to 1000, while the left side increases from 100 to 748. So 10^x is growing faster. So maybe no solution between 2 and 3. Wait, but at x=2, both are 100. Then for x>2, 10^x quickly overtakes. So maybe x=2 is the only solution. But wait, let's check if there's another solution between x=2 and x=3. At x=2.5: 3^{5} +19=243+19=262. 10^{2.5}=sqrt(10^5)=sqrt(100000)=approximately 316.23. So 262 <316.23. So left side is less than right side. So between x=2 and x=2.5, left side increases from 100 to 262, right side increases from 100 to 316.23. So there's a crossing point between x=2 and x=2.5. Wait, but at x=2, both are 100. At x=2.5, left is 262, right is ~316. So actually, the left side is increasing at a slower rate than the right side. Wait, but left side is 3^{2x} +19. The derivative of left side is 2*ln3*3^{2x}. The derivative of right side is ln10*10^x. Let's compute the derivatives at x=2: left derivative is 2*ln3*9 ≈2*1.0986*9≈19.7748. Right derivative is ln10*100≈2.3026*100≈230.26. So at x=2, right side is increasing much faster. So after x=2, 10^x is increasing faster than the left side. So maybe the left side never catches up again. So the only solution is x=2. Wait, but let's check again. At x=2, both are 100. At x=2.5, left is 262, right is 316.23. Then at x=3, left is 748, right is 1000. So 10^x is always above 3^{2x} +19 for x>2. So maybe x=2 is the only solution. But let's check if there's another solution between x=0 and x=2. Wait, at x=0, left is 20, right is 1. At x=1, left is 28, right is 10. At x=2, left is 100, right is 100. So left side is always above right side in [0,2). At x=2, they meet. Then for x>2, right side is above. For x<0, left side is 19 + something, right side is less than 1. So maybe x=2 is the only solution. But wait, let's check x=2. Let me verify again. 3^{2*2} +19=3^4 +19=81+19=100. 10^2=100. Correct. So x=2 is definitely a solution. Is there another solution? Let's check x= log_10(3^{2x} +19). Wait, but that's circular. Alternatively, maybe there's a solution where 3^{2x} +19 =10^x, but x is negative. Let's see. For x negative, 3^{2x} is 1/3^{-2x}, so very small. 10^x is 1/10^{-x}, which is also very small. So 3^{2x} +19 is approximately 19, while 10^x is small. So 19 is way larger than 10^x for negative x. So they can't be equal. Therefore, the only solution is x=2.

But wait, let me check if there's another solution between x=2 and x=3. Wait, at x=2, both are 100. At x=2.5, left is 262, right is ~316. So right is larger. Then, at x=3, left is 748, right is 1000. So right is still larger. So after x=2, right side is larger. So maybe x=2 is the only solution. But let's check if there's a solution between x=2 and x=3. Wait, but since at x=2, both are 100, and right side is increasing faster, so after that, right side is always larger. So maybe x=2 is the only solution.

Alternatively, maybe there's a solution between x=1 and x=2. Let's check x=1.5: 3^{3} +19=27+19=46. 10^{1.5}=sqrt(10^3)=sqrt(1000)=31.62. So left side is 46, right is ~31.62. So left is larger. So between x=1 and x=2, left is larger. At x=2, both are 100. So maybe x=2 is the only point where they meet. Then, after x=2, right side takes off. So yes, x=2 is the only solution. Therefore, the answer is x=2.

Wait, but let me verify this with a graph. If I plot y=3^{2x} +19 and y=10^x, they intersect at x=2. For x>2, 10^x is above. For x<2, 3^{2x}+19 is above. So x=2 is the only solution. Therefore, the solution is x=2.

**Final Answer**
The solution is \boxed{2}.
Therefore, the solution to the equation is $\boxed{2}$.
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