I have this equation to solve: $3^x=9$. I need to find the value of x. First, I know that 9 is a power of 3. Specifically, 3 squared is 9, because 3 times 3 is 9. So, if $3^x=9$ and $3^2=9$, then x should be 2. That seems straightforward.

But let me double-check. Is there any other way to solve this? Maybe using logarithms. If I take the logarithm of both sides, I can write $\log (3^x)=\log (9)$. Using the logarithm power rule, that becomes $x\log 3=\log 9$. Then, dividing both sides by log 3, I get $x=\frac{\log 9}{\log 3}$.

I know that log 9 can be written as log(3^2), which is 2 log 3. So, substituting that in, $x=\frac{2\log 3}{\log 3}$. The log 3 terms cancel out, leaving me with x = 2. So, that confirms it.

Since the bases are the same, I can set the exponents equal to each other. So, if $3^x=3^2$, then x must be 2. That's a basic rule of exponents.

I think that's solid. So, the solution is x = 2.