Solve $\log_2(3^x-1)=\log_3(2^x+1)$ 
Solve the following equation over the real number(preferably without calculus):
  $$\log_2(3^x-1)=\log_3(2^x+1).$$

 This problem is from a math contest held where I learn; I was unable to do much at all tinkering with it; I have observed the solution $x=1$ but haven't been able to prove there are no others or determine them if there are.

We'll use the following: $\log_a(b) = \frac{\log_c(b)}{\log_c(a)}$
So: $ \log_2(3^x - 1) = \log_3(3^x-1)  \iff \frac{\ln(3^x-1)}{\ln(2)}  = \frac{\ln(2^x+1)}{\ln(3)}$
That is:
$ \ln(3)\ln(3^x-1) = \ln(2)\ln(2^x+1) $
So we're left with a function $f:(0,+\infty) -> \Bbb R$ , $f(x) = \ln(3)\ln(3^x-1) -\ln(2)\ln(2^x+1) $
Looking at its derivative: 
$f'(x) = \ln(3) \frac{3^x\ln(3)}{3^x-1} - \ln(2) \frac{2^x\ln(2)}{2^x+1} = \frac{6^x(\ln^2(3) - \ln^2(2)) + 3^x\ln^2(3) - 2^x\ln^2(2)}{(3^x-1)(2^x+1)} $
We see, that the sign of it depends on the sign of the numerator.
Let $g(x) = 6^x(\ln^2(3) - \ln^2(2)) + 3^x\ln^2(3) - 2^x\ln^2(2) $
Which is clearly positive ( cause $3^x\ln^2(3) > 2^x\ln^2(2) $ and $6^x(\ln^2(3) - \ln^2(2)) > 0 $ )
So our function $f$ is increasing, and that means (because $\lim_{x \to 0^+} f(x) = -\infty$, $\lim_{x \to +\infty} f(x) = +\infty$),
that our function has only one root. However, it is a little bit of a guess to tell it's $x=1$.
EDIT:
In fact, it isn't that hard to find that solution. We have an equation involving $\ln$.
Let $a(x) = 3^x-1$, $b(x) = 2^x+1$
Then, we arrive with: $\ln(3)\ln(a(x)) = \ln(2)\ln(b(x))$
Which clearly has a solution when $a(x) = 2$ and $b(x) = 3$, and fortunately $3^x = 3$ and $2^x=2$ have a solution $x=1$
