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.
 A: If $t = \log_2(3^x-1) = \log_3(2^x+1)$, we have $2^t = 3^x - 1$ and $3^t = 2^x + 1$.
Thus $3^t + 2^t = 3^x + 2^x$.  It's easy to see that $3^x + 2^x$ is an increasing function of $x$, therefore we must have $t=x$. 
Now with $t=x$ the equation becomes $3^x - 2^x = 1$. Dividing by $2^x$, write it as
$(3/2)^x - 1 = (1/2)^x$.  Now the left side is an increasing function of $x$, while the right side is a decreasing function of $x$, so there can be only one $x$ where they are equal.  By inspection, that $x$ is $1$.
A: 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 $x=1$
A: The function $f: \mathbb{R}\to (0, \infty)$,
$$f(x)=\log_3(2^x+1)$$ 
is bijective and the inverse of this function is $f^{-1}:(0,\infty)\to \mathbb{R}$, 
$$f^{-1}(x)=\log_2(3^x-1)$$
The equation can be written as:
$$f(x)=f^{-1}(x)\Leftrightarrow f(x)=x\Leftrightarrow \log_3(2^x+1) = 3^x$$
or
$$\left(\frac{2}{3}\right)^x+\left(\frac{1}{3}\right)^x = 1$$
The left hand side is decreasing, so we can have only one solution and that's $x=1$.
