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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.

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    $\begingroup$ I think you will need a numerical method to solve this equation $\endgroup$ – Dr. Sonnhard Graubner Feb 22 at 16:28
  • $\begingroup$ that would be outrageous for this contest, there must be an at least somewhat elegant way of solving, or at the very least a non-numerical one $\endgroup$ – Luca Pana Feb 22 at 16:29
  • $\begingroup$ Or you have made a typo! $\endgroup$ – Dr. Sonnhard Graubner Feb 22 at 16:30
  • $\begingroup$ no sir , this is the exact problem given:) $\endgroup$ – Luca Pana Feb 22 at 16:31
  • $\begingroup$ I can do it, but with a little use of calculus. The only solution is x=1 $\endgroup$ – Presage Feb 22 at 16:32
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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 $1$.

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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 $x=1$

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