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.

  • 1
    $\begingroup$ 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

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 $1$.


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 $x=1$.


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 $x=1$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.