# 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.

• I think you will need a numerical method to solve this equation – Dr. Sonnhard Graubner Feb 22 at 16:28
• 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 – Luca Pana Feb 22 at 16:29
• Or you have made a typo! – Dr. Sonnhard Graubner Feb 22 at 16:30
• no sir , this is the exact problem given:) – Luca Pana Feb 22 at 16:31
• I can do it, but with a little use of calculus. The only solution is x=1 – 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 $$x=1$$.

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 $$x=1$$