Solve $\log_2(3^x-1)=\log_3(2^x+1)$

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 $1$.

Tags:

Algebra Precalculus

Logarithms

Exponential Function

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