Mathematical Reflections 2012, Issue 1 - Problem J219

Problem:
Trying to solve a problem, Jimmy used the following "`formula"': $\log_{ab} x = \log_a x \log_b x$,
where $a, b, x$ are positive real numbers different from $1$. Prove that this is correct only
if $x$ is a solution to the equation $\log_a x + \log_b x = 1$.

Proposed by Titu Andreescu.

Solution:

We have $\log_{ab} x = \dfrac{\log_a x}{1+\log_a b}$. So $$\log_{ab} x = \log_a x \log_b x \iff \dfrac{1}{1+\log_a b}=\log_b x \iff 1=\log_b x + \log_a b \log_b x.$$ Since $\log_b x = \dfrac{\log_a x}{\log_a b}$, we are done.