# Difference between revisions of "1988 AIME Problems/Problem 3"

## Problem

Find $(\log_2 x)^2$ if $\log_2 (\log_8 x) = \log_8 (\log_2 x)$.

## Solution

Raise both as exponents with base 8:

\begin{align*} 8^{\log_2 (\log_8 x)} &= 8^{\log_8 (\log_2 x)}\\ 2^{3 \log_2(\log_8x)} &= \log_2x\\ (\log_8x)^3 &= \log_2x\\ \left(\frac{\log_2x}{\log_28}\right)^3 &= \log_2x\\ (\log_2x)^2 &= (\log_28)^3 = \boxed{27}\\ \end{align*}

A quick explanation of the steps: On the 1st step, we use the property of logarithms that $a^{\log_a x} = x$. On the 2nd step, we use the fact that $k \log_a x = \log_a x^k$. On the 3rd step, we use the change of base formula, which states $\log_a b = \frac{\log_k b}{\log_k a}$ for arbitrary $k$.