# 1988 AIME Problems/Problem 3

## Problem

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

## Solution 1

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{027}\\ \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 $k$.

## Solution 2: Substitution

We wish to convert this expression into one which has a uniform base. Let's scale down all the powers of 8 to 2. \begin{align*} {\log_2 (\frac{1}{3}\log_2 x)} &= \frac{1}{3}{\log_2 (\log_2 x)}\\ {\log_2 x = y}\\ {\log_2 (\frac{1}{3}y)} &= \frac{1}{3}{\log_2 (y)}\\ {3\log_2 (\frac{1}{3}y)} &= {\log_2 (y)}\\ {\log_2 (\frac{1}{3}y)^3} &= {\log_2 (y)}\\ \end{align*} Solving, we get $y^2 = 27$, which is what we want. $\boxed{27}$

Just a quick note- In this solution, we used 2 important rules of logarithm: 1) $\log_a b^n=n\log_a b$. 2) $\log_{a^n} b=\frac{1}{n}\log_a b$.

## Solution 3

First we have \begin{align*} \log_2(\log_8x)&=\log_8(\log_2x)\\ \frac{\log_2(\log_8x)}{\log_8(\log_2x)}&=1 \end{align*} Changing the base in the numerator yields \begin{align*} \frac{3\log_8(\log_8x)}{\log_8(\log_2x)}&=1\\ \frac{\log_8(\log_8x)}{\log_8(\log_2x)}&=\frac{1}{3}\\ \end{align*} Using the property $\frac{\log_ab}{\log_ac}=\log_cb$ yields \begin{align*} \log_{\log_2x}(\log_8x)&=\frac{1}{3}\\ (\log_2x)^\frac{1}{3}&=\log_8x\\ \sqrt{\log_2x}&=\frac{\log_2x}{3} \end{align*} Now setting $y=\log_2x$, we have $$\sqrt{y}=\frac{y}{3}$$ Solving gets $y=\log_2x=3\sqrt{3}\Longrightarrow(\log_2x)^2=(3\sqrt{3})^2=\boxed{27}$.

~ Nafer

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