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:

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

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 $k$ .

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 $\log_{\log_2x}(\log_8x)=\frac{1}{3}$

1988 AIME (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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1988 AIME Problems/Problem 3

Contents

Problem

Solution 1

Solution 2: Substitution

Solution 3

See also