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{eqnarray*} 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{eqnarray*}$ (Error compiling LaTeX. Unknown error_msg)

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 $k$ .

1988 AIME (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

Page

Toolbox

Search

1988 AIME Problems/Problem 3

Problem

Solution

See also