1988 AIME Problems/Problem 3

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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:

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

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 • Resources)
Preceded by Problem 2	Followed by Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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1988 AIME Problems/Problem 3

From AoPSWiki

Problem

Solution

See also