1961 AHSME Problems/Problem 6

Problem

When simplified, $\log{8} \div \log{\frac{1}{8}}$ becomes:

$\textbf{(A)}\ 6\log{2} \qquad \textbf{(B)}\ \log{2} \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 0\qquad \textbf{(E)}\ -1$

Solution

First, note that $\frac{1}{8} = 8^{-1}$ . That means the expression can be rewritten as $\log{8} \div \log{8^{-1}}$ $\log{8} \cdot \frac{1}{\log{8^{-1}}}$ $\log{8} \cdot \frac{1}{-1 \cdot \log{8}}$ This simplifies to $-1$ , which is answer choice $\boxed{\textbf{(E)}}$ .

1961 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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1961 AHSME Problems/Problem 6

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