1954 AHSME Problems/Problem 15

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Problem 15

$\log 125$ equals:

$\textbf{(A)}\ 100 \log 1.25 \qquad \textbf{(B)}\ 5 \log 3 \qquad \textbf{(C)}\ 3 \log 25  \\ \textbf{(D)}\ 3 - 3\log 2 \qquad \textbf{(E)}\ (\log 25)(\log 5)$

Solution

$\log(1000)=\log(125)+\log(8)\implies 3=\log(125)+\log(2^3)\implies \log(125)=3-3\log(2)$, $\fbox{D}$