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1956 AHSME Problems/Problem 20

Revision as of 16:39, 21 June 2021 by Angrybird029 (talk | contribs) (→‎Solution)
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If $(0.2)^x = 2$ and $\log 2 = 0.3010$, then the value of $x$ to the nearest tenth is:

$\textbf{(A)}\ - 10.0 \qquad\textbf{(B)}\ - 0.5 \qquad\textbf{(C)}\ - 0.4 \qquad\textbf{(D)}\ - 0.2 \qquad\textbf{(E)}\ 10.0$

Solution

To be added.

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