1956 AHSME Problems/Problem 20

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

Taking reciprocals of the first condition yields $5^x=\frac{1}{2}$ . Then $5^x\cdot2^x=\frac{1}{2}\cdot2^x$ , so $10^x=2^{x-1}$ . Taking base $10$ log of both sides results in $x=(x-1)\log 2$ , so $x=\frac{\log2}{\log2-1}\approx{0.3}{-0.7}=-\frac{3}{7}\approx-0.429$ Thus $\boxed{\textbf{(C)}}$ is correct.

~ eevee9406

1956 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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1956 AHSME Problems/Problem 20

Solution

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