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1956 AHSME Problems/Problem 20

Revision as of 16:38, 21 June 2021 by Angrybird029 (talk | contribs) (Created page with "If <math>(0.2)^x = 2</math> and <math>\log 2 = 0.3010</math>, then the value of <math>x</math> to the nearest tenth is: <math>\textbf{(A)}\ - 10.0 \qquad\textbf{(B)}\ - 0.5 \...")
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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

$(0.2)^x = (\frac{1}{5})^x = 5^{-x}$

$\sqrt{5} \approx 2.236$

Thus, $-x \approx 0.5$, so the answer is $\boxed{\textbf{(B)}\ -0.5}$

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