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Difference between revisions of "1956 AHSME Problems/Problem 20"
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Revision as of 21:23, 19 March 2025

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 $\boxed{\textbf{(D)}}$ is correct.

~ eevee9406

See also

1956 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

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