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Difference between revisions of "1956 AHSME Problems/Problem 20"

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<math>\textbf{(A)}\ - 10.0 \qquad\textbf{(B)}\ - 0.5 \qquad\textbf{(C)}\ - 0.4 \qquad\textbf{(D)}\ - 0.2 \qquad\textbf{(E)}\ 10.0</math>
 
<math>\textbf{(A)}\ - 10.0 \qquad\textbf{(B)}\ - 0.5 \qquad\textbf{(C)}\ - 0.4 \qquad\textbf{(D)}\ - 0.2 \qquad\textbf{(E)}\ 10.0</math>
 
==Solution==
 
==Solution==
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To be added.
+
Taking reciprocals of the first condition yields <math>5^x=\frac{1}{2}</math>. Then <math>5^x\cdot2^x=\frac{1}{2}\cdot2^x</math>, so <math>10^x=2^{x-1}</math>. Taking base <math>10</math> log of both sides results in <math>x=(x-1)\log 2</math>, so
 +
<cmath>x=\frac{\log2}{\log2-1}\approx{0.3}{-0.7}=-\frac{3}{7}\approx-0.429</cmath>
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Thus <math>\boxed{\textbf{(D)}}</math> is correct.
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~ [https://artofproblemsolving.com/wiki/index.php/User:Eevee9406 eevee9406]
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==See also==
 +
{{AHSME box|year=1956|num-b=19|num-a=21}}
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{{MAA Notice}}

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 AHSME (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
All AHSME Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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