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Difference between revisions of "1962 AHSME Problems/Problem 28"

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==Solution==
 
==Solution==
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{{solution}}
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Taking the base-<math>x</math> logarithm of both sides gives <math>\log_{10}x=\log_x\frac{x^3}{100}</math>.
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This simplifies to
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<cmath>\log_{10}x=\log_x{x^3} - \log_x{100}</cmath>
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<cmath>\log_{10}x+\log_x{100}=3</cmath>
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<cmath>\log_{10}x+2 \log_x{10}=3</cmath>
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At this point, we substitute <math>u=\log_{10}x</math>.
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<cmath>u+\frac2u=3</cmath>
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<cmath>u^2+2=3u</cmath>
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<cmath>u^2-3u+2=0</cmath>
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<cmath>(u-2)(u-1)=0</cmath>
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<cmath>u\in\{1, 2\}</cmath>
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<cmath>x\in\{10, 100\}</cmath>
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The answer is <math>\boxed{\textbf{(D)}}</math>.

Latest revision as of 21:42, 16 April 2014

Problem

The set of $x$-values satisfying the equation $x^{\log_{10} x} = \frac{x^3}{100}$ consists of:

$\textbf{(A)}\ \frac{1}{10}\qquad\textbf{(B)}\ \text{10, only}\qquad\textbf{(C)}\ \text{100, only}\qquad\textbf{(D)}\ \text{10 or 100, only}\qquad\textbf{(E)}\ \text{more than two real numbers.}$


Solution

Taking the base-$x$ logarithm of both sides gives $\log_{10}x=\log_x\frac{x^3}{100}$. This simplifies to \[\log_{10}x=\log_x{x^3} - \log_x{100}\] \[\log_{10}x+\log_x{100}=3\] \[\log_{10}x+2 \log_x{10}=3\] At this point, we substitute $u=\log_{10}x$. \[u+\frac2u=3\] \[u^2+2=3u\] \[u^2-3u+2=0\] \[(u-2)(u-1)=0\] \[u\in\{1, 2\}\] \[x\in\{10, 100\}\] The answer is $\boxed{\textbf{(D)}}$.

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