Difference between revisions of "1962 AHSME Problems/Problem 28"
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==Solution== | ==Solution== | ||
− | {{ | + | Taking the base-<math>x</math> logarithm of both sides gives <math>\log_{10}x=\log_x\frac{x^3}{100}</math>. |
+ | This simplifies to | ||
+ | <cmath>\log_{10}x=\log_x{x^3} - \log_x{100}</cmath> | ||
+ | <cmath>\log_{10}x+\log_x{100}=3</cmath> | ||
+ | <cmath>\log_{10}x+2 \log_x{10}=3</cmath> | ||
+ | At this point, we substitute <math>u=\log_{10}x</math>. | ||
+ | <cmath>u+\frac2u=3</cmath> | ||
+ | <cmath>u^2+2=3u</cmath> | ||
+ | <cmath>u^2-3u+2=0</cmath> | ||
+ | <cmath>(u-2)(u-1)=0</cmath> | ||
+ | <cmath>u\in\{1, 2\}</cmath> | ||
+ | <cmath>x\in\{10, 100\}</cmath> | ||
+ | The answer is <math>\boxed{\textbf{(D)}}</math>. |
Latest revision as of 20:42, 16 April 2014
Problem
The set of -values satisfying the equation consists of:
Solution
Taking the base- logarithm of both sides gives . This simplifies to At this point, we substitute . The answer is .