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2010 AMC 12A Problems/Problem 11

Revision as of 14:55, 10 February 2010 by Rpond (talk | contribs) (Created page with '== Problem == The solution of the equation <math>7^{x+7} = 8^x</math> can be expressed in the form <math>x = \log_b 7^7</math>. What is <math>b</math>? <math>\textbf{(A)}\ \frac…')
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Problem

The solution of the equation $7^{x+7} = 8^x$ can be expressed in the form $x = \log_b 7^7$. What is $b$?

$\textbf{(A)}\ \frac{7}{15} \qquad \textbf{(B)}\ \frac{7}{8} \qquad \textbf{(C)}\ \frac{8}{7} \qquad \textbf{(D)}\ \frac{15}{8} \qquad \textbf{(E)}\ \frac{15}{7}$

Solution

This problem is quickly solved with knowledge of the laws of exponents and logarithms. $7^{x+7} = 8^x$ $7^x*7^7 = 8^x$ $\left(\frac{8}{7}\right)^x = 7^7$ $x = \log_{8/7}7^7$ Therefore, our answer is $\boxed{\textbf{(C)}\ \frac{8}{7}}$.

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