2010 AMC 12A Problems/Problem 11

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.

\begin{align*} 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 \end{align*}

Since we are looking for the base of the logarithm, our answer is $\boxed{\textbf{(C)}\ \frac{8}{7}}$.

See also

2010 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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
All AMC 12 Problems and Solutions

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