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 1

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 $\boxed{\textbf{(C)}\ \frac{8}{7}}$ .

Solution 2

First, take the $\log_{7}$ of both sides, which gives us $x+7=\log_{7}8^x=x\log_{7}8$ . We move the $x$ terms to 1 side. $x(\log_{7}8-1)=7$ . Isolate $x$ and manipulate the answer. $x=\frac{7}{\log_{7}\frac{8}{7}}=7\log_{\frac{8}{7}}7=\log_{\frac{8}{7}}7^7$ . Therefore, the answer is $\frac{8}{7}=\fbox{C}$

Video Solution1 (Clever Manipulations)

https://youtu.be/xGeL7864QLM

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2010 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2010 AMC 12A Problems/Problem 11

Contents

Problem

Solution 1

Solution 2

Video Solution1 (Clever Manipulations)

See also