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Difference between revisions of "2010 AMC 12A Problems/Problem 11"
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<math>\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}</math>
 
<math>\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}</math>
  
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== Solution ==
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== Solution 1 ==
 
This problem is quickly solved with knowledge of the laws of exponents and logarithms.
 
This problem is quickly solved with knowledge of the laws of exponents and logarithms.
  
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<math> 7^{x+7} = 8^x </math>
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<cmath>\begin{align*} 7^{x+7} &= 8^x \\
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7^x*7^7 &= 8^x \\
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\left(\frac{8}{7}\right)^x &= 7^7 \\
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x &= \log_{8/7}7^7 \end{align*}</cmath>
  
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<math> 7^x*7^7 = 8^x </math>
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Since we are looking for the base of the logarithm, our answer is <math>\boxed{\textbf{(C)}\ \frac{8}{7}}</math>.
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== Solution 2 ==
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First, take the <math>\log_{7}</math> of both sides, which gives us <math>x+7=\log_{7}8^x=x\log_{7}8</math>. We move the <math>x</math> terms to 1 side. <math>x(\log_{7}8-1)=7</math>. Isolate <math>x</math> and manipulate the answer. <math>x=\frac{7}{\log_{7}\frac{8}{7}}=7\log_{\frac{8}{7}}7=\log_{\frac{8}{7}}7^7</math>. Therefore, the answer is <math>\frac{8}{7}=\fbox{C}</math>
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==Video Solution1 (Clever Manipulations)==
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https://youtu.be/xGeL7864QLM
  
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<math> \left(\frac{8}{7}\right)^x = 7^7 </math>
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~Education, the Study of Everything
  
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<math> x = \log_{8/7}7^7 </math>
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== See also ==
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{{AMC12 box|year=2010|num-b=10|num-a=12|ab=A}}
  
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Since we are looking for the base of the logarithm, our answer is <math>\boxed{\textbf{(C)}\ \frac{8}{7}}</math>.
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[[Category:Introductory Number Theory Problems]]
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{{MAA Notice}}

Latest revision as of 10:31, 4 October 2023

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

~Education, the Study of Everything

See also

2010 AMC 12A (Problems • Answer Key • Resources)
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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