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> | ||
− | == Solution == | + | == 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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Since we are looking for the base of the logarithm, our answer is <math>\boxed{\textbf{(C)}\ \frac{8}{7}}</math>. | Since we are looking for the base of the logarithm, our answer is <math>\boxed{\textbf{(C)}\ \frac{8}{7}}</math>. | ||
+ | |||
+ | == Solution 2 == | ||
+ | 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> | ||
==Video Solution1 (Clever Manipulations)== | ==Video Solution1 (Clever Manipulations)== |
Latest revision as of 10:31, 4 October 2023
Problem
The solution of the equation can be expressed in the form . What is ?
Solution 1
This problem is quickly solved with knowledge of the laws of exponents and logarithms.
Since we are looking for the base of the logarithm, our answer is .
Solution 2
First, take the of both sides, which gives us . We move the terms to 1 side. . Isolate and manipulate the answer. . Therefore, the answer is
Video Solution1 (Clever Manipulations)
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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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.