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=xlog_{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}=7log_{\frac{8}{7}}7=log_{\frac{8}{7}}7^7. Therefore, the answer is </math>\frac{8}{7}=\fbox{C}$ | ||
==Video Solution1 (Clever Manipulations)== | ==Video Solution1 (Clever Manipulations)== | ||
Revision as of 14:11, 14 July 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. $x=\frac{7}{log_{7}\frac{8}{7}=7log_{\frac{8}{7}}7=log_{\frac{8}{7}}7^7. Therefore, the answer is$ (Error compiling LaTeX. Unknown error_msg)\frac{8}{7}=\fbox{C}$
Video Solution1 (Clever Manipulations)
~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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
