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Difference between revisions of "2010 AMC 12A Problems/Problem 11"

(Created page with '== Problem == The solution of the equation <math>7^{x+7} = 8^x</math> can be expressed in the form <math>x = \log_b 7^7</math>. What is <math>b</math>? <math>\textbf{(A)}\ \frac…')
 
Line 6: Line 6:
 
== Solution ==
 
== Solution ==
 
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.
 +
 
<math> 7^{x+7} = 8^x </math>
 
<math> 7^{x+7} = 8^x </math>
 +
 
<math> 7^x*7^7 = 8^x </math>
 
<math> 7^x*7^7 = 8^x </math>
 +
 
<math> \left(\frac{8}{7}\right)^x = 7^7 </math>
 
<math> \left(\frac{8}{7}\right)^x = 7^7 </math>
 +
 
<math> x = \log_{8/7}7^7 </math>
 
<math> x = \log_{8/7}7^7 </math>
−
Therefore, 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>.

Revision as of 15:57, 10 February 2010

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.

$7^{x+7} = 8^x$

$7^x*7^7 = 8^x$

$\left(\frac{8}{7}\right)^x = 7^7$

$x = \log_{8/7}7^7$

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

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