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> | ||
− | + | ||
+ | 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 can be expressed in the form
. What is
?
Solution
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 .