1962 AHSME Problems/Problem 17
Problem
If 
and 
, then 
, in terms of 
, is:
Solution
Using the change-of-base rule: 
and 
.
![\[\frac{a}{b} = \frac{\log 225 \log 2}{\log 8 \log 15}\]](http://latex.artofproblemsolving.com/d/b/e/dbef6cd33905fb2580c97a4b4287dcc3bc1edb66.png)
![\[a = b \cdot \frac{\log 225}{\log 15} \cdot \frac{\log 2}{\log 8}\]](http://latex.artofproblemsolving.com/8/8/c/88c43a50ec24325a896749b7ef1716b4ad14b0b6.png)
![\[a = b \log_{15} 225 \log_8 2\]](http://latex.artofproblemsolving.com/e/f/0/ef0ec26bd9555b37bbf1815b0e6b5f479ee1cd0b.png)
![\[a = \boxed{\frac{2b}3 \textbf{ (B)}}\]](http://latex.artofproblemsolving.com/a/4/e/a4e8bfc8cefcdc574fe5329a5f1868d1ff1980b2.png)
See Also
| 1962 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 16 |
Followed by Problem 18 | |
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| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
