Difference between revisions of "1962 AHSME Problems/Problem 17"
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==Solution== | ==Solution== | ||
| − | {{ | + | Using the change-of-base rule: <math>a = \frac{\log 225}{\log 8}</math> and <math>b = \frac{\log 15}{\log 2}</math>. |
| + | <cmath>\frac{a}{b} = \frac{\log 225 \log 2}{\log 8 \log 15}</cmath> | ||
| + | <cmath>a = b \cdot \frac{\log 225}{\log 15} \cdot \frac{\log 2}{\log 8}</cmath> | ||
| + | <cmath>a = b \log_{15} 225 \log_8 2</cmath> | ||
| + | <cmath>a = \boxed{\frac{2b}3 \textbf{ (B)}}</cmath> | ||
| + | |||
| + | |||
| + | ==See Also== | ||
| + | {{AHSME 40p box|year=1962|num-b=16|num-a=18}} | ||
| + | |||
| + | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 19:48, 31 May 2018
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 | |
| 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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
