Difference between revisions of "1983 AHSME Problems/Problem 25"
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==Problem 25== | ==Problem 25== | ||
| − | If <math>60^a=3</math> and <math>60^b=5</math>, then <math>12^{ | + | If <math>60^a=3</math> and <math>60^b=5</math>, then <math>12^{(1-a-b)/\left(2\left(1-b\right)\right)}</math> is |
<math>\textbf{(A)}\ \sqrt{3}\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \sqrt{5}\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 2\sqrt{3}\qquad</math> | <math>\textbf{(A)}\ \sqrt{3}\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \sqrt{5}\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 2\sqrt{3}\qquad</math> | ||
Revision as of 17:27, 27 January 2019
Problem 25
If 
and 
, then 
is
Solution
We have that 
. We can substitute our value for 5, to get
![\[12=\frac{60}{60^b}=60\cdot 60^{-b}=60^{1-b}.\]](http://latex.artofproblemsolving.com/3/e/9/3e9f343c772c6fcabb7c15c2b8e0ec5e56dfc51b.png)
Therefore we have
![\[12^{[(1-a-b)/2(1-b)]}=60^{[(1-b)(1-a-b)/2(1-b)]}=60^{(1-a-b)/2}.\]](http://latex.artofproblemsolving.com/0/b/c/0bcc4f489d4d41851740d1a9928b24a48ce08340.png)
Since 
, we have
![\[4=\frac{60}{60^a 60^b}=60^{1-a-b}.\]](http://latex.artofproblemsolving.com/d/2/f/d2f41a20a20ce8e7f41c6519afb08e70677589f6.png)
Therefore, we have
![\[60^{(1-a-b)/2}=4^{1/2}=\fbox{\textbf{(B)} 2}\]](http://latex.artofproblemsolving.com/a/8/1/a814bf7807cfe3405f92975dcd4b4be16a63a95b.png)
See Also
| 1983 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 24 |
Followed by Problem 26 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
