1983 AHSME Problems/Problem 25
Contents
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)
Hence
![\[12^{(1-a-b)/\left(2\left(1-b\right)\right)}=60^{(1-b)(1-a-b)/\left(2\left(1-b\right)\right)}=60^{(1-a-b)/2}.\]](http://latex.artofproblemsolving.com/6/e/b/6ebb5544f77ce86e1d129a0a929377b0a3a8f9e1.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
![\[60^{(1-a-b)/2}=4^{1/2}=\boxed{\textbf{(B)}\ 2}\]](http://latex.artofproblemsolving.com/2/6/9/26995b76464638b54f8741cf13123f3c606016a4.png)
Solution 2
This problem is easier if we turn the first two equations into logs: 
, 
. Then
\begin{align*} 12^{ \frac{1-a-b}{2(1-b)} } &= 12^{ \frac{1 - \log_{60} 3 - \log_{60} 5}{2(1 - \log_{60} 5)} } \end{align*}
Using the fact that 
, we can combine the linear combination of logs into one log. In fact, adding and subtracting logs in this way is how multiplication and division is done with log tables and slide rules.
\begin{align*} 12^{ \frac{1-a-b}{2(1-b)} } &= 12^{ \frac{\log_{60} 60/(3 \cdot 5)}{2(\log_{60} 60/5)} } \\
&= 12^{ \frac{\log_{60} 4}{2\log_{60} 12} } \\
&= 12^{\frac{1}{2} \log_{12} 4} \\
&= 4^{1/2} \\
&= \boxed{2}.
\end{align*}
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 | ||
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