# 1983 AHSME Problems/Problem 25

## Problem 25

If $60^a=3$ and $60^b=5$, then $12^{(1-a-b)/\left(2\left(1-b\right)\right)}$ is

$\textbf{(A)}\ \sqrt{3}\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \sqrt{5}\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 2\sqrt{3}\qquad$

## Solution

We have that $12=\frac{60}{5}$. We can substitute our value for 5, to get $$12=\frac{60}{60^b}=60\cdot 60^{-b}=60^{1-b}.$$ Hence $$12^{(1-a-b)/\left(2\left(1-b\right)\right)}=60^{(1-b)(1-a-b)/\left(2\left(1-b\right)\right)}=60^{(1-a-b)/2}.$$ Since $4=\frac{60}{3\cdot 5}$, we have $$4=\frac{60}{60^a 60^b}=60^{1-a-b}.$$ Therefore, we have $$60^{(1-a-b)/2}=4^{1/2}=\boxed{\textbf{(B)}\ 2}$$

## Solution 2

This problem is easier if we turn the first two equations into logs: $a = \log_{60} 3$, $b = \log_{60} 5$. Then

$12^{ \frac{1-a-b}{2(1-b)} } &= 12^{ \frac{1 - \log_{60} 3 - \log_{60} 5}{2(1 - \log_{60} 5)} }$ (Error compiling LaTeX. ! Misplaced alignment tab character &.)

Using the fact that $1 = \log_{60} 60$, 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*}