# Difference between revisions of "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 $$60^{(1-a-b)/2}=4^{1/2}=\boxed{\textbf{(B)}\ 2}$$