1983 AHSME Problems/Problem 25

Revision as of 05:31, 18 May 2016 by Quantummech (talk | contribs) (Solution)

Problem 25

If $60^a=3$ and $60^b=5$, then $12^[(1-a-b)/2(1-b)]$ 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

Since $12 = 60/5$, $12 = 60/(60^b)$= $60^{(1-b)}$

So we can rewrite $12^{[(1-a-b)/2(1-b)]}$ as $60^{[(1-b)(1-a-b)/2(1-b)]}$

this simplifies to $60^{[(1-a-b)/2]}$

which can be rewritten as $(60^{(1-a-b)})^{(1/2)}$

$60^{(1-a-b)} = 60^1/[(60^a)(60^b)] = 60/(3*5) = 4$

$4^{(1/2)} = 2$

Answer:$\fbox{\textbf{B}}$.