1983 AHSME Problems/Problem 25: Difference between revisions

Revision as of 19:13, 6 February 2016

Problem: 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: $B$

@@ Line 12: / Line 12: @@
 which can be rewritten as <math>(60^{(1-a-b)})^{(1/2)}</math>
-<math>60^(1-a-b) = 60^1/[(60^a)(60^b) = 60/(3*5) = 4</math>
+<math>60^{(1-a-b)} = 60^1/[(60^a)(60^b)] = 60/(3*5) = 4</math>
 <math>4^{(1/2)} = 2</math>
 Answer:<math>B</math>