Difference between revisions of "1983 AHSME Problems/Problem 25"

Revision as of 05:31, 18 May 2016

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

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

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

Revision as of 05:31, 18 May 2016 (view source) Quantummech (talk \| contribs) ← Older edit		Revision as of 05:31, 18 May 2016 (view source) Quantummech (talk \| contribs) (→‎Solution) Newer edit →
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	<math>4^{(1/2)} = 2</math>		<math>4^{(1/2)} = 2</math>

−	Answer:\fbox{~~<math>~~\textbf{B}</math>}.	+	Answer:<math>\fbox{\textbf{B}}</math>.