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

Revision as of 05:36, 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: $\fbox{\textbf{B}}$ .

1983 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Problem 26
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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 Answer:<math>\fbox{\textbf{B}}</math>.
 ==See Also==
-{{AHSME box|year=1983|ab=A|num-b=24 num-a=26}}
+{{AHSME box|year=1983|num-b=24|num-a=26}}
 {{MAA Notice}}