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

Revision as of 05:49, 18 May 2016

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

We have that $12=\frac{60}{5}$ . We can substitute our value for 5, to get: $12=\frac{60}{60^b}=60\cdot 60^{-b}=60^{(1-b)}.$ Therefore we have: $12^{[(1-a-b)/2(1-b)]}=60^{[(1-b)(1-a-b)/2(1-b)]}=60^{(1-a-b)/2}.$ Since $4=\frac{60}{3\cdot 5}$ , we have: $4=\frac{60}{60^a 60^b}=60^{(1-a-b)}.$ Therefore, we have: $60^{(1-a-b)/2}=4^{1/2}=\fbox{\textbf{(B)} 2}$

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 10: / Line 10: @@
 <cmath>12^{[(1-a-b)/2(1-b)]}=60^{[(1-b)(1-a-b)/2(1-b)]}=60^{(1-a-b)/2}.</cmath>
 Since <math>4=\frac{60}{3\cdot 5}</math>, we have:
-<cmath>4=\frac{60}{60^a 60^b}=60^(1-a-b).</cmath>
+<cmath>4=\frac{60}{60^a 60^b}=60^{(1-a-b)}.</cmath>
 Therefore, we have:
 <cmath>60^{(1-a-b)/2}=4^{1/2}=\fbox{\textbf{(B)} 2}</cmath>

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

Revision as of 05:49, 18 May 2016

Problem 25

Solution

See Also