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

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==Solution==   
 
==Solution==   
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Since <math>12 = 60/5</math>, <math>12 = 60/(60^b)</math>= <math>60^{(1-b)}</math>
+
We have that <math>12=\frac{60}{5}</math>. We can substitute our value for 5, to get:
 +
<cmath>12=\frac{60}{60^b}=60\cdot 60^{-b}=60^{(1-b)}.</cmath>
 +
Therefore we have:
 +
<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>
 +
Therefore, we have:
 +
<cmath>60^{(1-a-b)/2}=4^{1/2}=\fbox{\textbf{(B) 2}}</cmath>
  
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So we can rewrite <math>12^{[(1-a-b)/2(1-b)]}</math> as <math>60^{[(1-b)(1-a-b)/2(1-b)]}</math>
 
−
 
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this simplifies to <math>60^{[(1-a-b)/2]}</math>
 
−
 
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which can be rewritten as <math>(60^{(1-a-b)})^{(1/2)}</math>
 
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<math>60^{(1-a-b)} = 60^1/[(60^a)(60^b)] = 60/(3*5) = 4</math>
 
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<math>4^{(1/2)} = 2</math>
 
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Answer:<math>\fbox{\textbf{B}}</math>.
 
 
==See Also==
 
==See Also==
 
{{AHSME box|year=1983|num-b=24|num-a=26}}
 
{{AHSME box|year=1983|num-b=24|num-a=26}}
  
 
{{MAA Notice}}
 
{{MAA Notice}}

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: \[60^{(1-a-b)/2}=4^{1/2}=\fbox{\textbf{(B) 2}}\]

See Also

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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