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

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==Solution==   
 
==Solution==   
We have that <math>12=\frac{60}{5}</math>. We can substitute our value for 5, to get:
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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>
 
<cmath>12=\frac{60}{60^b}=60\cdot 60^{-b}=60^{(1-b)}.</cmath>
Therefore we have:
+
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>
 
<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:
+
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:
+
Therefore, we have
 
<cmath>60^{(1-a-b)/2}=4^{1/2}=\fbox{\textbf{(B)} 2}</cmath>
 
<cmath>60^{(1-a-b)/2}=4^{1/2}=\fbox{\textbf{(B)} 2}</cmath>
  

Revision as of 18:25, 27 January 2019

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 (ProblemsAnswer KeyResources)
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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