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

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We have that <math>12=\frac{60}{5}</math>. We can substitute our value for 5, to get
 
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
+
Hence
<cmath>12^{[(1-a-b)/2(1-b)]}=60^{[(1-b)(1-a-b)/2(1-b)]}=60^{(1-a-b)/2}.</cmath>
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<cmath>12^{(1-a-b)/\left(2\left(1-b\right)\right)}=60^{(1-b)(1-a-b)/\left(2\left(1-b\right)\right)}=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>
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<cmath>60^{(1-a-b)/2}=4^{1/2}=\boxed{\textbf{(B)}\ 2}</cmath>
  
 
==See Also==
 
==See Also==

Revision as of 14:37, 17 April 2021

Problem 25

If $60^a=3$ and $60^b=5$, then $12^{(1-a-b)/\left(2\left(1-b\right)\right)}$ 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 \[12=\frac{60}{60^b}=60\cdot 60^{-b}=60^{1-b}.\] Hence \[12^{(1-a-b)/\left(2\left(1-b\right)\right)}=60^{(1-b)(1-a-b)/\left(2\left(1-b\right)\right)}=60^{(1-a-b)/2}.\] Since $4=\frac{60}{3\cdot 5}$, we have \[4=\frac{60}{60^a 60^b}=60^{1-a-b}.\] Therefore, we have \[60^{(1-a-b)/2}=4^{1/2}=\boxed{\textbf{(B)}\ 2}\]

See Also

1983 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
Problem 26
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All AHSME Problems and Solutions


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