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

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(→‎Solution 2)
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Therefore, we have
 
Therefore, we have
 
<cmath>60^{(1-a-b)/2}=4^{1/2}=\boxed{\textbf{(B)}\ 2}</cmath>
 
<cmath>60^{(1-a-b)/2}=4^{1/2}=\boxed{\textbf{(B)}\ 2}</cmath>
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==Solution 2==
 
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This problem is easier if we turn the first two equations into logs: <math>a = \log_{60} 3</math>, <math>b = \log_{60} 5</math>. Then
 
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<cmath>12^{ \frac{1-a-b}{2(1-b)} } &= 12^{ \frac{1 - \log_{60} 3 - \log_{60} 5}{2(1 - \log_{60} 5)} }</cmath>
 
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Using the fact that <math>1 = \log_{60} 60</math>, we can combine the linear combination of logs into one log. In fact, adding and subtracting logs in this way is how multiplication and division is done with log tables and slide rules.
 
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\begin{align*}
 
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12^{ \frac{1-a-b}{2(1-b)} } &= 12^{ \frac{\log_{60} 60/(3 \cdot 5)}{2(\log_{60} 60/5)} } \\
 
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  &= 12^{ \frac{\log_{60} 4}{2\log_{60} 12} } \\
 
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  &= 12^{\frac{1}{2} \log_{12} 4} \\
 
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  &= 4^{1/2} \\
 
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  &= \boxed{2}.
 
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\end{align*}
 
  
 
==See Also==
 
==See Also==

Revision as of 13:37, 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 (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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