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

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<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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== Solution 2 ==
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We have <math>60^a = 3</math> and <math>60^b = 5</math>. We can say that <math>a = \log_{60} 3</math> and <math>b = \log_{60} 5</math>.
  
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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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<cmath>12^{(1-a-b)/[2(1-b)]} = 12^{(1-(a+b))/[2(1-b)]}</cmath>
  
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\begin{align*}
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We can evaluate (a+b) by the Addition Identity for Logarithms, <math>(a+b) = \log_{60} 15</math>. Also, <math>1 = \log_{60} 60</math>.
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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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<cmath> (1-(a+b) = \log_{60} 60 - \log_{60} 15 = \log_{60} 4 </cmath>
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  &= 12^{\frac{1}{2} \log_{12} 4} \\
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  &= 4^{1/2} \\
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Now we have to evaluate [2(1-b)], the denominator of the fractional exponent. Once again, we can say <math>1 = \log_{60} 60</math>
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  &= \boxed{2}.
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\end{align*}
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<cmath> 2(1-b) = 2(\log_{60} 12)</cmath>
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<cmath>12^{(\log_{60} 4)/[2(\log_{60} 12]} = 12^{\frac{1}{2} \cdot \log_{12} 4} = 4^{1/2} = 2</cmath>
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~YBSuburbanTea
  
 
==See Also==
 
==See Also==

Latest revision as of 11:12, 14 January 2022

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


Solution 2

We have $60^a = 3$ and $60^b = 5$. We can say that $a = \log_{60} 3$ and $b = \log_{60} 5$.

\[12^{(1-a-b)/[2(1-b)]} = 12^{(1-(a+b))/[2(1-b)]}\]

We can evaluate (a+b) by the Addition Identity for Logarithms, $(a+b) = \log_{60} 15$. Also, $1 = \log_{60} 60$.

\[(1-(a+b) = \log_{60} 60 - \log_{60} 15 = \log_{60} 4\]

Now we have to evaluate [2(1-b)], the denominator of the fractional exponent. Once again, we can say $1 = \log_{60} 60$

\[2(1-b) = 2(\log_{60} 12)\]

\[12^{(\log_{60} 4)/[2(\log_{60} 12]} = 12^{\frac{1}{2} \cdot \log_{12} 4} = 4^{1/2} = 2\]

~YBSuburbanTea

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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