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Difference between revisions of "1983 AHSME Problems/Problem 25"
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==Problem 25==
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If <math>60^a=3</math> and <math>60^b=5</math>, then <math>12^[(1-a-b)/2(1-b)]</math> is
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If <math>60^a=3</math> and <math>60^b=5</math>, then <math>12^{(1-a-b)/\left(2\left(1-b\right)\right)}</math> is
  
 
<math>\textbf{(A)}\ \sqrt{3}\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \sqrt{5}\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 2\sqrt{3}\qquad</math>
 
<math>\textbf{(A)}\ \sqrt{3}\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \sqrt{5}\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 2\sqrt{3}\qquad</math>
  
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Solution: Since <math>12 = 60/5</math>, <math>12 = 60/(60^b)</math>= <math>60^{(1-b)}</math>
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==Solution==  
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We have that <math>12=\frac{60}{5}</math>. We can substitute our value for 5, to get
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<cmath>12=\frac{60}{60^b}=60\cdot 60^{-b}=60^{1-b}.</cmath>
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Hence
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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>
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Since <math>4=\frac{60}{3\cdot 5}</math>, we have
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<cmath>4=\frac{60}{60^a 60^b}=60^{1-a-b}.</cmath>
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Therefore, we have
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<cmath>60^{(1-a-b)/2}=4^{1/2}=\boxed{\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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== 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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which can be rewritten as <math>(60^{(1-a-b)})^{(1/2)}</math>
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<cmath>12^{(1-a-b)/[2(1-b)]} = 12^{(1-(a+b))/[2(1-b)]}</cmath>
  
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<math>60^{(1-a-b)} = 60^1/[(60^a)(60^b)] = 60/(3*5) = 4</math>
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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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<math>4^{(1/2)} = 2</math>
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<cmath> (1-(a+b) = \log_{60} 60 - \log_{60} 15 = \log_{60} 4 </cmath>
  
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Answer:<math>B</math>
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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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<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
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==See Also==
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{{AHSME box|year=1983|num-b=24|num-a=26}}
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{{MAA Notice}}

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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