1983 AHSME Problems/Problem 25

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 $1 = log_{60} 60$

$2(1-b) = 2(log_{60} 12)$

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

1983 AHSME (Problems • Answer Key • Resources)
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1983 AHSME Problems/Problem 25

Contents

Problem 25

Solution

Solution 2

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