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

~YBSuburbanTea

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