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

See Also

1983 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
Problem 26
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