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 $12=\frac{60}{60^b}=60\cdot 60^{-b}=60^{1-b}.$ Therefore we have $12^{(1-a-b)/\left(2\left(1-b\right)\right)}=60^{(1-b)(1-a-b)/\left(2\left(1-b\right)\right)}=60^{(1-a-b)/2}.$ Since $4=\frac{60}{3\cdot 5}$ , we have $4=\frac{60}{60^a 60^b}=60^{1-a-b}.$ Therefore, we have $60^{(1-a-b)/2}=4^{1/2}=\fbox{\textbf{(B)} 2}$

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

1983 AHSME Problems/Problem 25

Problem 25

Solution

See Also