Difference between revisions of "1983 AHSME Problems/Problem 25"

Revision as of 05:31, 18 May 2016

If $60^a=3$ and $60^b=5$ , then $12^[(1-a-b)/2(1-b)]$ is

$\textbf{(A)}\ \sqrt{3}\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \sqrt{5}\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 2\sqrt{3}\qquad$

Since $12 = 60/5$ , $12 = 60/(60^b)$ = $60^{(1-b)}$

So we can rewrite $12^{[(1-a-b)/2(1-b)]}$ as $60^{[(1-b)(1-a-b)/2(1-b)]}$

this simplifies to $60^{[(1-a-b)/2]}$

which can be rewritten as $(60^{(1-a-b)})^{(1/2)}$

$60^{(1-a-b)} = 60^1/[(60^a)(60^b)] = 60/(3*5) = 4$

$4^{(1/2)} = 2$

Answer:\fbox{ $\textbf{B}$ }.

@@ Line 1: / Line 1: @@
-Problem:
+==Problem 25==
 If <math>60^a=3</math> and <math>60^b=5</math>, then <math>12^[(1-a-b)/2(1-b)]</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>
-Solution:  Since <math>12 = 60/5</math>, <math>12 = 60/(60^b)</math>= <math>60^{(1-b)}</math>
+==Solution==
+Since <math>12 = 60/5</math>, <math>12 = 60/(60^b)</math>= <math>60^{(1-b)}</math>
 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>
@@ Line 16: / Line 17: @@
 <math>4^{(1/2)} = 2</math>
-Answer:<math>B</math>
+Answer:\fbox{<math>\textbf{B}</math>}.