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

Revision as of 14:19, 28 October 2015

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

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

Solution: 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: $B$

@@ Line 1: / Line 1: @@
 Problem:
-If 60^a=3 and 60^b=5, then 12^[(1-a-b)/2(1-b)] is
+If <math>60^a=3</math> and <math>60^b=5</math>, then <math>12^[(1-a-b)/2(1-b)]</math> is
-(A):sqrt3   (B): 2     (C): sqrt5    (D): 3     (E): sqrt12
+<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 12 = 60/5, 12 = 60/(60^b) = 60^(1-b)
+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>
-So we can rewrite 12^[(1-a-b)/2(1-b)] as 60^[(1-b)(1-a-b)/2(1-b)]
+this simplifies to <math>60^{[(1-a-b)/2]}</math>
-this simplifies to 60^[(1-a-b)/2]
+which can be rewritten as <math>(60^{(1-a-b)})^{(1/2)}</math>
-which can be rewritten as (60^(1-a-b))^(1/2)
+<math>60^(1-a-b) = 60^1/[(60^a)(60^b) = 60/(3*5) = 4</math>
-^(1-a-b) = 60^1/[(60^a)(60^b) = 60/(3*5) = 4
+<math>4^{(1/2)} = 2</math>
-^(1/2) = 2
+Answer:<math>B</math>
-Answer:B