1983 AHSME Problems/Problem 25
Problem: If 60^a=3 and 60^b=5, then 12^[(1-a-b)/2(1-b)] is
(A):sqrt3 (B): 2 (C): sqrt5 (D): 3 (E): sqrt12
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