Difference between revisions of "1983 AHSME Problems/Problem 25"
(Created page with "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: We know that a=log 3 and b=log 5...") |
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| − | Solution: | + | 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) | ||
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| + | 60^(1-a-b) = 60^1/[(60^a)(60^b) = 60/(3*5) = 4 | ||
| + | |||
| + | 4^(1/2) = 2 | ||
Answer:B | Answer:B | ||
Revision as of 22:49, 9 July 2015
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
