Difference between revisions of "1983 AHSME Problems/Problem 25"
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− | 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 | 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> | <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 | + | ==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> | 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> | ||
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<math>4^{(1/2)} = 2</math> | <math>4^{(1/2)} = 2</math> | ||
− | Answer:<math>B</math> | + | Answer:\fbox{<math>\textbf{B}</math>}. |
Revision as of 06:31, 18 May 2016
Problem 25
If and
, then
is
Solution
Since ,
=
So we can rewrite as
this simplifies to
which can be rewritten as
Answer:\fbox{}.