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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− | Problem | + | ==Problem 25== |
− | If 60^a=3 and 60^b=5, then 12^ | + | If <math>60^a=3</math> and <math>60^b=5</math>, then <math>12^{(1-a-b)/\left(2\left(1-b\right)\right)}</math> is |
− | (A) | + | <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== | ||
+ | We have that <math>12=\frac{60}{5}</math>. We can substitute our value for 5, to get | ||
+ | <cmath>12=\frac{60}{60^b}=60\cdot 60^{-b}=60^{1-b}.</cmath> | ||
+ | Hence | ||
+ | <cmath>12^{(1-a-b)/\left(2\left(1-b\right)\right)}=60^{(1-b)(1-a-b)/\left(2\left(1-b\right)\right)}=60^{(1-a-b)/2}.</cmath> | ||
+ | Since <math>4=\frac{60}{3\cdot 5}</math>, we have | ||
+ | <cmath>4=\frac{60}{60^a 60^b}=60^{1-a-b}.</cmath> | ||
+ | Therefore, we have | ||
+ | <cmath>60^{(1-a-b)/2}=4^{1/2}=\boxed{\textbf{(B)}\ 2}</cmath> | ||
− | + | ==See Also== | |
− | + | {{AHSME box|year=1983|num-b=24|num-a=26}} | |
− | + | {{MAA Notice}} |
Revision as of 14:37, 17 April 2021
Problem 25
If and , then is
Solution
We have that . We can substitute our value for 5, to get Hence Since , we have Therefore, we have
See Also
1983 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Problem 26 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
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