Problem:
Positive real numbers x=1 and y=1 satisfy log2x=logy16 and xy=64. What is (log2yx)2?
Answer Choices:
A. 225
B. 20
C. 245
D. 25
E. 32
Solution:
Let log2x=logy16=k, so that 2k=x and yk=16⟹y=2k4. Then we have (2k)⎝⎛2k4⎠⎞= 2k+k4=26.
We therefore have k+k4=6, and deduce k2−6k+4=0. The solutions to this are k=3±5.
To solve the problem, we now find
(log2yx)2=(log2x−log2y)2
=(k−k4)2=(3±5−3±54)2
=(3±5−[3∓5])2
=(3±5−3±5)2
=(±25)2
=(B)20.
The problems on this page are the property of the MAA's American Mathematics Competitions