Problem:
Positive real numbers xξ =1 and yξ =1 satisfy log2βx=logyβ16 and xy=64. What is (log2βyxβ)2?
Answer Choices:
A. 225β
B. 20
C. 245β
D. 25
E. 32
Solution:
Let log2βx=logyβ16=k, so that 2k=x and yk=16βΉy=2k4β. Then we have (2k)βββ2k4ββ ββ= 2k+k4β=26.
We therefore have k+k4β=6, and deduce k2β6k+4=0. The solutions to this are k=3Β±5β.
To solve the problem, we now find
(log2βyxβ)2=(log2βxβlog2βy)2
=(kβk4β)2=(3Β±5ββ3Β±5β4β)2
=(3Β±5ββ[3β5β])2
=(3Β±5ββ3Β±5β)2
=(Β±25β)2
=(B)20β.
The problems on this page are the property of the MAA's American Mathematics Competitions