Problem:
Suppose x and y are positive real numbers such that
xy= 264 and (log2​x)log2​y= 27
What is the greatest possible value of log2​y ?
Answer Choices:
A. 3
B. 4
C. 3+2​
D. 4+3​
E. 7
Solution:
Taking the base-2 logarithm of both equations gives
ylog2​x=64 and (log2​y)(log2​log2​x)=7
Now taking the base-2 logarithm of the first equation again yields
log2​y+log2​log2​x=6
Let r=log2​y and s=log2​log2​x. Then r+s=6 and rs=7. Solving this system yields
{log2​y,log2​log2​x}∈{3−2​,3+2​}
Thus the greatest possible value of log2​y is 3+2​. This is achieved when x=223−2​≈8.0095 and y=23+2​≈21.3212.
The problems on this page are the property of the MAA's American Mathematics Competitions