Problem:
There is a positive real number x not equal to either 201​ or 21​ such that
log20x​(22x)=log2x​(202x)
The value log20x​(22x) can be written as log10​(nm​), where m and n are relatively prime positive integers. Find m+n.
Solution:
Let y=log20x​(22x). Then the given equations imply
(20x)y(2x)y​=22x=202x​
Thus
10y=(2x)y(20x)y​=10111​
Hence y=log10​10111​. The requested sum is 11+101=112. The value of x that satisfies the original equation is approximately equal to 0.047630 .
The problems on this page are the property of the MAA's American Mathematics Competitions