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