Problem:
There is a unique positive real number x such that the three numbers log8(2x), log4x, and log2x, in that order, form a geometric progression with positive common ratio. The number x can be written as nm, where m and n are relatively prime positive integers. Find m+n.
Solution:
By the Change of Base Formula the common ratio of the progression is
log4xlog2x=log24log2xlog2x=2
Hence x must satisfy
2=log8(2x)log4x=log24log2x÷log28log2(2x)=23⋅1+log2xlog2x
This is equivalent to 4+4log2x=3log2x. Hence log2x=−4 and x=161. The requested sum is 1+16=17.
The problems on this page are the property of the MAA's American Mathematics Competitions