Problem:
The value of x that satisfies log2xβ320=log2x+3β32020 can be written as nmβ, where m and n are relatively prime positive integers. Find m+n.
Solution:
We can simplify the given expression by converting everything to log2β.
Write the left hand side of the expression as x1ββ
log2β320 and the right hand side as x+31ββ
log2β32020.
Using the power rule for logarithms, we can simplify further to log2β3(20)β
x1β=log2β3(2020)β
x+31β.
Equating the exponents of 3 inside the logs, we get x20β=x+32020β.
Solving, x=1003β and the requested sum is 103β.
The problems on this page are the property of the MAA's American Mathematics Competitions