Problem:
Find (log2βx)2 if log2β(log8βx)=log8β(log2βx).
Solution:
Since log8βx=logxβ81β=3logxβ21β=31βlog2βx and log8β(log2βx)=31βlog2β(log2βx), the given equation is equivalent to
log2β(3yβ)=(31β)log2βy(1)
where y=log2βx. From (1), log2β(3yβ)3=log2βy, hence (3yβ)3=y; i.e.,
y(y2β27)=0(2)
Since yξ =0 (for otherwise, neither side of (1) would be defined), it follows from (2) that y2=(log2βx)2=27β.
The problems on this page are the property of the MAA's American Mathematics Competitions