Problem:
Determine the value of ab if log8a+log4b2=5 and log8b+log4a2=7.
Solution:
Adding the two equations and using standard 10 properties yields
log8a+log8b+log4a2+log4b2=log8(ab)+2log4(ab)=12
Moreover, since log8x=(log28)(log2x)=31log2x, and similar1y, log4x=(log24)(log2x)=21log2x, the above equation is equivalent to
34log2(ab)=12
It follows that log2(ab)=9, and hence ab=29=512.
The problems on this page are the property of the MAA's American Mathematics Competitions