Problem:
There exist real numbers x and y, both greater than 1, such that logx(yx)=logy(x4y)=10. Find xy.
Solution:
The given equations imply that
xlogxy=4ylogyx=10.
Multiplying the two expressions equaling 10 yields
100=10⋅10=(xlogxy)⋅(4ylogyx)=4xy⋅logxy⋅logyx=4xy.
Thus xy=25.
OR
Converting the given equations from logarithmic form into exponential form gives
x10y10=yx=x4y.
Raising each side of the second equation to the power x and using the first equation yields
x4xy=y10x=(yx)10=(x10)10=x100.
Thus 4xy=100, so xy=25.
Note: The equations are satisfied by the real numbers x≈1.535 and y≈16.291.
The problems on this page are the property of the MAA's American Mathematics Competitions