Problem:
Find the last three digits of the product of the positive roots of
1995βxlog1995βx=x2
Solution:
Taking logarithms of both sides of the equation, we find that
log(1995βxlogx)=logx2
where all the logarithms are to the base 1995. From this we obtain
21βlog1995+(logx)(logx)=2logx
which leads to
(logx)2β2logx+21β=0
Solving this last equation gives
logx=1Β±21ββ
Since these values for logx are both real, the original equation has two positive roots; call them r1β and r2β. Since logr1βr2β=logr1β+logr2β=2, the product of these roots is