Problem:
Positive real numbers bξ =1 and n satisfy the equations
logbβnβ=logbβnβ and bβ logbβn=logbβ(bn)
The value of n is kjβ, where j and k are relatively prime positive integers. Find j+k.
Solution:
Answer (881):
The first equation implies that logbβnβ=21ββ logbβn, from which (logbβn)2=4β logbβn, so logbβn is either 0 or 4 , which implies that n=1 or n=b4. If n=1, then the left side of the second equation is 0 , but the right side is logbβb=1. Therefore n=b4 and logbβn=4. Substituting these values into the second equation gives 4b=logbβb5=5, so b=45β and n=(45β)4=256625β. The requested sum is 625+256=881.