Problem:
There is a unique positive integer n such that
log2β(log16βn)=log4β(log4βn).
What is the sum of the digits of n ?
Answer Choices:
A. 4
B. 7
C. 8
D. 11
E. 13
Solution:
We can use the fact that logabβc=b1βlogaβc. This can be proved by using change of base formula to base a.
So, the original equation log2β(log24βn)=log22β(log22βn) becomes
log2β(41βlog2βn)=21βlog2β(21βlog2βn)
Using log property of addition, we expand both sides and then simplify:
log2β41β+log2β(log2βn)= 21β[log2β21β+log2β(log2βn)]
log2β41β+log2β(log2βn)= 21β[β1+log2β(log2βn)]
β2+log2β(log2βn)= β21β+21β(log2β(log2βn))
Subtracting 21β(log2β(log2βn)) from both sides and adding 2 to both sides gives us
21β(log2β(log2βn))= 23β
Multiplying by 2 , exponentiating, and simplifying gives us
log2β(log2βn)= 3
log2βn=8
n=256
Adding the digits together, we have 2+5+6= (E)13β .
OR
We will apply the following logarithmic identity:
logpkβqk=logpβq
which can be proven by the Change of Base Formula:
logpkβqk=logpβpklogpβqkβ=kklogpβqβ=logpβq
Note that log16βnξ =0, so we rewrite the original equation as follows:
log4β(log16βn)2=log4β(log4βn)
(log16βn)2=log4βn
(log16βn)2=log16βn2
(log16βn)2=2log16βn
log16βn=2
from which n=162=256. The sum of its digits is 2+5+6=(E)13β.
The problems on this page are the property of the MAA's American Mathematics Competitions