Suppose that logk is the arithmetic mean of log1,log2,log4,…,log(2n−1). Then which of these is true:
k=2(2n(n−1))
log2k=2n(n−1)
logk4=1−n4
k=(2)n−1
k=n22n2n−1
Solution
Using the laws of logs and exponents:
logk=n1(log1+log2+log4+…+log(2n−1))=n1(0+log21+log22+…+log(2n−1))=n1log(21⋅22⋅…⋅2n−1)=n1log(21+2+⋯+(n−1))=nlog2⋅(1+2+⋯+(n−1))=nlog2⋅2n(n−1)=2(n−1)log2
Therefore,
k=10(2(n−1)log2)=(10log2)2n−1=22n−1=(2)n−1
The answer is d.