2730.32 – Nothing but Logs

Suppose that log⁡k\log k is the arithmetic mean of log⁡1,log⁡2,log⁡4,…,log⁡(2n−1)\log 1, \log 2, \log 4, \ldots , \log (2^{n-1}). Then which of these is true:

  1. k=2(n(n−1)2)k = 2^{\left(\dfrac{n (n-1)}{2}\right)}

  2. log⁡2k=n(n−1)2\log_2 k = \dfrac{n (n-1)}{2}

  3. log⁡k4=41−n\log_k 4 = \dfrac{4}{1-n}

  4. k=(2)n−1k = \left( \sqrt{2} \right)^{n-1}

  5. k=2n−1n22nk = \dfrac{2^n - 1}{n^2 2^n}

Solution

Using the laws of logs and exponents:

log⁡k=1n(log⁡1+log⁡2+log⁡4+…+log⁡(2n−1))=1n(0+log⁡21+log⁡22+…+log⁡(2n−1))=1nlog⁡(21⋅22⋅…⋅2n−1)=1nlog⁡(21+2+⋯+(n−1))=log⁡2n⋅(1+2+⋯+(n−1))=log⁡2n⋅n(n−1)2=(n−1)log⁡22\begin{aligned}\log k &= \dfrac{1}{n} \left( \log 1 + \log 2 + \log 4 + \ldots + \log (2^{n-1}) \right) \\[1em]&= \dfrac{1}{n} \left( 0 + \log 2^1 + \log 2^2 + \ldots + \log (2^{n-1}) \right) \\[1em]&= \dfrac{1}{n} \log( 2^1 \cdot 2^2 \cdot \ldots \cdot 2^{n-1}) \\[1em]&= \dfrac{1}{n} \log \left(2^{1 + 2 + \cdots + (n-1)} \right) \\[1em]&= \dfrac{\log 2}{n} \cdot (1 + 2 + \cdots + (n-1)) \\[1em]&= \dfrac{\log 2}{n} \cdot \dfrac{n (n-1)}{2} = \dfrac{(n-1) \log 2}{2} \end{aligned}

Therefore,

k=10((n−1)log⁡22)=(10log⁡2)n−12=2n−12=(2)n−1k = 10^{\left(\dfrac{(n-1) \log 2}{2}\right)} = (10^{\log 2})^{\dfrac{n-1}{2}} = 2^{\dfrac{n-1}{2}} = \left( \sqrt{2} \right)^{n-1}

The answer is d.