2730.32 – Nothing but Logs


Suppose that logk\log k is the arithmetic mean of log1,log2,log4,,log(2n1)\log 1, \log 2, \log 4, \ldots , \log (2^{n-1}). Then which of these is true:

  1. k=2n(n1)2k = 2^{\frac{n (n-1)}{2}}
  2. log2k=n(n1)2\log_2 k = \frac{n (n-1)}{2}
  3. logk4=41n\log_k 4 = \frac{4}{1-n}
  4. k=(2)n1k = (\sqrt{2})^{n-1}
  5. k=2n1n22nk = \frac{2^n - 1}{n^2 2^n}

Solution

Using the laws of logs and exponents:

logk=1n(log1+log2+log4++log(2n1))=1n(0+log21+log22++log(2n1))=1nlog(21222n1)=1nlog(21+2++(n1))=log2n(1+2++(n1))=log2nn(n1)2=(n1)log22\begin{aligned} \log k &=& \frac{1}{n} \left( \log 1 + \log 2 + \log 4 + \ldots + \log (2^{n-1}) \right) \\ &=& \frac{1}{n} \left( 0 + \log 2^1 + \log 2^2 + \ldots + \log (2^{n-1}) \right) \\ &=& \frac{1}{n} \log( 2^1 \cdot 2^2 \cdot \ldots \cdot 2^{n-1}) \\ &=& \frac{1}{n} \log \left(2^{1 + 2 + \cdots + (n-1)} \right) \\ &=& \frac{\log 2}{n} \cdot (1 + 2 + \cdots + (n-1)) \\ &=& \frac{\log 2}{n} \cdot \frac{n (n-1)}{2} = \frac{(n-1) \log 2}{2} \end{aligned}

Therefore,

k=10(n1)log22=(10log2)n12=2n12=(2)n1.k = 10^{\frac{(n-1) \log 2}{2}} = (10^{\log 2})^{\frac{n-1}{2}} = 2^{\frac{n-1}{2}} = \left( \sqrt{2} \right)^{n-1}.

The answer is D.