# Find the minimum value of log_2^2 2 + log_ 2^3 2^2 + log_ 2^4 2^3.............log_2^n 2^(n-1)?

## Find the minimum value of ${\log}_{{2}^{2}} 2 + {\log}_{{2}^{3}} {2}^{2} + {\log}_{{2}^{4}} {2}^{3.} \ldots \ldots \ldots \ldots {\log}_{{2}^{n}} {2}^{n - 1}$

Apr 24, 2017

The Reqd. Minimum Value =$\left(n - 1\right) {\left(\frac{1}{n}\right)}^{\frac{1}{n - 1}} .$

#### Explanation:

We know the Change of Base Rule : ${\log}_{c} a = {\log}_{b} \frac{a}{\log} _ b c .$

Using $b \ne 1$ as common base to all $\log ,$ we have,

The Expression$= {\log}_{4} 2 + {\log}_{8} 4 + {\log}_{16} 8 + \ldots + {\log}_{{2}^{n}} {2}^{n - 1} ,$

$= {\log}_{b} \frac{2}{\log} _ b 4 + {\log}_{b} \frac{4}{\log} _ b 8 + {\log}_{b} \frac{8}{\log} _ b 16 + \ldots + {\log}_{b} {2}^{n - 1} / {\log}_{b} \left({2}^{n}\right) ,$

$= {\log}_{b} \frac{2}{\log} _ b {2}^{2} + {\log}_{b} {2}^{2} / {\log}_{b} {2}^{3} + {\log}_{b} {2}^{3} / {\log}_{b} {2}^{4} + \ldots + {\log}_{b} {2}^{n - 1} / {\log}_{b} \left({2}^{n}\right) ,$

$= {\log}_{b} \frac{2}{2 {\log}_{b} 2} + \frac{2 {\log}_{b} 2}{3 {\log}_{b} 2} + \frac{3 {\log}_{b} 2}{4 {\log}_{b} 2} + \ldots + \frac{\left(n - 1\right) {\log}_{b} 2}{n {\log}_{b} 2} ,$

$= \frac{1}{2} + \frac{2}{3} + \frac{3}{4} + \ldots + \frac{n - 1}{n} \ldots . \left(\ast\right)$

Now, we use the Arithmetic Mean-Geometric Mean Inequality, i.e.,

$A M \ge G M$

Observe that the AM of $\left(n - 1\right) + v e \text{ Nos. of } \left(\ast\right)$ is,

$\frac{1}{n - 1} \left\{\frac{1}{2} + \frac{2}{3} + \frac{3}{4} + \ldots + \frac{n - 1}{n}\right\} ,$ and, their GM is,

${\left\{\left(\frac{1}{2}\right) \left(\frac{2}{3}\right) \left(\frac{3}{4}\right) \ldots \left(\frac{n - 1}{n}\right)\right\}}^{\frac{1}{n - 1}} = {\left(\frac{1}{n}\right)}^{\frac{1}{n - 1}} .$

$\therefore , \frac{1}{n - 1} \left\{\frac{1}{2} + \frac{2}{3} + \frac{3}{4} + \ldots + \frac{n - 1}{n}\right\} \ge {\left(\frac{1}{n}\right)}^{\frac{1}{n - 1}} , i . e . ,$

$\left\{\frac{1}{2} + \frac{2}{3} + \frac{3}{4} + \ldots + \frac{n - 1}{n}\right\} \ge \left(n - 1\right) {\left(\frac{1}{n}\right)}^{\frac{1}{n - 1}} .$

Therefore, the Reqd. Minimum Value =$\left(n - 1\right) {\left(\frac{1}{n}\right)}^{\frac{1}{n - 1}} .$

Enjoy Maths.!

Apr 24, 2017

${\sum}_{k = 1}^{n} {\log}_{{2}^{k}} {2}^{k - 1} = n + {\sum}_{k = 1}^{n} {\left(- 1\right)}^{k} \frac{1}{k} \left(\begin{matrix}n \\ k\end{matrix}\right)$

#### Explanation:

${\log}_{{a}^{b}} {a}^{c} = \log {a}^{c} / \left(\log {a}^{b}\right) = \frac{c}{b}$

so

${\sum}_{k = 1}^{n} {\log}_{{2}^{k}} {2}^{k - 1} = {\sum}_{k = 1}^{n} \frac{k - 1}{k} = n - {\sum}_{k = 1}^{n} \frac{1}{k}$

we know that

$1 + x + {x}^{2} + \cdots + {x}^{n - 1} = \frac{{x}^{n} - 1}{x - 1}$

so

${\int}_{0}^{1} \frac{{x}^{n} - 1}{x - 1} \mathrm{dx} = - {\int}_{0}^{1} \frac{{\left(1 - y\right)}^{n} - 1}{y} \mathrm{dy} = - {\sum}_{k = 1}^{n} {\left(- 1\right)}^{k} \frac{1}{k} \left(\begin{matrix}n \\ k\end{matrix}\right)$

so finally

${\sum}_{k = 1}^{n} {\log}_{{2}^{k}} {2}^{k - 1} = n + {\sum}_{k = 1}^{n} {\left(- 1\right)}^{k} \frac{1}{k} \left(\begin{matrix}n \\ k\end{matrix}\right)$