The Easy Minimum!...#2

We have,

$\displaystyle 2\log _{ 10 }{ x } -\log _{ x }{ 0.01 }$

Which is equivalent to:

$\displaystyle \Rightarrow 2\log _{ 10 }{ x } +2\log _{ x }{ 10 }$

$\displaystyle \Rightarrow 2(\log _{ 10 }{ x } +\log _{ x }{ 10 })$

$\displaystyle \Rightarrow 2(\log _{ 10 }{ x } +\frac{1}{\log _{10}{x }})$

Set $\displaystyle \log _{ 10 }{ x }$ as t. Then:

$\displaystyle 2(t + \frac{1}{t})$

Appyling AM-GM Inequality to it, we get:

$\displaystyle \frac{(t+\frac{1}{t})}{2} \ge \sqrt{t/t}$

$\displaystyle (t+\frac{1}{t}) \ge 2$

Hence the minimum value of the given expression is $\displaystyle 2*2 = \boxed{4}$

$\large f(X)=\dfrac{2LnX}{Ln10}-\dfrac{Ln0.01}{LnX}\\ \large f'(X)=\dfrac{2}{X*Ln10}+\dfrac{Ln0.01}{X*(LnX)^2}= 0. ~~~~\implies~~X=10\\\large \therefore f(X)=\dfrac{2Ln10}{Ln10}-\dfrac{Ln0.01}{Ln10}= 4$