# AM-GM inequality in a specific question

If $k_1$ and $k_2$ are the least values of the functions $f(x)=2log_{10}(x)-log_x(0.01)$ and $g(x)=e^x+e^{-x}$.Then what is the value of $\frac{k_1}{k_2}$?

I found the solution as :-

This uses the concept that A.M is greater than or equal to G.M. But the problem is I do not know how to apply this concept and how has that been applied in the answer to this question. I'd be grateful if you explain me this concept.

• – StubbornAtom Oct 11 '16 at 7:52

If I understand your question correctly, you're asking how they get $f(x) \geq 4$ using AM-GM. AM-GM states that for non-negative real numbers $a,b$ $\frac{a+b}{2} \geq \sqrt{ab}$
$\frac{2\log x}{\log 10} + \frac{2\log 10}{\log x} \geq 2\sqrt{\frac{2\log x}{\log 10}\frac{2\log 10}{\log x}} = 2 \cdot 2$