Algebraic Rooting #5

We see that $2log_{10}x-log_x(10)^{(-2)}=2log_{10}x+2log_x10$ $=2log_{10}x+2\dfrac{1}{log_{10}}x$ $=2\bigg(log_{10}x+\dfrac{1}{log_{10}x}\bigg)$ Using $AM\ge{GM}$ ,we get :- $\dfrac{log_{10}x+\dfrac{1}{log_{10}x}}{2}\ge\bigg(log_{10}x.\dfrac{1}{log_{10}x}\bigg)^{1/2}$ $\implies log_{10}x+\dfrac{1}{log_{10}x}\ge{2}$ or $2log_{10}x-log_x(0.01)\ge{4}$ So ,least value is $\boxed{4}$ . $\ddot\smile$

$(a-1)^{2} \geq {0},\ \forall {a} \in \mathbb{R^{+}}$

$\implies a^{2}-2a+1 \geq {0} \implies a^{2}+1 \geq 2a$

$\implies \boxed{a+\frac{1}{a} \geq 2} \text{positive real numbers property}$

$E(x)=2 \cdot{log_{10}{x}}-log_{x}{0.01}=2 \cdot(log_{10}{x}+log_{x}{10})$

$\boxed{y=log_{x}{10}} \rightarrow x^{y}=10 \rightarrow (x^{y})^{\frac{1}{y}}=10^{\frac{1}{y}} \rightarrow x=10^{\frac {1}{y}} \rightarrow \boxed{log_{10}{x}=\frac{1}{y}}$

Substituting, comes:

$E(x)=2 \cdot {\displaystyle{[\frac{1}{y(x)}+y(x)]}}$

Hence, verified existential conditions, minimum is:

$min[E(x)]=2 \cdot{2}=\boxed{4}$