Question

The least value of the expression $2{\log }_{10}x-{\log }_x\left(0.01\right)$, for $x>1$, is

• A. $1$
• B. $2$
• C. $4$
• D. $8$

Answer 1

Answer: (C)

$4$

We know that $A.M.\ge G.M.$

$⟹\frac{a+b}{2}\ge \sqrt{ab}$

Let $a=2{\log }_{10}x$ and $b=-{\log }_{x}0.01$

Therefore, $\frac{2{\log }_{10}x-{\log }_{x}0.01}{2}\ge \sqrt{\left(2{\log }_{10}x\right)(-{\log }_{x}0.01)}$

$⟹\frac{2{\log }_{10}x-{\log }_{x}0.01}{2}\ge \sqrt{\left(2{\log }_{10}x\right)(-{\log }_x{10}^{-2})}$

$⟹\frac{2{\log }_{10}x-{\log }_{x}0.01}{2}\ge \sqrt{\left(2{\log }_{10}x\right)({\log }_x{10}^{-2}{)}^{-1}}$

$⟹\frac{2{\log }_{10}x-{\log }_{x}0.01}{2}\ge \sqrt{\left(2{\log }_{10}x\right)\left({\log }_x{10}^2\right)}$

$⟹\frac{2{\log }_{10}x-{\log }_{x}0.01}{2}\ge \sqrt{\left(2\frac{\log x}{\log 10}\right)\left(\frac{\log {10}^2}{\log x}\right)}$

$⟹\frac{2{\log }_{10}x-{\log }_{x}0.01}{2}\ge \sqrt{\left(2\frac{\log x}{\log 10}\right)\left(\frac{2\log 10}{\log x}\right)}$

$⟹2{\log }_{10}x-{\log }_{x}0.01\ge 4$

Therefore the minimum value of $3{\log }_{10}x-{\log }_{x}0.01$ is $4$