Question

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

• A. $10$
• B. $2$
• C. $-0.01$
• D. $4$

Answer 1

Answer: (D)

$4$

$2{\log }_{10}x-{\log }_x\left(0.01\right)$

$=2{\log }_{10}x-{\log }_x\left({10}^{-2}\right)$

$=2{\log }_{10}x-(-2){\log }_{x}10$

$=2{\log }_{10}x+2{\log }_{x}10$

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

$\therefore 2{\log }_{10}x-{\log }_x\left(0.01\right)$$=2({\log }_{10}x+\frac{1}{{\log }_{10}x})\cdots \cdots \left(1\right)$

Now, use $A.M.\ge G.M.$ that is arithmetic mean is greater than or equal to geometric mean of two numbers.

$\Rightarrow \frac{({\log }_{10}x+\frac{1}{{\log }_{10}x})}{2}\ge \sqrt{{\log }_{10}x\times \frac{1}{{\log }_{10}x}}$

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

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

$2({\log }_{10}x+\frac{1}{{\log }_{10}x})\ge 4$

So, from $1$ minimum value of $2{\log }_{10}x-{\log }_x\left(0.01\right)$ is $4$

Hence, option $D$ is correct.