Question

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

• A. $10$
• B. $2$
• C. $-0.01$
• D. None of these

Answer 1

The correct option is D

None of these

Step 1: Simplify the given expression

The given expression is,

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

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

$=2{\log }_{10}x-\left(-2\right){\log }_x\left(10\right)$ …$\left[\because lo{g}_a{m}^n=nlo{g}_{a}m\right]$

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

$=2{\log }_{10}x+2\left(\frac{{\log }_{10}10}{{\log }_{10}x}\right)$ …$[\because lo{g}_{a}x=\frac{{\log }_{c}a}{{\log }_{c}x}]$

$=2{\log }_{10}x+2\left(\frac{1}{{\log }_{10}x}\right)$ …$\left[\because lo{g}_{10}10=1\right]$

$=2\left[{\log }_{10}x+\left(\frac{1}{{\log }_{10}x}\right)\right]$

Step 2: Apply the relation between AM and GM

As we know, the AM (Arithmetic Mean) and the GM (Geometric Mean) of a set of data is related as,

$AM\ge GM$

Then, for ${\log }_{10}x$ and $\frac{1}{{\log }_{10}x}$, we have,

$\frac{{\log }_{10}x+\frac{1}{{\log }_{10}x}}{2}\ge {\left({\log }_{10}x\times \frac{1}{{\log }_{10}x}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

$\Rightarrow$ $\frac{{\log }_{10}x+\frac{1}{{\log }_{10}x}}{2}\ge {\left(1\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

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

$\Rightarrow$ $2\left[{\log }_{10}x+\frac{1}{{\log }_{10}x}\right]\ge 4$ [multiplying by $4$ on both sides]

Thus, the lease value of the given function is $4$.
Hence, option (D) is the correct option.