Question

80. If x > 1, then the least value of 2 log x base 10 - log 0.01 base x is
1)2
2)4
3) 6
4) 8​

Answer 1

$2\log(x)-\log_x(0.01)\ \implies\ 2\log(x)-\dfrac{\log(0.01)}{\log(x)}\\ \\ \\ \implies\ 2\log(x)-\dfrac{\log(10^{-2})}{\log(x)}\ \implies\ 2\log(x)-\dfrac{-2}{\log(x)}\\ \\ \\ \implies\ 2\log(x)+\dfrac{2}{\log(x)}\ \implies\ 2\left(\log(x)+\dfrac{1}{\log(x)}\right)$

Now it seems as twice the sum of a number and its reciprocal.

It is always true that, the minimum value of the sum of a positive real number and its reciprocal is 2.

$\forall n\in\mathbb{Z^+},\ \ n+\dfrac{1}{n}\geq 2$

The equality holds if  $n=1$  here.

So, is it possible that  $\log(x)=1\ \ ?$

$\log(x)=1\ \implies\ x=10>1$

Okay, it's possible. So,

$2\left(\log(x)+\dfrac{1}{\log(x)}\right)\ \geq\ 2\times 2\ =\ \mathbf{4}$

Hence the answer is (2) 4.