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

Solution :
$$\eqalign{ & {\text{Let }}y = 2{\log _{10}}x - {\log _x}0.01 \cr & = 2{\log _{10}}x - \frac{{{{\log }_{10}}0.01}}{{{{\log }_{10}}x}} = 2{\log _{10}}x + \frac{2}{{{{\log }_{10}}x}} \cr & = 2\left[ {{{\log }_{10}}x + \frac{2}{{{{\log }_{10}}x}}} \right] \cr & \left[ {{\text{Here }}x > 1\,\, \Rightarrow \,{{\log }_{10}}x > 0} \right] \cr} $$
Now since sum ofa real $$+ ve$$  number and its reciprocal is always greater than or equal to 2.
$$\eqalign{ & \therefore \,\,\,y \geqslant 2 \times 2 \cr & \Rightarrow \,y \geqslant 4 \cr} $$
∴ Least value of $$y$$ is $$4.$$