If the third term in the expansion of $${\left[ {x + {x^{{{\log }_{10}}x}}} \right]^5}$$   is $$10^6,\,$$ then $$x$$ may be

Solution :
Put $${\log _{10}}x = y,$$   the given expression becomes $${\left( {x + {x^y}} \right)^5}.$$
$$\eqalign{ & {T_3} = {\,^5}{C_2} \cdot {x^3}{\left( {{x^y}} \right)^2} = 10{x^{3 + 2y}} = {10^6}\left( {{\text{given}}} \right) \cr & \Rightarrow \left( {3 + 2y} \right){\log _{10}}x = 5{\log _{10}}10 = 5 \cr & \Rightarrow \left( {3 + 2y} \right)y = 5 \cr & \Rightarrow y = 1, - \frac{5}{2} \cr & \Rightarrow {\log _{10}}x = 1{\text{ or }}{\log _{10}}x = - \frac{5}{2} \cr & \therefore x = 10{\text{ or }}x = {\left( {10} \right)^{ - \frac{5}{2}}} \cr} $$