If the third term in the expansion of [x+ ^ _ 10 x ]^ 5 is 10^ 6 ,...

Put $\log _{10} x=y,$ the given expression becomes \left(x+x^{y}\right)^{5}\mathrm{T}_{3}={ }^{5} \mathrm{C}_{2} \cdot \mathrmx^{3}\left(\mathrmx^{\mathrm{y}}\right)^{2}=10 \mathrmx^{3+2 \mathrm{y}}=10^{6}$(given)$\Rightarrow(3+2 \mathrm{y}) \log _{10} \mathrm{x}=5 \log _{10} 10=5\Rightarrow(3+2 \mathrm{y}) \mathrm{y}=5\Rightarrow \mathrm{y}=1,-\frac{5}{2}\Rightarrow \log _{10} \mathrm{x}=1$or$\log _{10} \mathrm{x}=-\frac{5}{2}\therefore \mathrm{x}=10 or \mathrm{x}=(10)^{-5 / 2}