Question Number 63862 by gunawan last updated on 10/Jul/19
![If the 3rd term in the expansion of [x+x^(log_(10) x) ]^5 is 10^6 , then x may be](https://www.tinkutara.com/question/Q63862.png)
$$\mathrm{If}\:\mathrm{the}\:\mathrm{3rd}\:\mathrm{term}\:\mathrm{in}\:\mathrm{the}\:\mathrm{expansion}\:\mathrm{of} \\ $$$$\left[{x}+{x}^{\mathrm{log}_{\mathrm{10}} {x}} \right]^{\mathrm{5}} \mathrm{is}\:\mathrm{10}^{\mathrm{6}} ,\:\mathrm{then}\:{x}\:\mathrm{may}\:\mathrm{be} \\ $$
Commented by kaivan.ahmadi last updated on 10/Jul/19
$${First}\:{we}\:{notice}\:{that} \\ $$$${if}\:\:\:{y}={x}^{{log}_{\mathrm{10}} {x}} \Rightarrow{logy}={log}_{\mathrm{10}} {x}.{log}_{\mathrm{10}} {x}=\left({log}_{\mathrm{10}} {x}\right)^{\mathrm{2}} \Rightarrow \\ $$$${y}=\mathrm{10}^{\left({logx}\right)^{\mathrm{2}} } \\ $$$${T}_{\mathrm{3}} =\begin{pmatrix}{\mathrm{5}}\\{\mathrm{2}}\end{pmatrix}{x}^{\mathrm{3}} ×\left(\mathrm{10}^{\left({logx}\right)^{\mathrm{2}} } \right)^{\mathrm{2}} =\mathrm{10}^{\mathrm{6}} \Rightarrow \\ $$$$\mathrm{10}{x}^{\mathrm{3}} ×\mathrm{10}^{\mathrm{2}\left({logx}\right)^{\mathrm{2}} } =\mathrm{10}^{\mathrm{6}} \Rightarrow \\ $$$$ \\ $$$${x}=\mathrm{10}^{{k}} \Rightarrow\mathrm{10}^{\mathrm{3}{k}+\mathrm{1}} ×\mathrm{10}^{\mathrm{2}{k}^{\mathrm{2}} } =\mathrm{10}^{\mathrm{6}} \Rightarrow \\ $$$$\mathrm{2}{k}^{\mathrm{2}} +\mathrm{3}{k}+\mathrm{1}=\mathrm{6}\Rightarrow\mathrm{2}{k}^{\mathrm{2}} +\mathrm{3}{k}−\mathrm{5}=\mathrm{0} \\ $$$$\left({sum}\:{of}\:{coffecient}=\mathrm{0}\right) \\ $$$${k}=\mathrm{1}\:\:,\:\:{k}=\frac{−\mathrm{5}}{\mathrm{2}} \\ $$$$ \\ $$
Commented by kaivan.ahmadi last updated on 10/Jul/19
$${x}=\mathrm{10}\:\:,\:\:\mathrm{10}^{\frac{−\mathrm{5}}{\mathrm{2}}} =\frac{\mathrm{1}}{\mathrm{10}^{\frac{\mathrm{5}}{\mathrm{2}}} }=\frac{\mathrm{1}}{\mathrm{100}\sqrt{\mathrm{2}}}=\frac{\sqrt{\mathrm{2}}}{\mathrm{200}} \\ $$