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Question Number 63862 by gunawan last updated on 10/Jul/19

$$\mathrm{If}\mathrm{the}3\mathrm{rd}\mathrm{term}\mathrm{in}\mathrm{the}\mathrm{expansion}\mathrm{of}$$$${[x+x^{{\log }_{10}x}]}^5\mathrm{is}{10}^6,\mathrm{then}x\mathrm{may}\mathrm{be}$$

Commented by kaivan.ahmadi last updated on 10/Jul/19

$$Firstwenoticethat$$$$ify=x^{{log}_{10}x}\Rightarrow logy={log}_{10}x.{log}_{10}x={\left({log}_{10}x\right)}^2\Rightarrow$$$$y={10}^{{\left(logx\right)}^2}$$$$T_3=\left(\begin{array}{c}5\\ 2\end{array}\right)x^3\times {\left({10}^{{\left(logx\right)}^2}\right)}^2={10}^6\Rightarrow$$$$10x^3\times {10}^{2{\left(logx\right)}^2}={10}^6\Rightarrow$$$$$$$$x={10}^k\Rightarrow {10}^{3k+1}\times {10}^{2k^2}={10}^6\Rightarrow$$$$2k^2+3k+1=6\Rightarrow 2k^2+3k-5=0$$$$(sumofcoffecient=0)$$$$k=1,k=\frac{-5}{2}$$$$$$

Commented by kaivan.ahmadi last updated on 10/Jul/19

$$x=10,{10}^{\frac{-5}{2}}=\frac{1}{{10}^{\frac{5}{2}}}=\frac{1}{100\sqrt{2}}=\frac{\sqrt{2}}{200}$$

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