The value of x in the expression (x + x^{log_ | vedclass

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(A) The general term in the expansion of $(a + b)^n$ is $T_{r+1} = ^nC_r a^{n-r} b^r$.For the expression $(x + x^{\log_{10} x})^5$,the third term $(T_

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$10$

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(A) The general term in the expansion of $(a + b)^n$ is $T_{r+1} = ^nC_r a^{n-r} b^r$.For the expression $(x + x^{\log_{10} x})^5$,the third term $(T_3)$ corresponds to $r = 2$.$T_3 = ^5C_2 \cdot (x)^{5-2} \cdot (x^{\log_{10} x})^2 = 1,000,000$.$10 \cdot x^3 \cdot (x^{\log_{10} x})^2 = 10^6$.$x^3 \cdot x^{2 \log_{10} x} = 10^5$.Taking $\log_{10}$ on both sides:$\log_{10}(x^3 \cdot x^{2 \log_{10} x}) = \log_{10}(10^5)$.$3 \log_{10} x + 2(\log_{10} x)^2 = 5$.Let $y = \log_{10} x$,then $2y^2 + 3y - 5 = 0$.$(2y + 5)(y - 1) = 0$.So,$y = 1$ or $y = -2.5$.If $y = 1$,then $\log_{10} x = 1$,which implies $x = 10^1 = 10$.If $y = -2.5$,then $\log_{10} x = -2.5$,which implies $x = 10^{-2.5}$ (not an integer option).Thus,$x = 10$.

The value of $x$ in the expression $(x + x^{\log_{10} x})^5$,if the third term in the expansion is $1,000,000$.

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