If the fourth term in the Binomial expansion | vedclass

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Question

${T_4} = {T_{3 + 1}} = \left( {\frac{6}{3}} ight){\left( {\frac{2}{x}} ight)^3} \cdot {\left( {{x^{{{\log }_8}x}}} ight)^3}$

$20 	imes 8^{7}

Vedclass · Accepted Answer

$8^2$

Vedclass · Answer

${T_4} = {T_{3 + 1}} = \left( {\frac{6}{3}} ight){\left( {\frac{2}{x}} ight)^3} \cdot {\left( {{x^{{{\log }_8}x}}} ight)^3}$

$20 	imes 8^{7}=\frac{160}{x^{3}} \cdot x^{3000} x$

$8^{6}=x^{\log _{2} x}-3$

$2^{18}=x^{\log _{2} x-3}$

$\Rightarrow 18=\left(\log _{2} x-3ight)\left(\log _{2} xight)$

Let $\log _{2} x=t$

$\Rightarrow t^{2}-3 t-18=0$

$\Rightarrow(t-6)(t+3)=0$

$\Rightarrow \mathrm{t}=6,-3$

$\log _{2} x=6$

$ \Rightarrow x=2^{6}=8^{2}$

$\log _{2} x=-3$

$\Rightarrow x=2^{-3}=8^{-1}$

If the fourth term in the Binomial expansion of ${\left( {\frac{2}{x} + {x^{{{\log }_e}x}}} \right)^6}(x > 0)$ is $20\times 8^7,$ then a value of $x$ is

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