If fourth term in Binomial expansion of {left( {frac{2}{x} + {x^{{{log }ₑ}x}}} right)^6}(x &

English

Gujarati

Hindi

7.Binomial Theorem

hard

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

$8^3$

$8^{-2}$

$8$

$8^2$

(JEE MAIN-2019)

${T_4} = {T_{3 + 1}} = \left( {\frac{6}{3}} \right){\left( {\frac{2}{x}} \right)^3} \cdot {\left( {{x^{{{\log }_8}x}}} \right)^3}$

$20 \times 8^{7}=\frac{160}{x^{3}} \cdot x^{3000} x$

$8^{6}=x^{\log _{2} x}-3$

$2^{18}=x^{\log _{2} x-3}$

$\Rightarrow 18=\left(\log _{2} x-3\right)\left(\log _{2} x\right)$

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}$

$\log _{2} x=-3$

$\Rightarrow x=2^{-3}=8^{-1}$

Standard 11

Mathematics

normal

hard

(IIT-1993)

medium

medium

easy