If the fourth term in the Binomial expansion of ((2)/(x) + x ^ _ 8...

The general term in the binomial expansion ${\left(a + b\right)}^{n}$ is ${T}_{r + 1} = Cr n {a}^{n - r}{b}^{r} .$

Given, the fourth term in the expansion of ${\left(\dfrac{2}{x} + {x}^{\log_{8}x}\right)}^{6}$ is

${T}_{4} = 20 \times {8}^{7}$

$⇒ C3 6 {\left(\dfrac{2}{x}\right)}^{3}{\left({x}^{\log_{8}x}\right)}^{3} = 20 \times {8}^{7}$

$⇒ \dfrac{6 !}{3 ! \cdot 3 !}{\left(\dfrac{2}{x}\right)}^{3}{\left({x}^{\log_{8}x}\right)}^{3} = 20 \times {\left({2}^{3}\right)}^{7}$

$⇒ \left(\dfrac{6 \times 5 \times 4 \times 3 !}{3 \times 2 \times 1 \cdot 3 !}\right){\left(\dfrac{2}{x}\right)}^{3}{\left({x}^{\log_{8}x}\right)}^{3} = 20 \times {2}^{21}$

$⇒ \left(20\right){\left(\dfrac{2}{x}\right)}^{3}{\left({x}^{\log_{8}x}\right)}^{3} = 20 \times {2}^{21}$

$⇒ {\left[\left(\dfrac{2}{x}\right)\left({x}^{\log_{8}x}\right)\right]}^{3} = {\left({2}^{7}\right)}^{3}$

$⇒ \left(\dfrac{2}{x}\right)\left({x}^{\log_{8}x}\right) = {2}^{7}$

$⇒ \left(\dfrac{1}{x}\right)\left({x}^{\log_{8}x}\right) = {2}^{6}$

$⇒ {x}^{\log_{8}x - 1} = {8}^{2}$

Taking $log$ both side to the base $8$

$⇒ \log_{8}\left({x}^{\log_{8}x - 1}\right) = \log_{8}\left({8}^{2}\right)$

Using $\log_{a}{m}^{n} = n{\log}_{a}m,$ we get

$⇒ ( \log_{8}x - 1 )\left(\log_{8}x\right) = 2$

$⇒ \left(t - 1\right)\left(t\right) = 2$ (Let $\log_{8}\left(x\right) = t$)

$⇒ {t}^{2} - t - 2 = 0$

$⇒ t = 2$ or $- 1$

$⇒ \log_{8}x = 2$ or $- 1$

If $\log_{a}x = b, ⇒ x = {a}^{b},$

$⇒ x = {8}^{2}$ or ${8}^{- 1}$

$⇒ x = 64$ or $\dfrac{1}{8} .$