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

JEE Main · 2019 · 09 Apr 2019 · Shift 1 · Binomial Theorem

If the fourth term in the Binomial expansion of

(\frac{2}{x} + x^{l o g_{8} x})^{6} (x > 0)

20 \times 8^{7}

, then a value of

x

is:

Answer: (D)

The fourth term is

T_{4} = (3 6) (\frac{2}{x})^{3} (x^{l o g_{8} x})^{3} = 20 \cdot \frac{8}{x ^{3}} \cdot x^{3 l o g_{8} x} = \frac{160}{x ^{3}} x^{3 l o g_{8} x}

. Setting this equal to

20 \times 8^{7}

gives

x^{3 l o g_{8} x - 3} = 8^{6}

. Let

t = lo g_{2} x

, so

lo g_{8} x = \frac{t}{3}

; taking

lo g_{2}

of both sides turns the equation into

(t - 3) t = 18

, i.e.

t^{2} - 3 t - 18 = 0

, so

t = 6

t = - 3

. The value

t = 6

gives

x = 2^{6} = 8^{2}

JEE Main · 2019 · 09 Apr 2019 · Shift 1 · Single correct option