If the fourth term in the Binomial expansion | vedclass

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(D) The general term in the expansion of $(a+b)^n$ is $T_{r+1} = \binom{n}{r} a^{n-r} b^r$. For the given expansion,$n=6$,$a = \frac{2}{x}$,and $b = x

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$8^2$

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(D) The general term in the expansion of $(a+b)^n$ is $T_{r+1} = \binom{n}{r} a^{n-r} b^r$. For the given expansion,$n=6$,$a = \frac{2}{x}$,and $b = x^{\log_8 x}$. The fourth term is $T_4 = T_{3+1} = \binom{6}{3} \left( \frac{2}{x} ight)^{6-3} \left( x^{\log_8 x} ight)^3 = 20 \cdot \frac{8}{x^3} \cdot x^{3 \log_8 x} = 160 \cdot x^{3 \log_8 x - 3}$. Given $T_4 = 20 	imes 8^7$,we have $160 \cdot x^{3 \log_8 x - 3} = 20 \cdot 8^7$. $8 \cdot x^{3 \log_8 x - 3} = 8^7 \Rightarrow x^{3 \log_8 x - 3} = 8^6$. Taking $\log_8$ on both sides: $(3 \log_8 x - 3) \log_8 x = \log_8 (8^6) = 6$. Let $t = \log_8 x$. Then $(3t - 3)t = 6$ $\Rightarrow 3t^2 - 3t - 6 = 0$ $\Rightarrow t^2 - t - 2 = 0$. $(t-2)(t+1) = 0$,so $t=2$ or $t=-1$. If $t=2$,$\log_8 x = 2 \Rightarrow x = 8^2 = 64$. If $t=-1$,$\log_8 x = -1 \Rightarrow x = 8^{-1} = 1/8$.

If the fourth term in the Binomial expansion of ${\left( {\frac{2}{x} + {x^{\log_8 x}}} \right)^6}$ for $x > 0$ is $20 \times 8^7$,then a value of $x$ is

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