Binomial Theorem 1 Question 5
6. If the fourth term in the binomial expansion of $\frac{2}{x}+x^{\log _8 x^{6}}(x>0)$ is $20 \times 8^{7}$, then the value of $x$ is
(a) $8^{-2}$
(b) $8^{3}$
(c) 8
(d) $8^{2}$
(2019 Main, 9 April I)
Show Answer
Answer:
Correct Answer: 6. (d)
Solution:
- Given binomial is $\frac{2}{x}+x^{\log _8 x^{6}}$
Since, general term in the expansion of $(x+a)^{n}$ is $T _{r+1} \stackrel{n}{n} C _r x^{n-r} a^{r}$
$$ \begin{gathered} \therefore \quad T _4=T _{3+1}={ }^{6} C _3 \frac{2}{x}{ }^{6-3} \quad\left(x^{\log _8 x}\right)^{3}=20 \times 8^{7} \text { (given) } \\ \Rightarrow 20 \frac{2^{3}}{x} x^{3 \log _8 x}=20 \times 8^{7} \\ \left.\Rightarrow 2^{3} x^{\left[3\left(\log _8 x\right)-3\right]}=\left(2^{3}\right)^{7} \Rightarrow x^{\frac{3}{3} \log _2 x-3} C _3=20\right] \\ \because \log _{a^{n}}(x)=\frac{1}{n} \log _a x \text { for } x>0 ; a>0, \neq 1 \\ \Rightarrow x^{\left(\log _2 x-3\right)}=2^{18} \end{gathered} $$
On taking $\log _2$ both sides, we get
$$ \begin{aligned} & \quad\left(\log _2 x-3\right) \log _2 x=18 \\ & \Rightarrow \quad\left(\log _2 x\right)^{2}-3 \log _2 x-18=0 \\ & \Rightarrow \quad\left(\log _2 x\right)^{2}-6 \log _2 x+3 \log _2 x-18=0 \\ & \Rightarrow \quad \log _2 x\left(\log _2 x-6\right)+3\left(\log _2 x-6\right)=0 \\ & \Rightarrow \quad\left(\log _2 x-6\right)\left(\log _2 x+3\right)=0 \\ & \Rightarrow \quad \log _2 x=-3,6 \\ & \Rightarrow \quad x=2^{-3}, 2^{6} \Rightarrow x=\frac{1}{8}, 8^{2} \end{aligned} $$
