Question

In the expansion of $2{\log }_{e}x–{\log }_e(x+1)–{\log }_e(x-1)$ the coefficient of $x^{-4}$is:

• A. $\frac{1}{2}$
• B. $-1$
• C. $1$
• D. None of these

Answer 1

The correct option is A

$\frac{1}{2}$

Determine the coefficient of $x^{-4}$

We can rewrite the $(x+1)\mathrm{as}\left(1+\frac{1}{x}\right)x\mathrm{and}(x-1)\mathrm{as}\left(1-\frac{1}{x}\right)x$, therefore on substituting we have,

$$\begin{gathered} \Rightarrow 2{\log }_{e}x-{\log }_e\left(\left(1+\frac{1}{x}\right)x\right)–{\log }_e\left(\left(1-\frac{1}{x}\right)x\right) \\ \Rightarrow 2{\log }_{e}x–\left[{\log }_e\left(1+\frac{1}{x}\right)+{\log }_{e}x\right]–\left[{\log }_e\left(1-\frac{1}{x}\right)+{\log }_{e}x\right]\left[\because \log ab=\log a+\log b\right] \\ \Rightarrow -{\log }_e\left(1+\frac{1}{x}\right)-{\log }_e\left(1+\frac{-1}{x}\right) \\ \Rightarrow -\left[\left(\frac{1}{x}-\frac{1}{2x^2}+\frac{1}{3x^3}-\frac{1}{4x^4}+.......\right)+\left(-\frac{1}{x}-\frac{1}{2x^2}-\frac{1}{3x^3}-\frac{1}{4x^4}-.......\right)\right]\left[\because \log (1+a)=a-\frac{a^2}{2}+\frac{a^3}{3}-\frac{a^4}{4}+..........\right] \\ \Rightarrow \left(\frac{1}{4x^4}+\frac{1}{4x^4}\right)[\text{Neglecting other terms]} \\ \Rightarrow \frac{1}{2}{x}^{-4} \end{gathered}$$Hence, option A is the correct answer.