Question

The Coefficient of $x^n$ in the expansion of ${\log }_a\left(1+x\right)$ is

• A. $\frac{{\left(-1\right)}^{n-1}}{n}$
• B. $\left(\frac{-1^{n-1}}{n}\right){\log }_{a}e$
• C. $\left(\frac{-1^{n-1}}{n}\right){\log }_{e}a$
• D. $\left(\frac{-1^n}{n{\log }_{a}e}\right)$

Answer 1

The correct option is B

$\left(\frac{-1^{n-1}}{n}\right){\log }_{a}e$

Explanation for the correct answer:

Obtaining the coefficient of the given expression:

Given expression is ${\log }_a\left(1+x\right)$

Using the Logarithm Identity to identify the coefficient,

${\log }_a\left(1+x\right)=\frac{{\log }_e\left(1+x\right)}{{\log }_{e}a}$

Now, using the expansion series of ${\log }_e\left(1+x\right)=\left(x-\frac{x^2}{2}+\frac{x^3}{3}-....\right)$ we get,

${\log }_a\left(1+x\right)=\frac{\left(x-\frac{x^2}{2}+\frac{x^3}{3}-....\right)}{{\log }_{e}a}$

So, the coefficient of $x^n$ can be written as $\left(\frac{-1^{n-1}}{n}\right)\frac{1}{{\log }_{e}a}$ using the Logarithm Property $\frac{1}{{\log }_{e}a}={\log }_{a}e$

So, the coefficient of $x^n$ in the expansion of ${\log }_a\left(1+x\right)$ is $=\left(\frac{-1^{n-1}}{n}\right){\log }_{a}e$

Hence, option (B) is correct.