Question

The coefficient of ${{x}^{n}}$ in the expansion of ${{\log }_{a}}\left( 1+x \right)$ is
A. $\dfrac{{{\left( -1 \right)}^{n-1}}}{n}$
B. $\dfrac{{{\left( -1 \right)}^{n-1}}}{n}{{\log }_{a}}e$
C. $\dfrac{{{\left( -1 \right)}^{n-1}}}{n}{{\log }_{e}}a$
D. $\dfrac{{{\left( -1 \right)}^{n}}}{n}{{\log }_{a}}e$

Answer 1

Hint: We need to first convert the base of the expression of ${{\log }_{a}}\left( 1+x \right)$ to $e$. Then we can apply the expansion formula of ${{\log }_{e}}\left( 1+x \right)=x-\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{3}-.....+{{\left( -1 \right)}^{n-1}}\dfrac{{{x}^{n}}}{n}+.....\infty $. The coefficient of ${{x}^{n}}$ in that expansion gets divided by ${{\log }_{e}}a$. We find the final form.

Complete step by step solution:
We have to use the formula for infinite expansion series of logarithm and the base change for logarithm to find the coefficient of ${{x}^{n}}$ in the expansion of ${{\log }_{a}}\left( 1+x \right)$.
We know that the infinite series of logarithm gives
${{\log }_{e}}\left( 1+x \right)=x-\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{3}-.....+{{\left( -1 \right)}^{n-1}}\dfrac{{{x}^{n}}}{n}+.....\infty $.
We can understand that the base for the logarithmic expansion has to be $e$. For our given expression ${{\log }_{a}}\left( 1+x \right)$, the base is $a$. We need to first change the base.
We have the logarithmic formula of base change as ${{\log }_{n}}b=\dfrac{{{\log }_{m}}b}{{{\log }_{m}}n}$.
For our base change we take $m=e;b=\left( 1+x \right);n=a$ for the formula ${{\log }_{n}}b=\dfrac{{{\log }_{m}}b}{{{\log }_{m}}n}$.
The change gives ${{\log }_{a}}\left( 1+x \right)=\dfrac{{{\log }_{e}}\left( 1+x \right)}{{{\log }_{e}}a}$.
Now we can apply the expansion formula in the numerator for ${{\log }_{e}}\left( 1+x \right)$.
From the formula of ${{\log }_{e}}\left( 1+x \right)=x-\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{3}-.....+{{\left( -1 \right)}^{n-1}}\dfrac{{{x}^{n}}}{n}+.....\infty $, we have the coefficient of ${{x}^{n}}$ as $\dfrac{{{\left( -1 \right)}^{n-1}}}{n}$.
In the case of ${{\log }_{a}}\left( 1+x \right)=\dfrac{{{\log }_{e}}\left( 1+x \right)}{{{\log }_{e}}a}$, we have to divide every term in the expansion of ${{\log }_{e}}\left( 1+x \right)$ with ${{\log }_{e}}a$.
The coefficient of ${{x}^{n}}$ in the expansion of ${{\log }_{a}}\left( 1+x \right)$ also gets divided by ${{\log }_{e}}a$.
 The final coefficient is $\dfrac{{{\left( -1 \right)}^{n-1}}}{n{{\log }_{e}}a}$.
We can convert the logarithm in the multiplied form where we have the formula that ${{\log }_{n}}b=\dfrac{1}{{{\log }_{b}}n}$. Applying this we get $\dfrac{{{\left( -1 \right)}^{n-1}}}{n{{\log }_{e}}a}=\dfrac{{{\left( -1 \right)}^{n-1}}}{n}{{\log }_{a}}e$.
The correct option is B.

Note: We have to be careful about the base change. The formula can only be applied for the exponential base. We also have to check that the logarithm is defined in that change of base.