Question

Find ${{\log }_{e}}(x+1)-{{\log }_{e}}(x-1)=$
A. \[2\left[ x+\dfrac{{{x}^{3}}}{3}+\dfrac{{{x}^{5}}}{5}+.....\infty \right]\]
B. \[\left[ x+\dfrac{{{x}^{3}}}{3}+\dfrac{{{x}^{5}}}{5}+.....\infty \right]\]
C. \[2\left[ \dfrac{1}{x}+\dfrac{1}{3{{x}^{3}}}+\dfrac{1}{5{{x}^{5}}}+...\infty \right]\]
D. \[\left[ \dfrac{1}{x}+\dfrac{1}{3{{x}^{3}}}+\dfrac{1}{5{{x}^{5}}}+...\infty \right]\]

Answer 1

Hint: In this question, we are to find the subtraction of two logarithmic functions. For this, we need to know the logarithm properties. By applying them to the given logarithmic expression, we get the required value and its expansion.



Formula Used:Logarithmic series:
If $x\in R$ and $\left| x \right|<1$, then the expansion is
${{\log }_{e}}(1+x)=x-\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{4}}}{4}+...$
If $x\in R$ and $\left| x \right|<1$, then the expansion is
${{\log }_{e}}(1-x)=-x-\dfrac{{{x}^{2}}}{2}-\dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{4}}}{4}-...$
Some important formulae:
If $\left| x \right|<1$ then ${{\log }_{e}}\dfrac{1+x}{1-x}=2\left( x+\dfrac{{{x}^{3}}}{3}+\dfrac{{{x}^{5}}}{5}+... \right)$
If $x>1$ then ${{\log }_{e}}\dfrac{x+1}{x-1}=2\left( \dfrac{1}{x}+\dfrac{1}{3{{x}^{3}}}+\dfrac{1}{5{{x}^{5}}}+... \right)$



Complete step by step solution:Given that the required subtraction between two logarithmic functions is
${{\log }_{e}}(x+1)-{{\log }_{e}}(x-1)\text{ }...(1)$
We have the property of logarithm as
$\log a-\log b=\log \left( \dfrac{a}{b} \right)$
So, by applying this to (1), we get
${{\log }_{e}}(x+1)-{{\log }_{e}}(x-1)={{\log }_{e}}\left( \dfrac{x+1}{x-1} \right)$
On simplifying, we get
$\begin{align}
  & {{\log }_{e}}(x+1)-{{\log }_{e}}(x-1)={{\log }_{e}}\left( \dfrac{x+1}{x-1} \right) \\
 & \Rightarrow {{\log }_{e}}\left( \dfrac{1+\dfrac{1}{x}}{1-\dfrac{1}{x}} \right)\text{ }...(2) \\
\end{align}$
But we have,
${{\log }_{e}}\dfrac{1+x}{1-x}=2\left( x+\dfrac{{{x}^{3}}}{3}+\dfrac{{{x}^{5}}}{5}+... \right)$
Here, if we consider $x=\dfrac{1}{x}$, we get
$\begin{align}
  & {{\log }_{e}}\dfrac{1+x}{1-x}=2\left( x+\dfrac{{{x}^{3}}}{3}+\dfrac{{{x}^{5}}}{5}+... \right) \\
 & \Rightarrow {{\log }_{e}}\dfrac{1+\dfrac{1}{x}}{1-\dfrac{1}{x}}=2\left( \dfrac{1}{x}+\dfrac{{{\left( {}^{1}/{}_{x} \right)}^{3}}}{3}+\dfrac{{{\left( {}^{1}/{}_{x} \right)}^{5}}}{5}+... \right) \\
\end{align}$
On simplifying the above expansion, we get
${{\log }_{e}}\dfrac{1+\dfrac{1}{x}}{1-\dfrac{1}{x}}=2\left( \dfrac{1}{x}+\dfrac{1}{3{{x}^{3}}}+\dfrac{1}{5{{x}^{5}}}+... \right)\text{ }...(3)$
So, from (2) and (3), the given expression at (1) becomes
${{\log }_{e}}(x+1)-{{\log }_{e}}(x-1)={{\log }_{e}}\dfrac{1+\dfrac{1}{x}}{1-\dfrac{1}{x}}=2\left( \dfrac{1}{x}+\dfrac{1}{3{{x}^{3}}}+\dfrac{1}{5{{x}^{5}}}+... \right)$



Option ‘C’ is correct



Note: Here, the given question is a direct formula. For solving such formulae, we just need to know the basic properties of logarithms. Here we use the division property of the logarithm. That means the given subtraction operation got changed to the division operation. By applying the predefined formula by changing only the variable, we get the required answer. For this type of question, simply we need to apply the properties for changing the sign and the variables in the formulae we already have.