Question

Find ${{\log }_{e}}\sqrt{\dfrac{1+x}{1-x}}=$
A. ${{\log }_{e}}\dfrac{1}{2}$
B. $2\left[ x+\dfrac{{{x}^{3}}}{3}+\dfrac{{{x}^{5}}}{5}+...\infty \right]$
C. $2\left[ {{x}^{2}}+\dfrac{{{x}^{4}}}{4}+\dfrac{{{x}^{6}}}{6}+...\infty \right]$
D. None of these

Answer 1

Hint: In this question, we are to find the expansion of the given logarithmic function. For this, we need to use the logarithmic functions and expansions such as ${{\log }_{e}}(1+x)$ and ${{\log }_{e}}(1-x)$. By simplifying them, we get the required expansion of the given logarithmic function.



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 logarithmic function is
${{\log }_{e}}\sqrt{\dfrac{1+x}{1-x}}$
We can rewrite this as
${{\log }_{e}}\sqrt{\dfrac{1+x}{1-x}}={{\log }_{e}}{{\left( \dfrac{1+x}{1-x} \right)}^{\dfrac{1}{2}}}=\dfrac{1}{2}{{\log }_{e}}\left( \dfrac{1+x}{1-x} \right)$
{Here, we applied the logarithm property $\log \sqrt{x}=\dfrac{1}{2}\log x$}.
On simplifying, we get
$\dfrac{1}{2}{{\log }_{e}}\left( \dfrac{1+x}{1-x} \right)=\dfrac{1}{2}\left[ {{\log }_{e}}(1+x)-{{\log }_{e}}(1-x) \right]\text{ }...(1)$
{Here, we applied the logarithm property $\log (\dfrac{a}{b})=\log (a)-\log (b)$, which is said to be the division rule for logarithms.}
Since we know that
${{\log }_{e}}(1+x)=x-\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{4}}}{4}+...$
And
${{\log }_{e}}(1-x)=-x-\dfrac{{{x}^{2}}}{2}-\dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{4}}}{4}-...$
On substituting these two in (1), we get
$\begin{align}
  & \dfrac{1}{2}{{\log }_{e}}\left( \dfrac{1+x}{1-x} \right)=\dfrac{1}{2}\left[ {{\log }_{e}}(1+x)-{{\log }_{e}}(1-x) \right] \\
 & \Rightarrow \dfrac{1}{2}\left[ \left( x-\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{4}}}{4}+... \right)-\left( -x-\dfrac{{{x}^{2}}}{2}-\dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{4}}}{4}-... \right) \right] \\
\end{align}$
On simplifying the above sequences, we get
\[\begin{align}
  & \Rightarrow \dfrac{1}{2}\left[ \left( x-\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{4}}}{4}+... \right)-\left( -x-\dfrac{{{x}^{2}}}{2}-\dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{4}}}{4}-... \right) \right] \\
 & \Rightarrow \dfrac{1}{2}\left[ x-\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{4}}}{4}+...+x+\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{3}+\dfrac{{{x}^{4}}}{4}+... \right] \\
 & \Rightarrow \dfrac{1}{2}\left[ 2x+\dfrac{2{{x}^{3}}}{3}+\dfrac{2{{x}^{5}}}{5}+.... \right] \\
 & \Rightarrow x+\dfrac{{{x}^{3}}}{3}+\dfrac{{{x}^{5}}}{5}+.... \\
\end{align}\]
Therefore, the required logarithmic expansion for the given logarithmic function is
${{\log }_{e}}\sqrt{\dfrac{1+x}{1-x}}=x+\dfrac{{{x}^{3}}}{3}+\dfrac{{{x}^{5}}}{5}+....$



Option ‘D’ is correct



Note: Here, we need to remember that, the given logarithm function is in the form of $\log \sqrt{x}$. So, we can write it as $\log \sqrt{x}=\dfrac{1}{2}\log x$. By applying this property, the given logarithmic function is evaluated. Here we also applied another property of logarithm. I.e., $\log (\dfrac{a}{b})=\log (a)-\log (b)$. If we know the basic logarithmic properties and logarithmic series, we can evaluate such types of questions.