Question

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

Answer 1

Hint: In this question, we are to find the expression of a series. For this, we need to apply the logarithmic series formulae. By rewriting the expression into the form of a logarithmic series, we can evaluate the given expression.



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 series is
${{\log }_{e}}\left[ {{(1+x)}^{1+x}}{{(1-x)}^{1-x}} \right]$
By applying the properties of logarithm $\log {{(a)}^{n}}=n\log a$ and $\log xy=\log x+\log y$, we can rewrite the given series as
$\begin{align}
  & {{\log }_{e}}\left[ {{(1+x)}^{1+x}}{{(1-x)}^{1-x}} \right] \\
 & \Rightarrow (1+x){{\log }_{e}}(1+x)+(1-x){{\log }_{e}}(1-x)\text{ }...(1) \\
\end{align}$
But we have logarithmic series as
${{\log }_{e}}(1+x)=x-\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{4}}}{4}+...(2)$
$\begin{align}
  & {{\log }_{e}}(1-x)=-x-\dfrac{{{x}^{2}}}{2}-\dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{4}}}{4}-... \\
 & \Rightarrow -{{\log }_{e}}(1-x)=x+\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{3}+\dfrac{{{x}^{4}}}{4}...(3) \\
\end{align}$
Thus, on substituting these values (2) and (3) into the given logarithmic function at (1), we get
\[\begin{align}
  & {{\log }_{e}}\left[ {{(1+x)}^{1+x}}{{(1-x)}^{1-x}} \right]=(1+x){{\log }_{e}}(1+x)+(1-x){{\log }_{e}}(1-x) \\
 & \Rightarrow (1+x)\left[ x-\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{4}}}{4}+... \right]+(1-x)\left[ -x-\dfrac{{{x}^{2}}}{2}-\dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{4}}}{4}... \right] \\
 & \Rightarrow (1+x)\left[ x-\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{4}}}{4}+... \right]-(1-x)\left[ x+\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{3}+\dfrac{{{x}^{4}}}{4}+... \right] \\
\end{align}\]
On simplifying, we get
\[\begin{align}
  & (1+x)\left[ x-\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{4}}}{4}+... \right]-(1-x)\left[ x+\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{3}+\dfrac{{{x}^{4}}}{4}+... \right] \\
 & \Rightarrow \left[ x-\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{4}}}{4}+... \right]+\left[ {{x}^{2}}-\dfrac{{{x}^{3}}}{2}+\dfrac{{{x}^{4}}}{3}-\dfrac{{{x}^{5}}}{4}+... \right]-\left[ x+\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{3}+\dfrac{{{x}^{4}}}{4}+... \right]+\left[ {{x}^{2}}+\dfrac{{{x}^{3}}}{2}+\dfrac{{{x}^{4}}}{3}+\dfrac{{{x}^{5}}}{4}+... \right] \\
 & \Rightarrow 2\left( -\dfrac{{{x}^{2}}}{2}-\dfrac{{{x}^{4}}}{4}-\dfrac{{{x}^{6}}}{6}-.... \right)+2\left( {{x}^{2}}+\dfrac{{{x}^{4}}}{3}+\dfrac{{{x}^{6}}}{5}+... \right) \\
 & \Rightarrow 2\left( {{x}^{2}}\left( \dfrac{1}{1}-\dfrac{1}{2} \right)+{{x}^{4}}\left( \dfrac{1}{3}-\dfrac{1}{4} \right)+{{x}^{6}}\left( \dfrac{1}{5}-\dfrac{1}{6} \right)+.... \right) \\
 & \Rightarrow 2\left[ \dfrac{{{x}^{2}}}{1\cdot 2}+\dfrac{{{x}^{4}}}{3\cdot 4}+\dfrac{{{x}^{6}}}{5\cdot 6}+... \right] \\
\end{align}\]
Thus, the given expression is
${{\log }_{e}}\left[ {{(1+x)}^{1+x}}{{(1-x)}^{1-x}} \right]=2\left[ \dfrac{{{x}^{2}}}{1\cdot 2}+\dfrac{{{x}^{4}}}{3\cdot 4}+\dfrac{{{x}^{6}}}{5\cdot 6}+... \right]$



Option ‘C’ is correct



Note: In this question, we are to evaluate the given expression. Here, the expression is formed with the logarithmic functions. By writing their expansion and simplifying them, we get the required expansion of the given expression. For this, we need to remember the basic logarithmic series and logarithmic properties. By applying them to the given expression, we can expand it into the required series. Here, the properties of logarithms like multiplication property and power property are used for rewriting the given expression. Then we can substitute the series in place of a logarithmic function. Then, by simplifying, we get the required series.