Question

If $x,y,\mathrm{and}z$are three consecutive positive integers, then ${\log }_e\sqrt{x}+{\log }_e\sqrt{z}+\left(\frac{1}{2xz+1}\right)+\frac{1}{3}{\left(\frac{1}{2xz+1}\right)}^3+\frac{1}{5}{\left(\frac{1}{2xz+1}\right)}^5+.....$ is

• A. ${\log }_e\sqrt{y}$
• B. ${\log }_{e}y$
• C. ${\log }_e{y}^2$
• D. None of these

Answer 1

The correct option is B

${\log }_{e}y$

Explanation for the correct option.

Step 1: Solve $\left(\frac{1}{2xz+1}\right)+\frac{1}{3}{\left(\frac{1}{2xz+1}\right)}^3+\frac{1}{5}{\left(\frac{1}{2xz+1}\right)}^5+.....$

We know that $\frac{1}{2}\log \left(\frac{1+a}{1-a}\right)=a+\frac{a^3}{3}+\frac{a^5}{5}+....$

Here, let $a=\frac{1}{2xz+1}$

So,

$\begin{array}{rcl}\left(\frac{1}{2xz+1}\right)+\frac{1}{3}{\left(\frac{1}{2xz+1}\right)}^3+\frac{1}{5}{\left(\frac{1}{2xz+1}\right)}^5+.....& =& \frac{1}{2}\log \left(\frac{1+\frac{1}{2xz+1}}{1-\frac{1}{2xz+1}}\right)\\ & =& \frac{1}{2}\log \left(\frac{2xz+2}{2xz}\right)\\ & =& \frac{1}{2}\log \left(\frac{xz+1}{xz}\right)\end{array}$

Step 2: Solve ${\log }_e\sqrt{x}+{\log }_e\sqrt{z}$

$\begin{array}{rcl}{\log }_e\sqrt{x}+{\log }_e\sqrt{z}& =& {\log }_e{\left(x\right)}^{1/2}+{\log }_e{\left(z\right)}^{1/2}\\ & =& \frac{1}{2}{\log }_e\left(x\right)+\frac{1}{2}{\log }_e\left(z\right)\\ & =& \frac{1}{2}\log \left(xz\right)\end{array}$

Step 3: Solve ${\log }_e\sqrt{x}+{\log }_e\sqrt{z}+\left(\frac{1}{2xz+1}\right)+\frac{1}{3}{\left(\frac{1}{2xz+1}\right)}^3+\frac{1}{5}{\left(\frac{1}{2xz+1}\right)}^5+.....$

$\begin{array}{rcl}& & {\log }_e\sqrt{x}+{\log }_e\sqrt{z}+\left(\frac{1}{2xz+1}\right)+\frac{1}{3}{\left(\frac{1}{2xz+1}\right)}^3+\frac{1}{5}{\left(\frac{1}{2xz+1}\right)}^5+.....\\ & =& \frac{1}{2}\log \left(xz\right)+\frac{1}{2}\left(\frac{xz+1}{xz}\right)\\ & =& \frac{1}{2}\log \left(xz\times \frac{xz+1}{xz}\right)\\ & =& \frac{1}{2}\log \left(xz+1\right)\end{array}$

It is given that $x,y,\mathrm{and}z$are three consecutive positive integers, $x=y-1,\mathrm{and}z=y+1$.

So,

$\begin{array}{rcl}\frac{1}{2}\log \left(xz+1\right)& =& \frac{1}{2}\log \left[\left(y-1\right)\left(y+1\right)+1\right]\\ & =& \frac{1}{2}\log \left[y^2-1+1\right]\\ & =& \log y^{2\times 1/2}\\ & =& {\log }_{e}y\end{array}$

Hence, option B is correct.