Question

If x, y, z are three consecutive positive integers, then ${\log _e}\sqrt x + {\log _e}\sqrt z + \dfrac{1}{{2xz + 1}} + \dfrac{1}{3}{\left( {\dfrac{1}{{2xz + 1}}} \right)^3} + \dfrac{1}{5}{\left( {\dfrac{1}{{2xz + 1}}} \right)^5} + ...$ is
A.${\log _e}\sqrt y$
B.${\log _e}y$
C.${\log _e}{y^2}$
D.None of these

Answer 1

Hint : Use the various Logarithmic properties to simplify the terms given in the question. Apply the logarithmic expansion and solve it. You will be able to find the required coefficient. The Logarithmic Functions have some of the properties that allow you to simplify the logarithms such as follows:
$\log ab = \log a + \log b$ , $\log \dfrac{a}{b} = \log a - \log b$ , $a\log x = \log {x^a}$, ${\log _a}b = \dfrac{{\log b}}{{\log a}}$and$\log \left( {1 + x} \right) = x - \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{3} - \dfrac{{{x^4}}}{4} + \dfrac{{{x^5}}}{5} + ...$

Complete step-by-step answer :
Common logarithmic function: The logarithmic function with base $10$ is called the common logarithmic function and it is denoted by log.
Natural logarithmic function: The logarithmic function to the base e is called the natural logarithmic function.
We are given the expression ${\log _e}\sqrt x + {\log _e}\sqrt z + \dfrac{1}{{2xz + 1}} + \dfrac{1}{3}{\left( {\dfrac{1}{{2xz + 1}}} \right)^3} + \dfrac{1}{5}{\left( {\dfrac{1}{{2xz + 1}}} \right)^5} + ...$
We know that ${\log _e}\left( {\dfrac{{1 + x}}{{1 - x}}} \right) = 2\left( {x + \dfrac{{{x^3}}}{3} + \dfrac{{{x^5}}}{5} + ...} \right)$
Taking the value of $x = \dfrac{1}{{2xz + 1}}$
We get ${\log _e}\left( {\dfrac{{1 + \dfrac{1}{{2xz + 1}}}}{{1 - \dfrac{1}{{2xz + 1}}}}} \right) = 2\left( {\dfrac{1}{{2xz + 1}} + \dfrac{1}{3}{{\left( {\dfrac{1}{{2xz + 1}}} \right)}^3} + \dfrac{1}{5}{{\left( {\dfrac{1}{{2xz + 1}}} \right)}^5} + ...} \right)$
Hence we can rewrite the given equation in following way
${\log _e}\sqrt x + {\log _e}\sqrt z + \dfrac{1}{2}{\log _e}\left( {\dfrac{{1 + \dfrac{1}{{2xz + 1}}}}{{1 - \dfrac{1}{{2xz + 1}}}}} \right)$
Using the product rule of logarithmic function we get
$= {\log _e}\sqrt {xz} + \dfrac{1}{2}{\log _e}\left( {\dfrac{{1 + \dfrac{1}{{2xz + 1}}}}{{1 - \dfrac{1}{{2xz + 1}}}}} \right)$
On taking the LCM in second term we get
$= {\log _e}\sqrt {xz} + \dfrac{1}{2}{\log _e}\left( {\dfrac{{2xz + 2}}{{2xz}}} \right)$
Put $x = n,y = n + 1$ and $z = n + 2$
Therefore we get the expression
$= {\log _e}\sqrt {n(n + 2)} + \dfrac{1}{2}{\log _e}\left( {\dfrac{{n(n + 2) + 1}}{{n(n + 2)}}} \right)$
Using the power rule of logarithmic function we get the following
$= \dfrac{1}{2}{\log _e}\left( {n\left( {n + 2} \right)} \right) + \dfrac{1}{2}{\log _e}\left( {n\left( {n + 2} \right) + 1} \right) - \dfrac{1}{2}{\log _e}\left( {n\left( {n + 2} \right)} \right)$
Hence first and third term get cancelled and we are left with the term
$= \dfrac{1}{2}{\log _e}\left( {n\left( {n + 2} \right) + 1} \right)$
$= \dfrac{1}{2}{\log _e}\left( {{n^2} + 2n + 1} \right)$
Using the power rule of logarithmic functions we get the expression
$= {\log _e}\sqrt {{{(n + 1)}^2}}$
$= {\log _e}\left( {n + 1} \right)$
On putting the value of $n + 1 = y$ we get
$= {\log _e}y$
Hence we get our required answer.
So, the correct answer is “Option 2”.

Note : Keep in mind that Logarithmic functions are the inverses of exponential functions, and any exponential function can be expressed in logarithmic form. Use the suitable Logarithmic properties to simplify the terms given in the question as much as possible. Apply the logarithmic expansion correctly and solve it carefully to get the required coefficient. There are two types log functions
1. Natural logarithmic functions , it has base “e”
2. Common logarithmic functions, it has base “10”