Question

The sum of the series ${\log _9}3 + {\log _{27}}3 - {\log _{81}}3 + {\log _{243}}3 - .....$ is
$A)1 - lo{g_e}2$
$B)1 + lo{g_e}2$
$C)lo{g_e}2$
$D)1 + lo{g_e}3$

Answer 1

First. We need to know about the concepts of logarithm operations.
We will first understand what the logarithmic operator represents in mathematics. A logarithm function or log operator is used when we have to deal with the powers of a number, to understand it better which is $\log {x^m} = m\log x$
Also, we will make use of the binomial expansion of the logarithm function which is given below.
Formula used:

Using the logarithm law, ${\log _y}x = \dfrac{{\log x}}{{\log y}}$
${\log _e}(1 + x) = x - \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{3} - \dfrac{{{x^4}}}{4} + ....$
$\log {x^m} = m\log x$

Complete step-by-step solution:
Since given that we have the sum of the series ${\log _9}3 + {\log _{27}}3 - {\log _{81}}3 + {\log _{243}}3 - .....$ we will convert it into some form and then we will apply the logarithm formulas to obtain the result.
Since we know that ${\log _y}x = \dfrac{{\log x}}{{\log y}}$ and by using this formula we get ${\log _9}3 + {\log _{27}}3 - {\log _{81}}3 + {\log _{243}}3 - .....$
$= \dfrac{{\log 3}}{{\log 9}} + \dfrac{{\log 3}}{{\log 27}} - \dfrac{{\log 3}}{{\log 81}} + \dfrac{{\log 3}}{{\log 243}} - ....$
Since all the denominator terms are multiplications of the number $3$ then we are able to change all the numbers with respect to the square, cube, … of the number nine.
Thus, we have $9 = {3^2},27 = {3^3},81 = {3^4},243 = {3^5}$ and hence we get $\dfrac{{\log 3}}{{\log 9}} + \dfrac{{\log 3}}{{\log 27}} - \dfrac{{\log 3}}{{\log 81}} + \dfrac{{\log 3}}{{\log 243}} - ....$
$= \dfrac{{\log 3}}{{\log {3^2}}} + \dfrac{{\log 3}}{{\log {3^3}}} - \dfrac{{\log 3}}{{\log {3^4}}} + \dfrac{{\log 3}}{{\log {3^5}}} - ...$
Again, applying the logarithm formula of $\log {x^m} = m\log x$ then we get $\dfrac{{\log 3}}{{\log {3^2}}} + \dfrac{{\log 3}}{{\log {3^3}}} - \dfrac{{\log 3}}{{\log {3^4}}} + \dfrac{{\log 3}}{{\log {3^5}}} - ....$
$= \dfrac{{\log 3}}{{2\log 3}} + \dfrac{{\log 3}}{{3\log 3}} - \dfrac{{\log 3}}{{4\log 3}} + \dfrac{{\log 3}}{{5\log 3}} - ...$
Now cancel the common terms we have $\dfrac{{\log 3}}{{2\log 3}} + \dfrac{{\log 3}}{{3\log 3}} - \dfrac{{\log 3}}{{4\log 3}} + \dfrac{{\log 3}}{{5\log 3}} - ...$
$= \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + \dfrac{1}{5} - ....$
Now add and subtract $\dfrac{1}{2}$ on the above values, so that the value will not be changed and we are able to find its generalized form.
Thus, we get $\dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + \dfrac{1}{5} - ....$
$= \dfrac{1}{2} + \dfrac{1}{2} - \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + \dfrac{1}{5} - ....$ and since $\dfrac{1}{2} + \dfrac{1}{2} = 1$
Then we have $\dfrac{1}{2} + \dfrac{1}{2} - \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + \dfrac{1}{5} - ....$
$= 1 - \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + \dfrac{1}{5} - ....$
We know the binomial expansion of the logarithm is ${\log _e}(1 + x) = x - \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{3} - \dfrac{{{x^4}}}{4} + ....$ substitute the value $x = 1$ then we get ${\log _e}(1 + x) = x - \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{3} - \dfrac{{{x^4}}}{4} + ....$
$= {\log _e}(1 + 1) = 1 - \dfrac{{{1^2}}}{2} + \dfrac{{{1^3}}}{3} - \dfrac{{{1^4}}}{4} + ....$
Thus, we clearly see the expansion that we found $1 - \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + \dfrac{1}{5} - ....$ and hence we get $1 - \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + \dfrac{1}{5} - .... = {\log _e}2$
Therefore, the option $C)lo{g_e}2$ is correct.

Note: We simply substitute the value of $x = 1$ on the expansion then we get ${\log _e}(1 + x) = x - \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{3} - \dfrac{{{x^4}}}{4} + ....$
$\Rightarrow {\log _e}(1 + 1) = 1 - \dfrac{{{1^2}}}{2} + \dfrac{{{1^3}}}{3} - \dfrac{{{1^4}}}{4} + ....$ but we need to find this simplification so that only we can able to substitute the value of $x = 1$ and then we easily obtained the required result
The logarithm function we used $\log {x^m} = m\log x$ and logarithm derivative function can be represented as $\dfrac{d}{{dx}}\log x = \dfrac{1}{x}$