
\[{\log _e}x - {\log _e}(x - 1) = \]
A) \[\dfrac{1}{x} - \dfrac{1}{{2{x^2}}} + \dfrac{1}{{3{x^3}}} - .....\infty \]
B) \[\dfrac{1}{x} + \dfrac{1}{{2{x^2}}} + \dfrac{1}{{3{x^3}}} + .....\infty \]
C) \[2(\dfrac{1}{x} + \dfrac{1}{{3{x^3}}} + \dfrac{1}{{5{x^5}}} + .....\infty )\]
D) \[2(\dfrac{1}{x} - \dfrac{1}{{3{x^3}}} + \dfrac{1}{{5{x^5}}} - .....\infty )\]
Answer
164.1k+ views
Hint: in this question, we have to find the value of given expression. In order to find this first Rearrange the given expression to find standard pattern of the expression. Once we get type of standard function then by applying the formula of expansion, the required value is to be calculated.
Formula Used:Expansion of logarithm series is given as:
If\[x \in R\], \[\left| x \right| < 1\], then the expansion is given as
\[{\log _e}(1 + x) = x - \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{3} - \dfrac{{{x^4}}}{4} + \dfrac{{{x^5}}}{5} + ...\]
If\[x \in R\], \[\left| x \right| < 1\], then the expansion is given as
\[{\log _e}(1 - x) = - x - \dfrac{{{x^2}}}{2} - \dfrac{{{x^3}}}{3} - \dfrac{{{x^4}}}{4} - \dfrac{{{x^5}}}{5} - ...\]
Complete step by step solution:Given: \[{\log _e}x - {\log _e}(x - 1)\]
Now we have to rearrange the above expression in order to find standard form of expansion formula.
\[{\log _e}x - {\log _e}(x - 1) = {\log _e}(\dfrac{x}{{x - 1}})\]
Here we used \[{\log _e}a - {\log _e}b = {\log _e}\dfrac{a}{b}\]
Divide numerator and denominator by \[\dfrac{1}{x}\]
\[{\log _e}(\dfrac{x}{{x - 1}}) = {\log _e}(\dfrac{1}{{1 - \dfrac{1}{x}}})\]
\[{\log _e}(\dfrac{x}{{x - 1}}) = - {\log _e}(1 - \dfrac{1}{x})\]
We know that
Expansion of logarithm series is given as:
If\[x \in R\], \[\left| x \right| < 1\], then the expansion is given as
\[{\log _e}(1 - x) = - x - \dfrac{{{x^2}}}{2} - \dfrac{{{x^3}}}{3} - \dfrac{{{x^4}}}{4} - \dfrac{{{x^5}}}{5} - ...\]
Replace x with \[\dfrac{1}{x}\]
Now we get
\[ - {\log _e}(1 - \dfrac{1}{x}) = \dfrac{1}{x} + \dfrac{1}{{2{x^2}}} + \dfrac{1}{{3{x^3}}} + ....\]
Option ‘B’ is correct
Note: Here we have to rearrange the series in order to get standard pattern of series. Once we get type of series then by applying the formula of that series, the required sum is to be calculated.
In this question after rearrangement we found that series is written as a sum of expansion of\[{\log _e}(1 - x)\]After getting standard series we have to apply the expansion formula to get the sum of given series. In this type of question always try to find the pattern of the series and after getting pattern apply formula of that pattern.
Formula Used:Expansion of logarithm series is given as:
If\[x \in R\], \[\left| x \right| < 1\], then the expansion is given as
\[{\log _e}(1 + x) = x - \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{3} - \dfrac{{{x^4}}}{4} + \dfrac{{{x^5}}}{5} + ...\]
If\[x \in R\], \[\left| x \right| < 1\], then the expansion is given as
\[{\log _e}(1 - x) = - x - \dfrac{{{x^2}}}{2} - \dfrac{{{x^3}}}{3} - \dfrac{{{x^4}}}{4} - \dfrac{{{x^5}}}{5} - ...\]
Complete step by step solution:Given: \[{\log _e}x - {\log _e}(x - 1)\]
Now we have to rearrange the above expression in order to find standard form of expansion formula.
\[{\log _e}x - {\log _e}(x - 1) = {\log _e}(\dfrac{x}{{x - 1}})\]
Here we used \[{\log _e}a - {\log _e}b = {\log _e}\dfrac{a}{b}\]
Divide numerator and denominator by \[\dfrac{1}{x}\]
\[{\log _e}(\dfrac{x}{{x - 1}}) = {\log _e}(\dfrac{1}{{1 - \dfrac{1}{x}}})\]
\[{\log _e}(\dfrac{x}{{x - 1}}) = - {\log _e}(1 - \dfrac{1}{x})\]
We know that
Expansion of logarithm series is given as:
If\[x \in R\], \[\left| x \right| < 1\], then the expansion is given as
\[{\log _e}(1 - x) = - x - \dfrac{{{x^2}}}{2} - \dfrac{{{x^3}}}{3} - \dfrac{{{x^4}}}{4} - \dfrac{{{x^5}}}{5} - ...\]
Replace x with \[\dfrac{1}{x}\]
Now we get
\[ - {\log _e}(1 - \dfrac{1}{x}) = \dfrac{1}{x} + \dfrac{1}{{2{x^2}}} + \dfrac{1}{{3{x^3}}} + ....\]
Option ‘B’ is correct
Note: Here we have to rearrange the series in order to get standard pattern of series. Once we get type of series then by applying the formula of that series, the required sum is to be calculated.
In this question after rearrangement we found that series is written as a sum of expansion of\[{\log _e}(1 - x)\]After getting standard series we have to apply the expansion formula to get the sum of given series. In this type of question always try to find the pattern of the series and after getting pattern apply formula of that pattern.
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