Answer
455.1k+ views
Hint:- Use expansions of \[{\left( {1 + x} \right)^{ - n}}\] and \[{{\text{e}}^x}\].
As we know the expansion of \[{{\text{e}}^x}\] is,
\[ \Rightarrow {e^x} = 1 + x + \dfrac{{{x^2}}}{{2!}} + ........ + \dfrac{{{x^n}}}{{n!}}\] (1)
From equation 1. We can say that,
\[ \Rightarrow {{\text{e}}^x} > 1 + x\]
Now, taking log both sides of the above equation. It becomes,
\[ \Rightarrow \log {e^x} > \log (1 + x)\] (2)
Solving above equation. It becomes,
\[ \Rightarrow 1 + x = {\left( {1 - \dfrac{x}{{1 + x}}} \right)^{ - 1}}\] (3)
As we know the expansion of \[{(1 + y)^{ - 1}}\].
\[ \Rightarrow {(1 + y)^{ - 1}} = 1 - y + {y^2} - {y^3} + .......{\text{ }}\] (4)
Now, putting the value of \[{\text{y = }}\left( { - \dfrac{x}{{1 + x}}} \right)\] in equation 4. We get,
\[ \Rightarrow {\left( {1 - \dfrac{x}{{1 + x}}} \right)^{ - 1}} = 1 + \dfrac{x}{{1 + x}} + {\left( {\dfrac{x}{{1 + x}}} \right)^2} + .....\] (5)
And now putting the value of \[x = \dfrac{x}{{1 + x}}\] in equation 1. We get,
\[ \Rightarrow {e^{\dfrac{x}{{1 + x}}}} = 1 + \dfrac{x}{{1 + x}} + \dfrac{{{{\left( {\dfrac{x}{{1 + x}}} \right)}^2}}}{{2!}} + ........ + \dfrac{{{{\left( {\dfrac{x}{{1 + x}}} \right)}^n}}}{{n!}}\] (6)
From equation 3, 5 and 6. We can say that,
\[ \Rightarrow 1 + x > {e^{\dfrac{x}{{1 + x}}}}\]
Taking log both sides of the above equation. We get,
\[ \Rightarrow \log (1 + x) > \log {e^{\dfrac{x}{{1 + x}}}}\]
As we know that, \[\log (e) = 1\].
So, we can write above equation as,
\[ \Rightarrow \log (1 + x) > \dfrac{x}{{1 + x}}\] (7)
Therefore, from equation 2 and 7. We can say that,
\[ \Rightarrow x > \log (1 + x) > \dfrac{x}{{1 + x}}\]
Hence Proved.
Note:- Whenever we came up with this type of problem where log is
Involved, then we should use the expansion of \[{{\text{e}}^x}\], \[{\left( {1 + x} \right)^n},{\left( {1 + x} \right)^{ - n}}\] and
\[\log (1 + x)\] and then try to manipulate their expansions to get the required
result. As this will be the easiest and efficient way to prove the result.
As we know the expansion of \[{{\text{e}}^x}\] is,
\[ \Rightarrow {e^x} = 1 + x + \dfrac{{{x^2}}}{{2!}} + ........ + \dfrac{{{x^n}}}{{n!}}\] (1)
From equation 1. We can say that,
\[ \Rightarrow {{\text{e}}^x} > 1 + x\]
Now, taking log both sides of the above equation. It becomes,
\[ \Rightarrow \log {e^x} > \log (1 + x)\] (2)
Solving above equation. It becomes,
\[ \Rightarrow 1 + x = {\left( {1 - \dfrac{x}{{1 + x}}} \right)^{ - 1}}\] (3)
As we know the expansion of \[{(1 + y)^{ - 1}}\].
\[ \Rightarrow {(1 + y)^{ - 1}} = 1 - y + {y^2} - {y^3} + .......{\text{ }}\] (4)
Now, putting the value of \[{\text{y = }}\left( { - \dfrac{x}{{1 + x}}} \right)\] in equation 4. We get,
\[ \Rightarrow {\left( {1 - \dfrac{x}{{1 + x}}} \right)^{ - 1}} = 1 + \dfrac{x}{{1 + x}} + {\left( {\dfrac{x}{{1 + x}}} \right)^2} + .....\] (5)
And now putting the value of \[x = \dfrac{x}{{1 + x}}\] in equation 1. We get,
\[ \Rightarrow {e^{\dfrac{x}{{1 + x}}}} = 1 + \dfrac{x}{{1 + x}} + \dfrac{{{{\left( {\dfrac{x}{{1 + x}}} \right)}^2}}}{{2!}} + ........ + \dfrac{{{{\left( {\dfrac{x}{{1 + x}}} \right)}^n}}}{{n!}}\] (6)
From equation 3, 5 and 6. We can say that,
\[ \Rightarrow 1 + x > {e^{\dfrac{x}{{1 + x}}}}\]
Taking log both sides of the above equation. We get,
\[ \Rightarrow \log (1 + x) > \log {e^{\dfrac{x}{{1 + x}}}}\]
As we know that, \[\log (e) = 1\].
So, we can write above equation as,
\[ \Rightarrow \log (1 + x) > \dfrac{x}{{1 + x}}\] (7)
Therefore, from equation 2 and 7. We can say that,
\[ \Rightarrow x > \log (1 + x) > \dfrac{x}{{1 + x}}\]
Hence Proved.
Note:- Whenever we came up with this type of problem where log is
Involved, then we should use the expansion of \[{{\text{e}}^x}\], \[{\left( {1 + x} \right)^n},{\left( {1 + x} \right)^{ - n}}\] and
\[\log (1 + x)\] and then try to manipulate their expansions to get the required
result. As this will be the easiest and efficient way to prove the result.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)