Answer

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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.

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