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If x is positive, show that \[\log (1 + x) < x\] and \[ > \dfrac{x}{{1 + x}}\].

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Last updated date: 27th Feb 2024
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IVSAT 2024
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.