Question

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

Answer 1

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.