prove $\log(1+x)0$ Prove $\log(1+x) \lt x$ for $x\gt0$
my attempt:
I show $e^{x}\gt 1+x$ for $x\gt0$
since
$e^{x}=1+x+\frac{n(n-1)}{2}\frac{x^2}{n^2}+...$
so if $x\gt0$ then all terms are positive 
so $e^{x}\gt 1+x$ for $x\gt0$
now given $e^{x}\gt 1+x$ for $x\gt0$, can I take $\log$ on both sides and show 
$\log(1+x)\lt x$ for $x\gt0$
or do I have to prove firstly that $\exp(x)=e^x$ for $x\gt 0$ then I can take $\log$..
 A: Let $$f(x) = \ln(1+x)-x\;,$$ Where $x>0$
Now $$\displaystyle f'(x) = \frac{1}{1+x}-1 = -\frac{x}{1+x}<0$$ for all $x>0$
So function $f(x)$ is Strictly Decreasing function
So Here $$x>0\Rightarrow f(x)<f(0)\;,$$ bcz function $f(x)$ is Strictly Decreasing function.
So $$\ln(1+x)-x<0\Rightarrow \ln(1+x)<x\;,$$ for $x>0$
A: Your derivation is fine. In particular note that when $x > 0$ then we have
$$ \forall x > 0 : e^x > 1, \ 1 + x > 1$$
Now when we take the logarithm of both sides the inequality sign does not turn around.
A: Let $f: x \mapsto x - \log (1+x)$ on $]0, +\infty[$. Since $f'(x) = 1 - \frac{1}{1+x} > 0$ for all $x > 0$, so $f$ is strictly increasing on $]0, +\infty[$. Can you finish it?
A: The function $\ln (1+x)$ is strictly concave on  $(-1,\infty).$ Thus this function is below any of its tangent lines in this domain (except for the point of tangency). Since $y=x$ is the line tangent to $y=\ln (1+x)$ at $(0,0),$ we have the desired result (and more).
A: Your inequality can be improved. The $\exp(x)$ is strictly convex.
(Because the second derivative is positive).
If you no nothing about convexity, you can take the argument that if Hessian is a positive definite of the function then you can use Taylor (1685-1731) series and use a reminder in a Lagrange form and lower bound value of quadratic form from the Hessian by zero.
In any case you should come in some way to notation: $f(x) > f(x_0)+\langle \nabla f(x),x-x_0\rangle$, or in case when function image is $\mathbb{R}$  you will have: $f(x)>f(x_0) + f'(x_0)(x-x_0)$. Now you use exponent as $f$ and $x_0=0$ and so:
$exp(x)>1+1(x-0)=1+x \implies \forall x>-1,\ln(1+x)<x$
#!/usr/bin/env python

# You can visualize the functions using a python interpreter
import numpy as np
import matplotlib.pyplot as plt

x = np.arange(-1, 10, 0.01) 
plt.plot (x, x) 
plt.plot (x, np.log(1+x)) 
plt.grid ( True ) 
plt.show ( )

