Finding all $x$ such that $\log{(1+x)}\leq x$ Find the set of all $x$ for which $\log{(1+x)}$ is lesser than or equal to $x$. 
I am new to such problem, so any help?
 A: $f(x)=\log(1+x)$ is a concave function over $(-1,+\infty)$, since its second derivative equals:
$$f''(x) = -\frac{1}{(1+x)^2}<0.$$
Concavity implies that the graphics of $f(x)$ always lies under the graphics of any tangent line. 
So, consider the equation of the tangent line in $x=0$ in order to have:
$$\forall x\in(-1,+\infty),\qquad \log(x+1)\leq x.$$

Equivalently, you can exploit the convexity of $e^x$ to prove that:
$$\forall y\in\mathbb{R},\qquad e^{y}\geq y+1,$$
then take the logarithm of both members.
A: So the inequality is:
$$\log(1+x) \le x$$
First, notice that $x > -1$ for all $x$ in the function $f(x) = \log(1+x)$
Since, start plugging in values, and see that for any $x$, where $x > -1$, $x > \log(1+x)$, except when $x=0$, which in that case $x = \log(1+x)$
So the solution set is $$x > -1$$
EDIT
For a calculus approach.
Use the derivative.
$$f'(x) = \frac{1}{1+x}$$
And 
$$f'(x) = 1$$
Now, when $x > -1$, both function are increasing. however, as $x\to\infty$, the fraction $\frac{1}{1+x} \to 0$ and will never be $>1$. Therefore, since the logarithm function will never be increasing as fast as $f(x) = x$, you can conclude that the inequality will be satisfied.
A: A start: Let $f(x)=x-\ln(1+x)$. Note that $f(0)=0$. Use the derivative of $f(x)$ to conclude that $f(x)$ reaches a minimum at $x=0$.
A: Hint : go by calculus (increasing or decreasing)
See the domain ant that's all
A: We define $$f(x)=\log{(1+x)}-x$$
The domain of $f$ is $(-1, +\infty)$.
$$f'(x)=\frac{1}{1+x}-1=\frac{1-1-x}{1+x}=\frac{-x}{1+x}$$
$$f'(x)=0 \Rightarrow x=0$$
$$\text{ For } x \geq 0 : f'(x)\leq 0$$
$$\text{ For } -1<x\leq 0 : f'(x)\geq 0$$
That means that $f$ is decreasing on $[0, +\infty)$ and increasing at $(-1,0]$.
Therefore, $$x\geq 0 \Rightarrow  f(x) \leq f(0)\Rightarrow f(x)\leq 0 \\ x\leq 0 \Rightarrow f(x)\leq f(0) \Rightarrow f(x)\leq 0$$
So, $f(x)\leq 0 \ \ \  \forall x \in (-1, +\infty) \Rightarrow \log{(1+x)}-x\leq 0 \ \ \ \forall x \in (-1, +\infty) \\ \Rightarrow \log{(1+x)} \leq x \ \ \ \forall x \in (-1, +\infty)$
