Finding all $x$ such that $\log{(1+x)}\leq x$

Question

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?

Answer 1

$f(x)=\log(1+x)$ 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: $$\forall x\in(-1,+\infty),\qquad \log(x+1)\leq x.$$

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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.

Answer 2

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.

Answer 3

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 $x=0$.

Answer 4

Hint : go by calculus (increasing or decreasing)

See the domain ant that's all

Answer 5

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, $f(x)\leq 0 \ \ \ \forall x \in (-1, +\infty) \Rightarrow \log{(1+x)}-x\leq 0 \ \ \ \forall x \in (-1, +\infty) \\ \Rightarrow \log{(1+x)} \leq x \ \ \ \forall x \in (-1, +\infty)$