Question

show that
ln(1+x)<x

Answer 1

Answer on Question #50334 – Math – Other

Question.

Show that

$$\ln (1 + x) < x$$

Solution.

Proof by contradiction.

Let us use the Mean Value Theorem: if a function $f$ is continuous on the closed interval $[a, b]$, where $a < b$, and differentiable on the open interval $(a, b)$, then there exists a point $c$ in $(a, b)$ such that

$$f'(c) = \frac{f(b) - f(a)}{b - a}$$

In our case,

$$f(x) = \ln(1 + x) - x$$

For $x = 0$: $f(x) = 0$

For $x = 1$: $f(x) = \ln(2) - 1 < 0$

For $x \in (0; \infty)$:

$$f'(c) = \frac{f(x) - f(0)}{x - 0} = \frac{f(x)}{x} = \frac{\ln(1 + x)}{x} - 1 < 0$$

But, it can be negative. It's contradiction. So,

$$\ln(1 + x) < x$$

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