fofopausiomiava

2022-10-08

How to prove that $\mathrm{log}(x)<x$ when $x>1$?

It's very basic but I'm having trouble to find a way to prove this inequality

$\mathrm{log}(x)<x$

when $x>1$

($\mathrm{log}(x)$ is the natural logarithm)

I can think about the two graphs but I can't find another way to prove it, and, besides that, I don't understand why should it not hold if $x<1$

Can anyone help me?

Thanks in advance.

It's very basic but I'm having trouble to find a way to prove this inequality

$\mathrm{log}(x)<x$

when $x>1$

($\mathrm{log}(x)$ is the natural logarithm)

I can think about the two graphs but I can't find another way to prove it, and, besides that, I don't understand why should it not hold if $x<1$

Can anyone help me?

Thanks in advance.

Helena Bentley

Beginner2022-10-09Added 10 answers

You may just differentiate

$$f(x):=\mathrm{log}x-x,\phantom{\rule{1em}{0ex}}x\ge 1,$$

giving

$${f}^{\prime}(x)=\frac{1}{x}-1=\frac{1-x}{x}<0\phantom{\rule{1em}{0ex}}\text{for}\phantom{\rule{1em}{0ex}}x>1$$

since

$$f(1)=-1<0$$

and $f$ is strictly decreasing, then

$$f(x)<0,\phantom{\rule{1em}{0ex}}x>1,$$

that is

$$\mathrm{log}x-x<0,\phantom{\rule{1em}{0ex}}x>1.$$

$$f(x):=\mathrm{log}x-x,\phantom{\rule{1em}{0ex}}x\ge 1,$$

giving

$${f}^{\prime}(x)=\frac{1}{x}-1=\frac{1-x}{x}<0\phantom{\rule{1em}{0ex}}\text{for}\phantom{\rule{1em}{0ex}}x>1$$

since

$$f(1)=-1<0$$

and $f$ is strictly decreasing, then

$$f(x)<0,\phantom{\rule{1em}{0ex}}x>1,$$

that is

$$\mathrm{log}x-x<0,\phantom{\rule{1em}{0ex}}x>1.$$

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