Show that $\ln (x) \leq x-1 $

Question

I'm not really sure how to show this, it's obvious if we draw a graph of it but that won't suffice here. Could we somehow use the fact that $e^x$ is the inverse? I mean, if $e^{x-1} \geq x$ then would the statement be proved?

Try proving $F(x)=\ln{x}-x+1$ is negative, by using first derivative test to conclude about where $F$ is increasing and where it's decreasing. — Poppy, Apr 06 '15 at 20:27
Do you know the derivative of the logarithm? Use that and the Mean Value Theorem. — hmakholm left over Monica, Apr 06 '15 at 20:28
These older posts are also related: Intuition behind logarithm inequality: $1 - \frac1x \leq \log x \leq x-1$, Showing $\frac{x}{1+x}<\log(1+x)<x$ for all $x>0$ using the mean value theorem and Prove by Mean Value Theorem $\frac{x}{1+x}<\ln(1+x)<x$ for $x>0$. — Martin Sleziak, Aug 31 '19 at 23:17

score 12 · Answer 1 · answered Apr 06 '15 at 20:33

12

Yes, one can use $$\tag1e^x\ge 1+x,$$ which holds for all $x\in\mathbb R$ (and can be dubbed the most useful inequality involving the exponential function). This again can be shown in several ways. If you defined $e^x$ as limit $\lim_{n\to\infty}\left(1+\frac xn\right)^n$, then $(1)$ follows from Bernoullis inequality: $(1+t)^n>1+nt$ if $t>-1$ and $n>0$.

To show that $\ln(x)\le x-1$ for all $x>0$, just substitute $\ln x$ for $x$ in $(1)$.

answered Apr 06 '15 at 20:33

Hagen von Eitzen

374,180

I keep using that inequality, and have no idea what makes it so useful... – djechlin Nov 13 '15 at 00:15
+1 This is a really nice answer. – user577215664 Sep 25 '20 at 13:28

score 9 · Answer 2 · edited Apr 23 '23 at 22:59

9

$y=x-1$ is the equation of the tangent to the $\ln$ curve at $(1,0)$ and the function is concave, hence its graph is under the tangent.

edited Apr 23 '23 at 22:59

user577215664

40,625

answered Apr 06 '15 at 20:31

Bernard

175,478

score 8 · Accepted Answer · edited May 13 '18 at 16:12

8

Define for $\;x>0\;$

$$f(x)=\ln x-x+1\implies f'(x)=\frac1x-1=0\iff x=1$$

and since $\;f''(x)=-\dfrac1{x^2}<0\quad \forall x>0\;$ , we get a maximal point.

But also

$$\lim_{x\to 0+}f(x)=-\infty=\lim_{x\to\infty}f(x)$$

Thus, the above is a global maximal point and

$$\forall\,x>0\;,\;\;\;f(x)\le f(1)=0$$

edited May 13 '18 at 16:12

imedadel

177

answered Apr 06 '15 at 20:29

Timbuc

34,191

1

Are you that divergence at $0+$ and $\infty$ is enough to conclude that some maxima is a global one? The function might quite well rise before diverging to $-\infty$... – StefanH Apr 27 '17 at 15:39

score 4 · Answer 4 · answered Jun 04 '18 at 19:58

4

There are at least two possible methods, the first is by studying the functions' variation (see Timbuc's answer above) and the second is by Integration: $$\forall x>0,\qquad \frac 1x \leq 1 \iff \int_1^x\frac 1x\; \mathrm dx \leq \int_1^x 1\; \mathrm dx \iff \left[\ln x\right]_1^x \leq \left[x\right]_1^x \\ \iff \ln x - \ln 1 \leq x -1 \iff \boxed {\ln x \leq x -1}.$$ Note: I started from $\frac 1x \leq 1$ because it is the derivative of $\ln x \leq x -1$.

answered Jun 04 '18 at 19:58

imedadel

177

5

but your first step is wrong, no? for $x > 0$, you can't say $\frac{1}{x} < 1$ (for $x \rightarrow 0$). What am I missing? – math2001 Sep 04 '19 at 09:09
1

I think it still holds. For $0<x<1$, $\frac{1}{x}>1$, then $\int_x^1\frac{1}{x}dx>\int_x^1dx$, which implies that $-ln(x)>1-x$ or $ln(x)<x-1$. – Heatconomics Apr 18 '22 at 13:21

marwalix · Answer 5 · 2023-04-24T15:18:16.147

2

Denote $f(x)=\log{x}-x+1$. We have $f'(x)=\frac{1}{x}-1$.

For $0\lt x\leq 1$ we have $f'(x)\geq 0$ so $f$ is increasing for $x\in ]0,1]$. $f(1)=0$ so on this interval $-\infty\lt f(x)\leq 0$

For $x\gt 1$ we have $f'(x)\lt 0$ so $f$ is decreasing for $x\in [1, +\infty]$. $f(1)=0$ so on this interval $-\infty\lt f(x)\leq 0$

And we have proven $f(x)\leq 0$ Q.E.D

edited Apr 24 '23 at 15:18

answered Apr 06 '15 at 20:35

marwalix

16,773

score 1 · Answer 6 · answered Jun 06 '18 at 11:32

1

You may want to read the following note by E. R. Love: The Mathematical Gazette Vol. 64, No. 427 (Mar., 1980), pp. 55-57

answered Jun 06 '18 at 11:32

Jerome Benoit

175

Show that $\ln (x) \leq x-1 $

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