How to prove $\ln x By Daniel Rodriguez •
How can we prove that the inequality $$\ln~x<x$$
It is trivial in the cases $0<x\le 1$. I couldn't do anything for $x>1$.
$\endgroup$ 88 Answers
$\begingroup$This is one way to see it among others.
The function $x\mapsto\ln(1+x)$ is a concave function (it's twice differentiable and its second derivative is strictly negative). Thus it's below all its tangents.
The tangent at the point $(0,0)$ is the line $y=x$. Hence $$\forall x>0,\, \ln(1+x)\leq x$$
We deduce from this that $$\forall x>0,\, \ln x<x$$
$\endgroup$ $\begingroup$One possibility is to make use of the fact that $\frac{\partial}{\partial x} \ln(x)=\frac{1}{x}<1$ for $x>1$. Thus, $\ln(x)$ increases slower than $x$ when $x>1$.
For $x>1$ we have
$$\ln(x)=\int_{1}^x \frac{1}{\tau} d\tau$$$$x=1+\int_{1}^x 1 d\tau $$
thus
$$\ln(x)-x=\underbrace{\int_{1}^x \left(\frac{1}{\tau}-1 \right) d\tau}_{<0\text{ because integrand }<0}-1 < 0$$
$\endgroup$ $\begingroup$Your inequality is equivalent to $x < e^x$ for any $x$. (Substitute $x = \log t$.) It is obvious for $x\le 0$. Moreover, if $x>0$ then the Taylor expansion of $e^x$ gives $$x < 1+x \le 1+x+\frac{x^2}{2}+\cdots = e^x.$$
$\endgroup$ $\begingroup$The inequality is true at $x=1$ and it holds between the derivatives for $x>1$ (and the inverse inequality holds between the derivatives for $0<x<1,$ so that gives another proof for that interval).
$\endgroup$ $\begingroup$Consider $$f(x)=\frac{e^x}x$$ $$\implies f'(x)=\frac{(x-1)e^x}{x^2}$$ So $x=1$ is the turning point. And obviously, $\forall x>1$, $f'(x)>0$. So $f(x)$ is increasing in $(1,\infty)$.
Meanwhile, $\forall x\in(0,1)$, $f'(x)<0$ and $f(x)$ is decreasing.
So equivalently, we say $\forall x , f(x)>f(1)=e>1$.
Here
$$f(x)=\frac{e^x}{x}>1\iff e^x>x \iff x>\ln x$$
$\endgroup$ $\begingroup$If you start with $e^x > 1 + x$ and apply the logarithmic function, which is increasing, to both sides then you get $\log(x + 1) < x \implies \log(u) < u -1$. This gives you even a tighter bound.
$\endgroup$ $\begingroup$This is the graphical representation of your inequality-
$\endgroup$ 3 $\begingroup$$$x-\ln(x)=\int 1\space dx-\int \frac{1}{x}\space dx>0$$
Well, graphically, you can easily prove that. Plotting $\color {blue}{y = e^x}$ vs $\color{red}{y = x}$ graph, we can see:
$e^x > x$
Take log on both sides (Considering only +ve values of $x$).
$x > \ln x$
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