How to prove that $\log(x)1$? It's very basic but I'm having trouble to find a way to prove this inequality
$\log(x)<x$ 
when $x>1$
($\log(x)$ 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. 
 A: Define $f(x) = \log x - x$. Now $f'(x) = \frac{1}{x}-1$ which is negative if $x > 1$. Thus $f$ is strictly decreasing on the interval $(1, \infty)$.
Now since $f(1) = \log 1 - 1 = 0-1 = -1$, we must have $f(x) < -1$ on $(1, \infty)$. Thus $\log x - x < -1  < 0$ on $(1, \infty)$. This implies $\log x < x$ when $x > 1$.
A: You may just differentiate
$$
f(x):=\log x-x, \quad x\geq1,
$$ giving
$$
f'(x)=\frac1x-1=\frac{1-x}x<0 \quad \text{for}\quad x>1
$$ since
$$
f(1)=-1<0
$$ and $f$ is strictly decreasing, then
$$
f(x)<0, \quad x>1,
$$ that is 
$$
\log x -x <0, \quad x>1.
$$
A: I thought it might be instructive to present a proof that relies on standard tools only.  We begin with the limit definition of the exponential function
$$e^x=\lim_{n\to \infty}\left(1+\frac xn\right)^n$$
It is easy to show that the sequence $e_n(x)=\left(1+\frac xn\right)^n$ increases monotonically for $x>-1$.  To show this we simply analyze the ratio
$$\begin{align}
\frac{e_{n+1}(x)}{e_n(x)}&=\frac{\left(1+\frac x{n+1}\right)^{n+1}}{\left(1+\frac xn\right)^n}\\\\
&=\left(1+\frac{-x}{(n+x)(n+1)}\right)^{n+1}\left(1+\frac xn\right) \tag 1\\\\
&\ge \left(1+\frac{-x}{n+x}\right)\left(1+\frac xn\right)\tag 2\\\\
&=1
\end{align}$$
where in going from $(1)$ to $(2)$ we used Bernoulli's Inequality.  Note that $(2)$ is valid whenever $n>-x$ or $x>-n$.
Since $e_n(x)$ monotonically increases and is bounded above by $e^x$, then
$$e^x\ge \left(1+\frac xn\right)^n \tag 3$$
for all $n\ge 1$.  And therefore, for $x>-1$ we have
$$e^x\ge 1+x \tag 4$$
Since $e^x>0$ for all $x$, then $(4)$ is true for $x\le -1$ also.  Therefore, $e^x\ge 1+x$ for all $x$.

ASIDE:
From $(4)$ we note that $e^{-x}\ge 1-x$.  If $x<1$, then since $e^x\,e^{-x}=1$, $e^x\le \frac{1}{1-x}$.  Thus, for $x<1$ we can write
$$1+x\le e^x\le \frac{1}{1-x}$$

Taking the logarithm of both sides of $(4)$ produces the coveted inequality
$$\log(1+x)\le x  \tag 5$$
Interestingly, setting $x=-z/(z+1)$ into $(4)$ reveals
$$\log(1+z)\ge \frac{z}{z+1}$$
for $z>-1$.  Putting it all together we have for $x>0$
$$\frac{x-1}{x}\le \log x\le x-1<x$$
A: When $x=1$, $\log x=0<1=x$. Further, for $x>1$ we have $\frac{d}{dx}\log x=\frac{1}{x}<1=\frac{d}{dx}x$.
This shows that $x$ is larger than $\log x$ at $x=1$ and that $x$ grows faster than $\log x$ for $x>1$. Hence $x>\log x$ for $x\ge 1$.
A: different ways of doing this exercise certainly depend on what you wish to assume. suppose we take $\log x$ to be a continuous non-constant map $f:\mathbb{R}^+ \to \mathbb{R}$ satisfying
$$
f(xy) = f(x)+f(y) \tag{1}
$$
this immediately gives $f(1)=0, f(x)+f(\frac1{x})=0$ so $f$ is a non-trivial abelian group homomorphism with $\exists c\dot f(c) \ne 0$
(1) implies that for any integers $m,n \ne 0$ we have 
$$
\log \sqrt{[n]c^m}=\log c^{\frac{m}{n}}= \frac{m}{n} \log c \tag{2}
$$
since for $\mathbb{R}^+\ni x \ne 1$ the set $\{c^{\frac{m}{n}}\}_{m,n \in \mathbb{Z}\setminus \{0\}}$ is dense in $\mathbb{R}^+$ we have, by continuity,
$$
f(c^r)=r\log c
$$
for any $r \in \mathbb{R}^+$
(2), together with the density of $\text{Im}(f)$ in $\mathbb{R}^+$ (1) implies that $f$ is order-preserving or order-inverting depending on the sign of $\log c$ and whether $c \gt 1$. thus to rule out the order anti-isomorphisms we require one further assumption, that $f((1,\infty)) \subseteq (0,\infty)$
suppose $f$ had a fixed point $\zeta \gt 1$. i.e a point for which as real numbers 
$$
f(\zeta) = \zeta
$$
we will show this leads to a contradiction.
since $f(1)=0$ and $f$ is strictly monotonic and continuous the equation $f(x)=1$ has a unique solution, let us say $x=e \gt 1$. 
since $\text{Im}(f)\subset \text{Domain}(f)$ we may define a sequence of functions $f_n$ with
$\text{Domain}(f_{n+1})=F_{n+1} = \text{Im}(f_n)$ and $f_{n+1}=f_{|F_{n+1}}$
renaming $f$ as $f_0$ we have a sequence $F_n$ with
$$
F_{n}=(e^n,\infty) \\
\bigcap F_n = \emptyset
$$
but $\forall n \zeta \in \text{Image} (f_n)$, contradiction
since $f$ has no fixed point and $f(1) \lt 1$ we have our result 
A: $\log_{10}x<x$ implies $x<10^x$ We can directly see it by observation and is true for all $x$. Or directly go for derivatives!
A: Note that the second derivative of $\ln(x)$ is $-\frac{1}{x^2}$, which is always negative. This means that any tangent line to the graph $y=\ln(x)$ will be greater than or equal to $\ln(x)$, equality only being achieved at the tangent point. We can then conclude that the tangent line $x-1$ is greater than or equal to $ln(x)$. Since $x>x-1$, $x>\ln(x)$ for any value of $x$.
A: If you defined the logarithm as $$\log(x)=\int_{1}^{x}{\frac{1}{t}dt},$$
$$\frac{1}{x} \le 1 \; \text{ for }x\ge 1.$$ Hence, $$ \log(x)=\int_{1}^{x}{\frac{1}{t}\,dt} \le \int_{1}^{x}\!{1}\,dt =x-1 \le x.$$
If $0< x\le 1\;$ then you simply get $$\log(x)=\int_{1}^{x}{\frac{1}{t}\,dt}=- \int_{x}^{1}{\frac{1}{t}\,dt}\le 0 < x.$$
A: I am assuming you know the derivative of $\log$.
Let $f(x)=\log x -x$. Then
$$f'(x) = \frac 1x -1<0\ \ \forall x>1.$$
Moreover, $f(1) = -1<0$. So you have a function that starts negative at $x=1$, and decreases afterwards since its derivative is always negative. This means that
$$f(x) = \log(x) - x <0\ \ \forall x>1,$$
which is what you wanted to show.
A: Taylor series give $$e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots$$
Hence $e^x > 1+x > x$ for $x\geq0$, so $\log(e^x) > \log(x)$ since $\log$ is increasing. Hence $x > \log(x)$ for $x\geq0$.
A: You even have $\;\log x \le x-1$, because $\log$ is a concave function, and the line with equation $y=x-1$ is the tangent to the graph of $\log$ at $(1,0)$.
Hence:
$$\log x \le x-1 <x. $$
A: If you're familiar with the Taylor Theorem, this is a short proof of the inequality:
Define $f(x)=\log(x)$ for $x>0$. I'm going to assume that the natural logarithm is the (only) function which verifies that $f(1)=0$ and $f'(x)=\dfrac{1}{x}$ (this is one of the many definitions we can give, others would be to say it's the inverse function of the exponential function, $e^{x}$, or saying it is the integral $\int_{1}^{x}\dfrac{1}{t}dt$, but that's not the problem now).
The second derivative of $f$ is $f''(x)=-\dfrac{1}{x^{2}}$, continuous in $(0,\infty)$.
Let's take a point $x>1$, since $f$ is a $\mathcal{C}^{2}$ function (its first and second derivatives exist and are continuous), the Taylor Theorem guarantees that there exists $c\in (1,x)$ which satisfies 
$f(x)=f(1)+f'(1)(x-1)+f''(c)(x-1)^{2}$
Then
$\log(x)=\log(1)+\dfrac{1}{1}(x-1)-\dfrac{1}{c^{2}}(x-1)^{2}=x-1-\dfrac{1}{c^{2}}(x-1)^{2}$
But since $c^{2}>0$ and $(x-1)^{2}>0$, we know that
$\log(x)<x-1<x$, which is what we wanted to prove.
Even though the question was made only considering $x>1$, the result is also true when $x\leq 1$. The case $x<1$ would be solved in the same manner, whereas doing it while $x=1$ is just a matter of checking directly. 
A: You can just use Lagrange's theorem in the interval $[1,x]$:
$$
f(x) - f(1) = f'(\xi) (x-1), \quad \xi \in [1,x],
$$
which yields $ \log x = \frac{1}{\xi} (x-1) \leq x - 1 < x$.
A: This is equivalent to prove that $e^x>x$.
It is obvious that $e^x>x$ if $x<0$ since the LHS is positive and the RHS is negative.
Suppose that for some $a\ge 0$, the inequality $e^a\le a$ holds.
Then $a\ge e^a\ge 1$ since $e^a\ge e^0$ because $a\ge 0$. But now we can see that $a\ge 1$ and again, $a\ge e^a\ge e^1$ and so $a\ge e$. We continue applying the same observation and conclude that $a\ge e^{^{e}}$ and so on, which means that $a$ is unbounded which is a contradiction.
