How to use other method to prove $2^x>2x-1$. let $x\in R$,show $$2^x>2x-1$$
My try:
 take a function $f(x)=2^x-2x+1$, then  we have
$$f'(x)=2^x\ln{2}-2,~~~~f''(x)=2^x\ln^2{2}>0$$
if let $x_{0}$ such   $f'(x_{0})=0\Longrightarrow 2^{x_{0}}=\dfrac{2}{\ln{2}}$ is  minimum the point (since $f''(x)>0$).
so we have
$$f(x)\ge f(x_{0})= \dfrac{2}{\ln{2}}-1+\dfrac{\ln{\ln{2}}}{\ln{2}}+1=\dfrac{2+\ln{\ln{2}}}{\ln{2}}>0$$
because $$2+\ln{\ln{2}}>0\Longleftrightarrow 2^{e^2}>e$$
have other  methods?Thanks 
 A: Let's see if we can prove the result with resorting to logs or calculus.
We'll begin with a rewrite of $2^x\gt2x-1$, letting $x=2u+{1\over2}$, as
$$4^u\gt2\sqrt2u$$
Since the inequality is obvious for $u\le0$ (since $u^u$ is always positive), it suffices to prove the stronger inequality 
$$u\le{4^u\over3}$$
for $u\ge0$. Toward this end, it will help to know that
$${3\over2}\lt4^{1/3},\quad2=4^{1/2},\quad{5\over2}\lt4^{2/3},\quad3\lt4^{5/6},\quad4=4^1,\quad\text{and}\quad6\lt4^{4/3}$$
(The first three inequalities amount to ${27\over8}\lt4$, ${125\over8}\lt16$, and $27\lt32$; the last inequality essentially repeats the first.) We now see that
$$\begin{align}
0\le u\le{1\over3}&\implies u\le{4^0\over3}\le{4^u\over3}\\
{1\over3}\le u\le{1\over2}&\implies u\le{3/2\over3}\lt{4^{1/3}\over3}\le{4^u\over3}\\
{1\over2}\le u\le{2\over3}&\implies u\le{4^{1/2}\over3}\le{4^u\over3}\\
{2\over3}\le u\le{5\over6}&\implies u\le{5/2\over3}\lt{4^{2/3}\over3}\le{4^u\over3}\\
{5\over6}\le u\le1&\implies u\le{3\over3}\lt{4^{5/6}\over3}\le{4^u\over3}\\
1\le u\le{4\over3}&\implies u\le{4^1\over3}\le{4^u\over3}\\
{4\over3}\le u\le2&\implies u\le{6\over3}\lt{4^{4/3}\over3}\le{4^u\over3}
\end{align}$$
This establishes the inequality for all $u\in[0,2]$. To show it for $u\gt2$, it's enough to prove that $n+1\le4^n/3$ for all $2\le n\in\mathbb{N}$, since that inequality tells us that for $u\ge2$, 
$$u\le\lfloor u\rfloor+1\le{4^{\lfloor u\rfloor}\over3}\le{4^u\over3}$$
The proof of $n+1\le 4^n/3$ is by induction: the inequality is easily checked for the base case $n=2$ after which we see that
$$n+1\le{4^n\over3}\implies(n+1)+1\le{4^n\over3}+1\le{4^n\over3}+{3\cdot4^n\over3}={4^{n+1}\over3}$$
And this completes the proof of $2^x\gt2x-1$.
Remark: I did not expect the proof to require as many incremental steps (breaking $[0,2]$ into seven pieces) as it turned out to use. I'd be keen to see an alternative that uses fewer.
A: For $x<\frac{1}{2}, LHS>0>RHS.$
For $\frac{1}{2}<x<1, LHS>\sqrt{2}>1>RHS$.
For $1<x<2$,
$$2^x = e^{x\ln 2}$$
$$=\sum_{i=0}^\infty\frac{(x\ln2)^i}{i!}$$
$$>1+x\sum_{i=1}^\infty\frac{(\ln2)^i}{i!}$$
$$=1+x(e^{\ln2}-1)$$
$$=1+x$$
$$>2x-1$$
For $x\geq 2$, $$2^x\geq2x>2x-1$$
Equality holds at $x=2$, and further rate of growth (i.e. derivative) of LHS is more than that of RHS. So, we are done.
A: The following 'argument' is guided by following a geometric/intuitive path. Every claim can be justified using calculus, but it would be of interest to see how much of that firepower is really required.
We have two functions,
$\tag 1 f(x) = 2^x$
$\tag 2 g(x) = 2x - 1$
To say that the OP's inequality is false is equivalent to saying the graphs of these two functions have a nonempty intersection. This is equivalent to finding a $k \le -1$ and a linear function,
$\tag 3 h(x) = 2x + k$
such that the graph of $h(x)$ intersects the graph of $f(x)$ at exactly one point.
Let $(x_0,y_0)$ be this point of intersection.
Using calculus/algebra it can be shown that
$\tag 4 x_0 = \displaystyle{\frac{\ln(2) - \ln(\ln(2))}{\ln(2)}}$
The value of $k$ is the $y\text{-intercept}$, and so, using algebra,
$\tag 5 k = 2^{x_0} - 2 x_0 \approx -0.172 $
But this contradicts the premise that $k \le -1$. In conclusion, the graph of a function of the form $\text{(3)}$ is disjoint from the graphs of $f(x)$ if and only if: 
$\tag 6 k \lt  2^{x_0} - 2 x_0$
A: $\require{begingroup} \begingroup$
$\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}$
\begin{align} 
2^x&>2x-1
\tag{1}\label{1}
\end{align}
Let's consider
\begin{align} 
2^x&=2x-1
\tag{2}\label{2}
.
\end{align} 
It is known, that solution to equation \eqref{2}
can be found in terms of the
Lambert W function.
Let $2x-1=y$, $x=\tfrac12y+\tfrac12$:
\begin{align} 
2^{\tfrac12y+\tfrac12} &= y
\tag{3}\label{3}
,\\
\sqrt2\cdot 2^{\tfrac12y} &= y
,\\
\sqrt2\,\exp(\ln(2^{\tfrac12y})) &= y
,\\
\sqrt2\,\exp(\tfrac12y\cdot \ln2) &= y
,\\
y\,\exp(-\tfrac{\ln2}2\cdot y) &= \sqrt2
,\\
-\tfrac{\ln2}2\cdot y\,\exp(-\tfrac{\ln2}2\cdot y) &= -\tfrac{\ln2}2\cdot\sqrt2
,\\
\W\left(-\tfrac{\ln2}2\cdot y\,\exp(-\tfrac{\ln2}2\cdot y)\right) 
&=\W\left( -\tfrac{\ln2}2\cdot\sqrt2\right)
,\\
-\tfrac{\ln2}2\cdot y
&=\W\left( -\tfrac{\sqrt2}2\,\ln2\right)
\tag{4}\label{4}
,\\
y
&=
-\tfrac2{\ln2}\cdot 
\W\left( -\tfrac{\sqrt2}2\,\ln2\right)
\tag{5}\label{5}
,\\
x
&=
\tfrac12
-\tfrac1{\ln2}\cdot 
\W\left( -\tfrac{\sqrt2}2\,\ln2\right)
\tag{6}\label{6}
.
\end{align} 
The argument of $\W\left( -\tfrac{\sqrt2}2\,\ln2\right)$
is $-\tfrac{\sqrt2}2\,\ln2<-\tfrac1{\mathrm e}$,
hence there are no real solutions to \eqref{2}, and the function
\begin{align} 
2^x-(2x-1)
\tag{7}\label{7}
\end{align} 
never cross the $x$-axis. Evaluation of \eqref{7} at $x=0$
gives 
\begin{align} 
2^0-(2\cdot0-1)
&=2> 0
\tag{8}\label{8}
,
\end{align} 
hence 
\begin{align} 
2^x-(2x-1)>&0
\quad\text{ for all } x\in\mathbb R
\tag{9}\label{9}
\end{align}
and \eqref{1} follows.
$\endgroup$
