How to use other method to prove $2^x>2x-1$.

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.