Equivalent metrics to Euclidean meter I wanna show two following meterics on $\mathbb{R}$ are equivalent:

$d_1(x,y) = | x - y |$
$d_2 (x,y) = | e^x - e^y| $ 

Could anyone help me ?
 A: Mainlines of the proof
The map $f : x \mapsto e^x$ is continuously differentiable from $\mathbb R$ onto $(0,\infty)$. It is also strictly increasing. Therefore, it is a bijection.
Using the mean value theorem for $f$ and the inverse map (namely $\ln$), we get that for any finite interval $I \subseteq \mathbb R$, it exists $k,K >0$ such that
$$kd_2(x,y) \le d_1(x,y) \le K d_2(x,y)$$ for $x,y \in I$. This implies that the two metrics are equivalent.
A: Well,  Let $B(a,r) = \{x\in \mathbb R| |x-a|< r\} = (a-r, a+r)$.
Let $E(a,s) = \{x\in \mathbb R| |e^x-e^a| < s\}=\{x| e^a-s < e^x<e^a+s\}=\{x|\log(e^a-s)< x < \log(e^a+s)\}=(\log(e^a-s), \log(e^a+s))$
.....
Fix $r$.
Let $t= e^a-e^{a-r}$ so $e^a-t=e^{a-r}$ and $\log(e^a-t)=a-r$.
Let $t' = e^{a+r} - e^a$ so $\log(e^a+t)=a+r$.
Let $s'=\min(t,t')$ so $E(a,s')=(\log(e^a-s), \log(e^a+s'))\subset (\log(e^a-t),\log(a^a + t') =(a-r,a+r) =B(a,r)$.
.......
Likewise fix $s$.
Let $r' = \min(a-\log(e^a-s), \log(e^a+s)-a)$.
So $B(a,r')=(a-r', a+r')\subset (\log(e^a-s),\log(e^a+s)) = E(a,s)$.
......
And that's the definition of equivalence.
