The same open sets How to prove that the two distances $d(x,y)=|x-y|$ and $d'(x,y)= 
|\exp(x)-\exp(y)|$ induced the same topology on R ? 
I know that for $a\in R, r>0$ 
$B_d(a,r)=]a-r, a+r[$, 
$B_{d'}(a,r)=]\ln(\exp(a)-r), \ln(\exp(a)+r)[$
How to do ?
 A: I think that you can use this fact:
If $X$ is a first countable space then a function $f: X\to Y$ is continuos if and only if for each convergent sequence $\{x_n\}_n$ convergent to a point $x$ then $\{f(x_n)\}_n$ is convergent to $f(x)$
Now you can observe that if $X$ has two Topology $\tau_1$ and $\tau_2$ such that $(X,\tau_1)$ is first countable and for  each $\tau_1-$convergent sequence $\{x_n\}_n$ to a point $x$ is $\tau_2-$ convergent sequence to the point $x$ then $\tau_2\subseteq \tau_1$ infact by first lemma you have that the identity map 
$Id: (X,\tau_1)\to (X,\tau_2)$ 
is a countinuos function so for each open set $A\in \tau_2$ then $Id^{-1}(A)=A$ is open in $\tau_1$. So $\tau_2\subseteq \tau_1$. 
In your case $\mathbb{R}$ is a first countable set with respect both Topology because they are induced by a distance. You have also that the function $exp$ is a countinuos function from $\mathbb{R}$ to $\mathbb{R}$ so if $\{x_n\}_n$ is convergent to $x$ with respect $d_1$ then
$\lim_{n\to \infty}|exp(x_n)-exp(x)|=|exp(x)-exp(x)|=0$
so $\{x_n\}$ it is convergent to $x$ with respect to $d_2$ so you have that 
$\tau_2\subseteq \tau_1$ 
Now you can hypothesize that a sequence  $\{x_n\}_n$ is convergent with respect $d_2$ to a point $x$ then 
$\{x_n\}_n$ is convergent with respect $d_1$ to a point $x$ because $log$ is a countinuos function and so 
$\lim_{n\to \infty}|x_n-x|=\lim_{n\to\infty}| log(e^{x_n})-log(e^x)|=0$ 
So you have that $\tau_1\subseteq \tau_2$ 
It is obvious that you can generalize your statement:
If $f:\mathbb{R}\to \mathbb{R}$ is an injective function then it is possible define a distance $d_2$ on $\mathbb{R}$ in the following way:
$d_2(x,y):=|f(x)-f(y)|$ 
If $f:(\mathbb{R},\tau_{eu})\to (\mathbb{R},\tau_{eu})$ is a continuos function then $\tau_2\subseteq \tau_{eu}$.
If $f$ in bijective and your inverse it is continuos then 
$\tau_{eu}\subseteq \tau_2$
A: Suppose that $(a, b)$ is an open ball in the ordinary topology. (That is, it is the ball
$$
B_d\left(\frac{a+b}2, \frac{b-a}2\right), 
$$
but that doesn't matter for our argument.) Then take
$$
x = \log\left(\frac{\exp(a)+\exp(b)}2\right), r = \frac{\exp(b) - \exp(a)}{2}.
$$
We have that
$$
B_{d'}(x, r) = \left(\log(\exp(x) - r), \log(\exp(x) + r)\right) = (a, b),
$$
where I've used your expression for $B_{d'}(x, r)$ in your question. Conversely, consider $B_{d'}(x, r)$ for arbitrary $x, r > 0$. We consider two cases. If $\exp(x) - r$ is positive, then we have your expression again, namely
$$
B_{d'}(x, r) = \left(\log(\exp(x) - r), \log(\exp(x) + r)\right),
$$
and then we already have an open interval in the normal sense. On the other hand, if $\exp(x) - r \leq 0$, then we have that
$$
B_{d'}(x, r) = (-\infty, \log(\exp(x) + r)).
$$
Can you prove that this set is open in the ordinary topology (induced by $d$)?
