$d$ is finer than $d'$ I have $E$ endowed with two metrics $d$ and $d'=\ln(1+d)$ 
How to prove that $d$ is finer than $d'$; that is how to prove that:
For all $x\in E$ any open ball for $d'$ contains an open ball for $d$?

It also can be easily proved that for $x>0$, $ln(1+x)\leq x$, which shows that for all $x,y\in E$, $$d'(x,y)\leq d(x,y)$$. This implies that $$B_d(x,\epsilon) \subseteq B_{d'}(x,\epsilon)$$ for all $x,\epsilon \in R$.
There is another way to find that. Using the result you found:
$$B_{d'}(x,\epsilon) = B_d(x, e^{\epsilon} - 1)$$ So we also trivially get that for any $x,\epsilon \in R$, $$B_d(x, e^{\epsilon} - 1) \subseteq B_{d'}(x,\epsilon) $$
Thus every open $d'$-ball contains an open $d$-ball.
