Solve $\arctan(x)-\ln(x)=0$ $f:(0, \infty)\rightarrow \mathbb{R}; f(x)=\arctan(x)-\ln(x).$
An interval $I$ in which the equation $f(x)=0$ has a real solution is: a. $(0,1)$; b.$(1,e)$; c.$(e,e^{2})$; d.$(e^{2}, \infty)$;
I tried to solve $f(x)=0$ with Rolle's Theorem but the derrivative has no real solutions.
Also, I tried with graphic method but I get stuck with intersection point of the functions.

$f$ is obviously continuous.
Recall the following fact about continuous functions : if a continuous function $g$ has the propery that $g(a)<0$ and $g(b)>0$ or $g(a)>0$ and $g(b)<0$ for some $a$ and $b$ ($a<b$), then $\exists c \in (a,b)$ such that $g(c)=0$.
Since you are faced with a multiple choice question, you only have to compute the values of the function at the points that appear in the answer choices.
Note that even if your function is not defined at all those points, the above statement still works for limits.
i.e. If for a continuous function $h$ and for some $a$ and $b$ ($a<b$) we have that $\lim\limits_{x\to a^{+}} h(x)<0$ and $\lim\limits_{x\to b^{-}} h(x)>0$ or $\lim\limits_{x\to a^{+}} h(x)>0$ and $\lim\limits_{x\to b^{-}} h(x)<0$, then $\exists c\in(a,b)$ such that $h(c)=0$.
Also observe that since $f$ is strictly decreasing on its domain the equation will have only one real solution.
