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.
 A: For $x>0$, the function is indeed growing. Then
$$f(0)>0,$$
$$f(1)>0,$$
$$f(e)>0,$$
$$f(e^2)<0,$$
$$f(\infty)<0$$ gives you the answer.
A: $f$ 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.
A: If you want the solution, consider that you look for the zero of function
$$f(x)=\tan ^{-1}(x)-\log (x)$$ for which
$$f'(x)=\frac{1}{x^2+1}-\frac{1}{x} \,\, < 0 \,\,\forall x \qquad \text{and}\qquad f''(x)=\frac{1}{x^2}-\frac{2 x}{\left(x^2+1\right)^2}\,\, > 0 \,\,\forall x$$
So, by Darboux theorem, starting iterating at any $x_0$ such that $f(x_0)>0$ will lead to the solution without any overshoot of the solution.
Being very lazy, starting with $x_0=1$ would give the following iterates
$$\left(
\begin{array}{cc}
 n & x_n \\
 0 & 1.0000000 \\
 1 & 2.5707963 \\
 2 & 3.5632065 \\
 3 & 3.6909683 \\
 4 & 3.6925854 \\
 5 & 3.6925857
\end{array}
\right)$$ The solution is not recognized by inverse symbolic calculators.
If you want to provide a fancy approximation, remembering that $\tan \left(\frac{5\pi }{12}\right)=2+\sqrt 3$, make a Taylor expansion around this value to get locally
$$\tan ^{-1}(x)-\log (x)=\left(\frac{5 \pi }{12}-\log \left(2+\sqrt{3}\right)\right)-\frac{3
   \left(x-\sqrt{3}-2\right)}{4
   \left(2+\sqrt{3}\right)}+O\left(\left(x-\sqrt{3}-2\right)^2\right)$$ Neglecting the higher order terms, this gives, as a better estimate,
$$x=\frac{1}{9} \left(2+\sqrt{3}\right) \left(9+5 \pi -12 \log
   \left(2+\sqrt{3}\right)\right) \approx 3.69244 $$ for which $f(x)$ is still positive.
A: Use Intermediate value Theorem. 
$$f(x) = \arctan x - \log(x)$$
$$f(1) = \pi/4$$
$$f(e) = \arctan e - 1 > 0$$
This holds because $\arctan \sqrt 3 = \pi/3 > 1$ and since $\arctan x$ is strictly increasing, $\arctan e > \arctan \sqrt 3  
> 1$
$$f(e^2) = \arctan e^2 - 2 < 0$$
This is because the range of $\arctan x$ is $(-\pi/2,\pi/2)$
