How many solutions have $ 6 \ln(x^2+1) = e^x $ How many solutions   have $$ 6 \ln(x^2+1) =  e^x $$
I can use there derivatives.
my try
Let $$ f(x) = 6 \ln(x^2+1) -  e^x $$
$$ f'(x) = \frac{12x}{x^2-1} - e^x $$
I think that it is useful to find there extremes: 
So when $f'(x) = 0$?
$$\frac{12x}{x^2-1} = e^x$$
But I don't know how to solve that using calculus
I have seen this post but there was only link to relative post where was method which wasn't on my lecture.
 A: Too long for a comment.
You are looking for the zero's of 
$$f(x)=6 \ln(x^2+1) -  e^x$$
Since there is no problem with the logarithms, it is the same as looking for the intersections of
$$g(x)=\log \left(6 \log \left(x^2+1\right)\right)\qquad \text{and} \qquad h(x)=x$$ Function $g(x)$ has an infinite branch at $x=0$ and $g(x)=g(-x)$ so there is at least two roots.
When $x$ is "large" $f(x)\sim 12\log(x)-e^x$ varies faster; so there will be another one.
Edit
@Barry Cipra's comment reveals that I went to fast. If the negative root will always exist, the existence of the other roots depend on the value assigned to the constant in
$$g(x)=\log \left(\color{red}{k} \log \left(x^2+1\right)\right)$$ There will be a double root if, at the same time, $g(x)=x$ and $g'(x)=1$.
The second condition write
$$\frac{2 x}{\left(x^2+1\right) \log \left(x^2+1\right)}=1$$ which cannot solve explicitely. The numerical solution is $x=1.2801$ and, from the first condition,  $k=3.7072$.
So, 


*

*if $k <3.7072$, there is only one root (the negative one)

*if $k=3.7072$, there are a double positive root $x=1.2801$

*if $k>3.7072$, there are two distinct positive roots


Update
The second condition consists in finding the zero of function 
$$h(x)={2 x}-{\left(x^2+1\right) \log \left(x^2+1\right)}$$ By inspection, it is clear that the solution is between $1$ and $2$.
We can approximate the function by its simplest $[1,1]$ Padé approximant built at $x=1$ and solve for $x$ the linear equation given by its numerator. Applied to the present case, this would lead to
$$x=\frac{2-\log ^2(2)+\log (2)}{2+\log ^2(2)-\log (2)}\approx 1.23801$$
The first condition write
$$k=\frac{e^x}{\log \left(x^2+1\right)}$$
Replacing gives a nasty expression for $k$ the numerical value of which being $3.71123$.
A: We can see that the function $g(x)=6\ln(1+x^2)-e^x$ is decreasing on $(-\infty,0)$ and $\lim_{x\to-\infty}g(x)=\infty$, $g(0)=-1$. By intermediate value theorem, there is a unique negative root of $g$. To investigate positive roots, set $h(x)=6e^{-x}\ln(1+x^2)$ and observe that $h(1)=\frac{6\ln 2}e>1$ and that $h'(x)=6e^{-x}(\frac{2x}{1+x^2}-\ln(1+x^2)).$ We can show for all $c\in (0,1]$,  $\ln(1+c^2)<\frac{2c}{1+c^2}$ holds because mean value theorem implies the existence of $d\in (0,c)$ such that
$$
\frac{\ln(1+c^2)-\ln(1)}c=\frac{2d}{1+d^2}=\frac{2}{\frac1{d}+d}<1 \le\frac{2}{1+c^2}.
$$Thus $h'(c)>0$ on $(0,1]$. On the other hand, $\frac{2x}{1+x^2}-\ln(1+x^2)$ is decreasing on $[1,\infty)$ with $\lim_{x\to\infty } \frac{2x}{1+x^2}-\ln(1+x^2)=-\infty$. By IVP, there is a unique $x_0>1$ such that $h'(x_0)=0$, and  we conclude from the previous observations that $h(x)$ is increasing on $x<x_0$ and decreasing on $(x_0,\infty)$. As was already noted, we have $h(x_0)\ge h(1)>1$, and this in turn implies there are exactly 2 roots of the equation $h(x)=1$, i.e. $g(x)=0$. To sum up, there is one negative root and two positive roots of the equation $g(x)=0$.
A: Rearrange to the form $x=g(x)$ and apply $x_0=1, x_{n+1}=g(x_n)$, iterating to see where it goes:
For example, we take the $\ln$ of both sides and get:
$$x_{n+1}=\ln\bigg[6\ln(x_n^2+1)\bigg]$$
Which leads to $x\approx 2.461953976$ (use a calculator with the $Ans$ function)
Now try rearranging to form a different expression of the form $x=h(x)$, and see where that leads.
