TITLE: Number of solutions of $e^x=x^3$
QUESTION [6 upvotes]: How can I analytically obtain the number of solutions of the equation $e^x=x^3$?
I know Lambert's $W$ function, but even using that, we need to know the value of $W(-1/3)$ which I think we cannot calculate without a calculator/graph/etc.
I do not need the exact solutions of the equation, I just need to know the number of solutions it has. Is there any good approximation?
I tried using a method of comparing slopes of the two functions on the L.H.S and R.H.S , but it was turning out to be too lengthy.

REPLY [2 votes]: Not using Lambert function, consider that you look for the zero's of 
$$f(x)=e^x-x^3$$But easier would be
$$g(x)=x-3\log(x)$$for which $$g'(x)=1-\frac 3x \qquad \text{and} \qquad g''(x)=\frac 3{x^2} > 0 \qquad \forall x$$
The first derivative cancels when
$$x_*=3 \implies g(3)=3-3\log(3) <0$$ So, two roots.
To approximate them, use a Taylor series around $x=3$ and get
$$g(x)=(3-3 \log (3))+\frac{1}{6} (x-3)^2+O\left((x-3)^3\right)$$ Ignoring the higher order terms, the solutions are given by
$$x_\pm=3\pm 3 \sqrt{2 (\log (3)-1)}$$ that is to say $1.6677$ and $4.3323$; not too bad for a first approximation.
If you want to polish the roots, using Newton method we should get the following iterates
$$\left(
\begin{array}{cc}
n & x_n \\
 0 & 1.667700786 \\
 1 & 1.834637542 \\
 2 & 1.856830114 \\
 3 & 1.857183772 \\
 4 & 1.857183860
\end{array}
\right)$$
$$\left(
\begin{array}{cc}
n & x_n \\
 0 & 4.332299214 \\
 1 & 4.546901400 \\
 2 & 4.536427193 \\
 3 & 4.536403655
\end{array}
\right)$$
Edit
Sooner or later, you will learn that, better than with Taylor series, functions can be locally approximated using Padé approximants. Using the $[2,2]$ one (not too complex to build), we shoulf get
$$g(x)=\frac{(3-3 \log (3))-\frac{2(\log (3)-1)}{3} t +\frac{(8+\log
   (3))}{54} t^2  } {1+\frac{2 }{9}t-\frac{1}{162}t^2}$$ where $t=(x-3)$. 
Solving the quadratic in numerator will lead to
$$t=\frac{9 \left(2 \log
   (3)-2\pm\sqrt{6 \left(-2+\log ^2(3)+\log (3)\right)}\right)}{8+\log (3)}$$ given then, as estimates , $x_1=1.85574$ and $x_2=4.53443$.