Number of solutions of $e^x=x^3$ 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.
 A: I understand that you are not looking for an exact solution. Nevertheless it is for interest to show the analytical solution :
$$e^x=x^3$$
$$e^{x/3}=x$$
$$xe^{-x/3}=1$$
$$\frac{-x}{3}e^{-x/3}=-\frac13$$
According to the definition of the Lambert W function :
$\quad \frac{-x}{3}=W\left(-\frac13 \right)$
The Lambert W function is multivalued. In the real domain the two branches are noted $W_0$ and $W_{-1}$. So they are two real solutions :
$$x=-3W_0\left(-\frac13  \right)\simeq 1.857183$$
$$x=-3W_{-1}\left(-\frac13  \right)\simeq 4.536403$$
Approximates :
$$W_0(x)\simeq-\sum_{k=1}^n \frac{(-1)^k k^{k-1}}{k!}x^k$$
The series is very slowly convergent. For enough accuracy $n$ must be large.
$$W_{-1}(x)\simeq -\theta-\ln(\theta+\ln(\theta+\ln(\theta+...)))$$
with $\theta=-\ln(-x)$ .
The number of terms must be large for enough accuracy.
The above recursive formulas are not recommended (except with computer). Better use numerical calculus with iterative method such as Newton-Raphson.
A: 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 $x_2=4.53443$.
A: Since $e^x>0$, the equation is equivalent to $x=3\log x$. Consider $f(x)=x-3\log x$ (for $x>0$. The limit at $0$ and $\infty$ are both $\infty$.
The derivative is $f'(x)=1-3/x=(x-3)/x$. So the function has an absolute minimum at $x=3$ and $f(3)=3-3\log 3=3(1-\log3)$. Thus the equation has exactly two solutions.
In order to approximate them you can use any numerical method. One is $\approx1.8571$.

Suppose you have $e^x=x^2$; then the function to study is $f(x)=x-2\log\lvert x\rvert$. The limits are
$$
\lim_{x\to-\infty}f(x)=-\infty,\quad
\lim_{x\to0^-}f(x)=\infty,\quad
\lim_{x\to0^+}f(x)=\infty,\quad
\lim_{x\to\infty}f(x)=\infty
$$
and
$$
f'(x)=1-\frac{2}{x}=\frac{x-2}{x}
$$
The function is increasing over $(-\infty,0)$ and $[2,\infty)$; decreasing over $(0,2]$.
Since $f(2)=2-2\log2>0$ ( $e>2$ ), we have one solution in $(-\infty,0)$ and no solutions in $(0,\infty)$.
