At what values of $x$ do $e^{3x}$ and $x^3$ have parallel tangent lines? I was given this problem:  

Let $f$ be the function defined by $f(x)=e^{3x}$, and let $g$ be the function defined by $g(x)=x^3$. At what value of $x$ do the graphs of $f$ and $g$ have parallel tangent lines?    

To solve this problem, I set the derivatives of $f$ and $g$ equal to each other: $3x^2=3e^{3x}$. Now, I just need to solve for $x$, but for some reason I am completely at a loss of how to do that. I divided the problem by $3$, which got me $x^2=e^{3x}$. But, now what? 
I also plugged it into my calculators solver, which got me the correct answer: $-.484$. But, I don't know how to find that answer without solver.
 A: This problem can be solved using the Lambert-W function as follows.
$$x^2=e^{3x}$$
$$\sqrt{x^2}=\sqrt{e^{3x}}$$
$$x=\pm e^{3x/2}$$
$$-\frac32 x=\mp\frac32 e^{3x/2}$$
$$-\frac32 xe^{-3x/2}=\mp\frac32$$
$$-\frac32 x=W_k\left(\mp\frac32\right)$$
$$x=-\frac23W_k\left(\pm\frac32\right)$$
for any branch of the function $k\in\mathbb{Z}$. The only real solution would be when $k=0$ and $+$ is taken for the $\pm$ giving
$$x=-\frac23W_0\left(\frac32\right)\approx-0.4839075718\dots$$
which is the value provided by your calculator.
A: Since the other answer provides the exact solution using the Lambert $W$ function, let me show you how you can prove that the equation $x^2 = e^{3x}$ has one and only one solution using elementary methods.
Let $F(x) = x^2$ and $G(x) = e^{3x}$.
If you plot $F$ and $G$ on the same plane, you can see that there is exactly one point $x$ such that $F(x) = G(x)$. Here $y = F(x)$ is in red and $y = G(x)$ is in blue:

Now, if you want to prove that there is a unique solution to the equation, you can reason as follows.
Consider first the interval $(-\infty, 0]$.


*

*Since $F$ is decreasing and $G$ is increasing, there is at most one $x \in (-\infty, 0]$ such that $F(x) = G(x)$.

*Since $F(-1) > G(-1)$ and $F(0) < G(0)$, there is at least one $x \in (-1, 0)$ such that $F(x) = G(x)$.


Combining the two results, we know that there is exactly one $x \in (-\infty, 0]$ such that $F(x) = G(x)$. Moreover, $x \in (-1, 0)$.
For the interval $(0, \infty)$ the situation is a bit trickier, because both functions are increasing. But if you know that $x < e^x$ for any $x > 0$, you can prove that $F(x) < G(x)$ as follows:
$$x^2 < (e^x)^2 = e^{2x} < e^{3x}$$
Therefore there are no solutions in $(0, \infty)$.
A: You can use Fixed-point iteration method to compute the root up to desired accuracy, which goes like this
$x^2=e^{3x}\implies x=-e^{3x/2}=\phi (x)$.....(1)
As $\phi(x)$ is continuous in $[-1,0]$ with $\phi(x)\in [-1,0]$ and the condition
|$\phi'(x)$|$<1$ is satisfied in $[-1,0]$, hence the iterative scheme from (1)
$x_{n+1}=-e^{3x_n/2}$ , $n=0,1,2,...$
is guaranteed to converge to the root $-0.4839...$ starting with any choice of $x_0\in[-1,0]$.
