Mfcttrf

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.

This problem can be solved using the Lambert-W function as follows.
$$x^2=e^3x$$
$$sqrtx^2=sqrte^3x$$
$$x=pm e^3x/2$$
$$-frac32 x=mpfrac32 e^3x/2$$
$$-frac32 xe^-3x/2=mpfrac32$$
$$-frac32 x=W_kleft(mpfrac32right)$$
$$x=-frac23W_kleft(pmfrac32right)$$
for any branch of the function $kinmathbbZ$. The only real solution would be when $k=0$ and $+$ is taken for the $pm$ giving
$$x=-frac23W_0left(frac32right)approx-0.4839075718dots$$
which is the value provided by your calculator.

Since the other answers provide 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 $(-infty, 0]$.

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


2. 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 $(0, infty)$.