Solve for $x$: $2^x = x^3$ What category of equation is this?
What methods are available to solve it?
$2^x -x^3 = 0$ where $x\in\Bbb R$
 A: *

*it is an exponential equation or an exponential diophantine equation if only integers (or fractions) are allowed

*you may find a real or imaginary solution :

*

*using iterations starting with $x_0=2$ ($x_{n+1}=2^{x_n/3}$, the converse $x_{n+1}=\frac {3\ln(x_n)}{\ln 2}$ or faster Newton-Raphson iterations)

*with the LambertW function (since $n e^{-\frac {n\ln 2}3}=1$ is wanted) 

*graphically

*...


A: This is an equation without solution, if $n$ is supposed to be integer. In order for $2^n=n^3$ to hold, $n$ should be


*

*positive, since $2^n$ is,

*a power of $2$, since $n^3$ would otherwise have an odd prime factor,

*divisible by $3$ in order for $2^n$ to be a cube.


Obviously the last two conditions are contradictory.
Since $n$ has now been renamed $x$ and has become real, there are more solutions. Probably just the two indicated by @avatar, but some effort is required to show that this is all, since $f: x\to 2^x-x^3$ is not a convex function: it has two inflexion points, one near $0.08$ and another one near $6.3$.
A: You can find a solution in terms of the Lambert W Function. Rewrite as:
$$
1 = \frac{x^3}{2^x} = x^3 \exp(-x\log 2)
$$
and take the real cube root:
$$
1 = x \exp \left(-\frac{x\log 2}{3}\right)
$$
Now multiply by $-\log 2/3$:
$$
-\frac{\log 2}{3} = -\frac{x\log 2}{3} \exp\left(-\frac{x\log 2}{3}\right)
$$
Hence:
$$
x = -\frac{3W_0\left( -\frac{\log 2}{3}\right)}{\log 2}
$$
where $W_0$ is the principal branch of Lambert's W. The value is about 1.37.
There is another real root between 9 and 10, which is found on the second branch of the Lambert W function:
$$x = -\frac{3W_{-1}\left(-\frac{\log 2}{3}\right)}{\log 2}$$
and whose value is about 9.94.
A: Consider $f(n)=2^n-n^3$. $f(9)=-217$ and $f(10)=24$,therefore,there exists a root between $9$ and $10.$ You can solve $f(n)=0$ numerically using Newton-Raphson method taking $x_0=9$. 
Also $f(1)=1$ and $f(2)=-6$, therefore there will be a root between 1 and 2.
For $n<1$, $f(n)$ is always positive and for $n>10$, $f(n)$ is always positive,so there are no more roots other than between $9$ and $10, 1$ and $2.$
