Solve $n^4>3^n$ $n^4>3^n$
I'm trying to solve this inequality problem, but everything I can find online is either how to solve log inequality problems or exponent inequality problems.
I think this may be a combination of both of them, but I can't figure out how.
 A: Set $f(n) = n^4/3^n$. Then 
$$\frac{f(n+1)}{f(n)} = \frac{1}{3}\left(1 + \frac{1}{n}\right)^4 < 1$$
for $n\geq 4$. Thus $f(n)$ is decreasing for $n \geq 4$. Since
$$f(8) = \frac{8^4}{3^8} = \left(\frac{\sqrt{8}}{3}\right)^8 < 1,$$
it follows that $f(n) < 1$ for all $n\geq 8$. The remaining cases can be checked by hand.
A: Not an answer (in view of the other ones), just a suggestion for future work.
Most questions of the form:
$$m^n<?>n^m$$
are quickly addressed as equivalent to:
$$m^{\frac{1}{m}}<?>n^{\frac{1}{n}}$$
which is addressed by the behavior of:
$$f(x)=x^{1/x}$$
in its domain $(0,+\infty)$, with care to be taken at $x_0=e$.
A: $$n^4>3^n\\
n^{\frac 1n}>3^\frac 14\\$$
Solving numerically, or better still, plotting graphs of LHS and RHS, will show that the inequality holds for $1.517<n<7.175$. 
If $n\in\Bbb{Z}$ then 
$$2\leq n\leq 7$$
A: If you solve the places where the equations are equal, you get three points.
I'm not sure how to find this, I'll edit the question if I find a way, but these 3 points are: (source)
$$n=-\frac{4\text W\left(-\frac{\log 3}4\right)}{\log 3}\approx1.5167764122831706197$$
$$n=-\frac{4\text W\left(\frac{\log 3}4\right)}{\log 3}\approx-0.80224643154616987684$$
$$n=-\frac{4\text W_{-1}\left(-\frac{\log 3}4\right)}{\log 3}\approx7.1747558273891581697$$
Where $\text W(x)$ is the lambert W function and $\text W_{-1}(x)$ is the analytic continuation of the lambert W function.
This gives rise to the two cases where the original equation is true.
$$n^4>3^n$$
$$\text{is true if}$$
$$-\frac{4\text W\left(-\frac{\log 3}4\right)}{\log 3} < n < -\frac{4\text W_{-1}\left(-\frac{\log 3}4\right)}{\log 3}$$
$$\text{or}$$
$$n<-\frac{4\text W\left(\frac{\log 3}4\right)}{\log 3}$$
