My question is how to solve this problem specifically, but this type of problem generally?
$$ 2^{n/8} - n < 0 $$
EDIT I should have probably been more specific. This is a computer science algorithms running time question: "For inputs of size $n$, insertion sort runs in $8n^2$ steps, while merge sort runs in $64nlog_2n$ steps. For which values of $n$ does insertion sort beat merge sort?"
So I made this inequality $$8n^2 < 64n\,log_2\,n$$
In the solutions manual they go from $$2^n < n^8$$ to $$2 \leq n \leq 43$$
And I guess I am wondering how they got to there?
So I am not sure if you are just supposed to approximate this with like a graph or if there is a 'simple' way to solve this algebraically?
Consider that you look for the zero's of function $$f(n)=2^{n/8}-n$$ for which $$f'(n)=2^{\frac{n}{8}-3} \log (2)-1\qquad \text{and} \qquad f''(n)=2^{\frac{n}{8}-6} \log ^2(2)\quad > 0 \quad \forall n$$ The first derivative cancels at $$n_*=24-\frac{8 \log (\log (2))}{\log (2)}\approx 28.2301$$ This point is a minimum by the second derivative test and $$f(n_*)=\frac{8 \left(1+\log \left(\frac{\log (2)}{8}\right)\right)}{\log (2)}\approx -16.6886$$ So, two roots.
To approximate the roots, build a Taylor expansion around $n_*$ to get $$f(n)=\frac{8 \left(1+\log \left(\frac{\log (2)}{8}\right)\right)}{\log (2)}+\frac{\log (2)}{16} \left(n-n_*\right)^2+O\left(n-n_*)^3\right)$$ This gives two solutions $8.60300$ and $47.8573$.
Now, let us start Newton method with these estimates. The iterates would be
$$\left( \begin{array}{cc} k & n_k \\ 0 & 8.6029995 \\ 1 & 0.6563652 \\ 2 & 1.0991250 \\ 3 & 1.0999970 \end{array} \right)$$
$$\left( \begin{array}{cc} k & n_k \\ 0 & 47.857262 \\ 1 & 44.427279 \\ 2 & 43.601473 \\ 3 & 43.559365 \\ 4 & 43.559260 \end{array} \right)$$
As said in comments, there is explicit solution in terms of Lambet function $$-\frac{8 W_0\left(-\frac{\log (2)}{8}\right)}{\log (2)} \qquad \text{and} \qquad -\frac{8 W_{-1}\left(-\frac{\log (2)}{8}\right)}{\log (2)}$$