Solve $4 \log_2 (n) \leq \frac{n}{2}$ I have to find the $n \in \mathbb{N}$ whenever $4 \log_2 (n) \leq \frac{n}{2}$. So
$$4 \log_2 (n) \leq \frac{n}{2}$$
$$ \iff \log_2 (n) \leq \frac{n}{8}$$
$$ \iff \frac{1}{n}\log_2 (n) \leq \frac{1}{8}$$
$$ \iff \log_2 (n^\frac{1}{n}) \leq \frac{1}{8}$$
$$ \iff n^\frac{1}{n} \leq 2^\frac{1}{8}$$
I am not sure how I can use that to find the appropriate $n$. Is there a better technique to find the right $n$?
EDIT
Not sure I can use induction here. I have to find the minimum $n_0 \in \mathbb{N}$ where $4 \log_2 (n) \leq \frac{n}{2}$ for $n \geq n_0$. How to find that $n_0$?

 A: As a real, the solution of $$4 \log_2 (n) = \frac{n}{2}$$ is given in terms of Lambert function
$$n=-\frac{8 }{\log (2)}W_{-1}\left(-\frac{\log (2)}{8}\right)\sim 43.5593$$ so $\lceil n \rceil=44$.
Checking
$$4 \log_2 (43) - \frac{43}{2} \sim +0.205059$$
$$4 \log_2 (44) - \frac{44}{2} \sim -0.162274$$
Edit
If you do not know about Lambert function (or if you cannot use it), consider the function and derivatives
$$f(n)=\frac{4 \log (n)}{\log (2)}-\frac{n}{2}$$
$$f'(n)=\frac{4}{n \log (2)}-\frac{1}{2}$$
$$f''(n)=-\frac{4}{n^2 \log (2)} \quad < 0 \quad \forall n$$
The first derivative cancels at
$$n_*=\frac{8}{\log (2)}\implies f(n_*)=-\frac{4 \left(1+\log \left(\frac{\log (2)}{8}\right)\right)}{\log (2)} <0$$ Develop $f(n)$ as a Taylor series around $n_*$ to get
$$f(n)=f(n_*)+\frac 1 2 f''(n_*)(n-n_*)^2+O((n-n_*)^3$$
Solve the quadratic to get a starting point
$$n_0=n_*+\sqrt{-2\frac{f(n_*)}{f''(n_*)}}\sim 31.1687$$ Now, start Newton method and get the following iterates
$$\left(
\begin{array}{cc}
k & n_k \\
 0 & 31.1687 \\
 1 & 44.7107 \\
 2 & 43.5645 \\
 3 & 43.5593
\end{array}
\right)$$
A: A hint:
You can use the fact that $2^{x}$ is an increasing function.
$$8log_{2}(n)\leq n$$
whenever
$$2^{8log_{2}(n)} \leq 2^{n} $$
simplifying
$$n^{8} \leq 2^{n} $$
Whenever
$$0 \leq 2^{n} - n^{8} $$
Either solve the equation 2^{x} - x^{8} = 0 and take the next largest integer value which follows the largest solution. If this is within the methods and techniques that you have been taught. The solution looks quite messy so it may be too complex in your course to find the exact value of such x.
Otherwise try solutions of the form $n=2^{k}$ find the first such $k > 0$ which satisfies the inequality, which is $k=6$ and you can look for $2^{5}=32<n<64=2^{6}$. Plotting the graph also helps narrow down where to look.
$n=44$ is the first such value.
Maybe there is some slick way to finish the computation that I am missing.
A: Iterate: $$k_0=1, k_{n+1}=8\log_2(k_n)\implies \lim_{j\to\infty} k_j\approx 43.55926044$$
