Determine the points where polynomial function intersects logarithmic function

Question

Given the function $n^k$ where $k$ is a constant such that $0<k \leq k_{max}$ where $k_{max}$ is the point at which $n^k$ first intersects with $log_2n$ determine:

1. $k_{max}$
2. For a given $k$, the point/points at which $n^k$ intersects $log_2n$

Finding $k_{max}$ I have no clue. Here's what I tried for part 2:

$log_2n = n^k$

$2^{n^k}= n$

$\sqrt[n]{2^{k^n}} = \sqrt[n]{n} $

$2^k = \sqrt[n]{n}$

And I don't know how to proceed from here.

PD: As you probably guessed, that is not a real question from a book, but a question I wrote myself. If there is an inaccuracy in the question, please let me know and I'll fix it. If there is something fundamentally wrong about it, then I apologize and delete it.

Answer 1

There isn't a closed form for this using elementary functions.

The solution to $x^k \leq \log x$ would appear in terms of the Lambert W function (product log function).

Answer 2

For $k \ne 0$, $n^k = \log_2(n)$ is equivalent to $$n = \exp \left( -\dfrac{- W(-k \ln(2))}{k}\right)$$ where $W$ is one of the branches of the Lambert W function. If $ k > 1/(e\ln(2))$, $-k \ln(2) < -1/e$ and there is no real solution. If $0 < k < 1/(e \ln(2))$, there are two real solutions, and the least is given by the principal branch of $W$. At $k = 1/(e\ln(2))$, there is one real solution $n = 2^e$.

Thus $k_{max} = 1/(e \ln(2))$.

A more elementary way of doing (1): with $n = e^t$ write $n^k = \log_2(n) = \dfrac{\ln(n)}{\ln(2)}$ as $$ k = \dfrac{\ln(t/\ln(2))}{t}$$ and find the maximum of this for $t > 0$.