Asymptotic estimate of the form $k=C\log\log n$. Suppose that we have
$$
n=k^{\frac{1}{\beta}k^22^{k+12}\log\frac 1{\alpha}}
$$
where $\alpha,\beta\in(0,1)$. It is claimed that
$$
k\sim C\log\log n
$$
for some constant $C=C(\alpha,\beta)$. How do one deduce such an estimate?
I am totally lost here, any help is very much appreciated.

I assume logarithms in base $2$. This would not change anything anyway, besides constants.
First, note that
$$
n=k^{\frac{1}{\beta}k^22^{k+12}\log\frac{1}{\alpha}}
= 2^{\frac{1}{\beta}k^22^{k+12}\log\frac{1}{\alpha} \log k}
= 2^{\frac{2^{12}\log\frac{1}{\alpha}}{\beta}k^22 ^{k} \log k}
$$
so
$$
\log n = \frac{2^{12}\log\frac{1}{\alpha}}{\beta}\cdot k^22 ^{k} \log k
= C'\cdot 2 ^{k + 2\log k+\log\log k}
$$
setting $C'\stackrel{\rm def}{=} \frac{2^{12}\log\frac{1}{\alpha}}{\beta}$.
Taking the logarithm again, we get
$$
\log\log n = k + 2\log k+\log\log k + O(1)
$$
and, when $k\to\infty$, the right-hand side satisfies $$k + 2\log k+\log\log k + O(1)\operatorname*{\sim}_{k\to\infty} k$$
which implies
$$
\log\log n\operatorname*{\sim}_{k\to\infty} k.
$$
