minimizing the maximum between a degree of a tree and its height
Yes, it is true that $\dfrac{\log n}{\log \frac {\log n}{\log \log n}} = \Theta(\dfrac{\log n}{\log \log n}).$ In fact, we have the following more accurate approximation, 
$$\frac{\log n}{\log \frac {\log n}{\log \log n}} \sim \frac{\log n}{\log \log n}.\tag{app}$$

Proof. $$\begin{aligned}
\lim_{n\to\infty}\frac{\ \frac{\log n}{\log \frac {\log n}{\log \log n}}\ }{\ \frac{\log n}{\log \log n}\ }&amp;=\lim_{n\to\infty}\frac{\log\log n}{\log \frac {\log n}{\log \log n}}
=\lim_{n\to\infty}\frac{\log\log n}{\log\log n-\log\log \log n}\\
&amp;=\frac{1}{1-\lim_{n\to\infty}\frac{\log\log \log n}{\log\log n}}
=\frac{1}{1-0}=1.
\end{aligned}$$

So, if we set $k = \frac {\log n}{\log \log n}$, then the height of the tree,  $\frac{\log n}{\log \frac {\log n}{\log \log n}}$ is $\frac{\log n}{\log \log n}$ asymptotically and, hence, so is the larger one between the height of the tree and $k$. 



Now, the question is what is the asymptotic behavior of the minimum of the larger one between $\log_kn $ and $k$, i.e. $h(n)=\min_k\max(\log_kn,k)$. 

Since $k$ is increasing with respect to $k$ while $\log_k n$ is decreasing with respect to $k$, the minimum of the larger one between $k$ and $\log_k n$ for a given $n$ is obtained when $k=\log_k n$, assuming $k,n&gt;1$.

What is the solution to $k=\log_k n$ for a given $n$? 
$$k=\log_kn\Leftrightarrow k= \frac{\log n}{\log k}\Leftrightarrow \log k\cdot k= \log n\Leftrightarrow \log k\cdot e^{\log k}= \log n$$
By the definition of the Lambert $W$ function, we should have
$$\log k = W_0(\log n),$$
where $W_0(\cdot)$ is the principal branch of the Lambert $W$ function, as we are only interested in the real positive values. So, 
$$k = e^{W_0(\log n)}.$$
Accordingly,
$$h(n)=\left.\max(\log_kn,k)\right|_{k=e^{W_0(\log n)}}=e^{W_0(\log n)}.$$

The same article on Lambert $W$ function tells that
$$ W_0(x) = \log x - \log\log x + o(1).$$
Substituting $\log n$ for $x$, we have 
$$ W_0(\log n) = \log\log n - \log\log\log n + o(1).$$
So,
$$ h(n) =e^{W(\log n)}= e^{\log\log n} e^{-\log\log\log n} e^{ o(1)}=\frac{\log n}{\log\log n}\ e^{o(1)},$$
Since $e^{o(1)}$ goes to 1 when $n$ goes to infinity,
$$h(n)\sim{\frac{\log n}{\log\log n}}.$$





Combining the above conclusions, we see that if we set $k = \frac {\log n}{\log \log n}$, then the larger one between the height of the tree and $k$ approaches its minimum $h(n)$ as well as ${\frac{\log n}{\log\log n}}$ asymptotically.



To be nit picky, it should  be pointed out that we have been ignoring the expected requirement that $k$ should be an integer. It would require cumbersome analysis to adapt the above analysis and conclusion for integer $k$. In the end, we have still
$$\min_{1\lt k\le n,\ k\in\mathbb N}\max(k, \log_kn)\sim\frac{\log n}{\log\log n}.$$