If $a=b\log b$, how does $b$ grow asymptotically?

Question

If $a=b\log b$, how does $b$ grow asymptotically in terms of $a$?

I think the answer should be $b=\Theta\left(\frac{a}{\log a}\right)$. I tried taking logs to get $\log a=\log b+\log\log b$, but it's not clear how to separate $b$.

Answer 1

Write $a=(\log b) e^{\log b}$ to see that $$\log b=W(a)\Rightarrow b=e^{W(a)}=\frac a {W(a)}=\frac a {\log a - \log\log a+o(1)}$$ where $W$ is Lambert $W$-function.

Answer 2

From $a = b\log b$ we see that $b \to \infty$ as $a \to \infty$. So, taking logs we deduce that

$$ \log a = \log b + \log\log b \sim \log b $$

as $a \to \infty$, and hence

$$ b = \frac{a}{\log b} \sim \frac{a}{\log a} $$

as $a \to \infty$.