We have $a=b^b$, so $$\log(a)=b\log(b)$$ $$x=\frac{x}{W(x)}\log\left(\frac{x}{W(x)}\right)$$ $$b=\frac{\log(a)}{W(\log(a))}$$ Next we have $c=d^{(d^{d})}$, so $$\log(c)=d^{d}\log(d)$$ In general $^{k}m=n$, $$^{k-1}m\log(m)=\log(n)$$ Is there a way to solve it as $m=f_{k}(n)$ using Lambert $W(x)$?


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