How to derive $y^{y^n}=x$ explicit form with Lambert $W$ function

Question

I think the answer is $y = \left( \frac{n \cdot \ln(x)}{W(n \cdot \ln(x))} \right)^{\frac{1}{n}}$, seems tricky. I'm a noob!

Answer 1

We have $$y^{y^{n}}=x\Leftrightarrow n\log\left(y\right)+\log\left(\log\left(y\right)\right)=\log\left(\log\left(x\right)\right)\Leftrightarrow e^{n\log\left(y\right)}n\log\left(y\right)=n\log\left(x\right)\Leftrightarrow $$ and if we take $$u=n\log\left(y\right) $$ we get $$ue^{u}=n\log\left(x\right) $$ and so $$n\log\left(y\right)=W\left(n\log\left(x\right)\right)\Leftrightarrow y=e^{W\left(n\log\left(x\right)\right)/n}=\left(\frac{n\log\left(x\right)}{W\left(n\log\left(x\right)\right)}\right)^{1/n} $$ since $$e^{W\left(x\right)}=\frac{x}{W\left(x\right)}. $$

Answer 2

$$y^{y^n}=x$$

Power both sides by $n$.

$$y^{ny^n}=x^n$$

Use $u=y^n$ to get

$$u^u=x^n$$

$$\ln(u^u)=\ln(x^n)$$

$$u\ln(u)=n\ln(x)$$

We see that $u\ln(u)=\ln(u)e^{\ln(u)}$

$$\ln(u)e^{\ln(u)}=n\ln(x)$$

$$\ln(u)=W(n\ln(x))$$

$$u=e^{W(n\ln(x))}$$

Remember $u=y^n$?

$$y^n=e^{W(n\ln(x))}$$

$$y=e^{\frac{W(n\ln(x)}n}=\left(\frac{n\ln(x)}{W(n\ln(x))}\right)^{\frac1n}$$