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Question Number 73909 by Pk1167156@gmail.com last updated on 16/Nov/19

Commented by Pk1167156@gmail.com last updated on 16/Nov/19

$$slutionplz...$$

Answered by arkanmath7@gmail.com last updated on 16/Nov/19

$${it}^{\prime }sjust\mathrm{\infty }Ithink$$$$bczurqst=$$$$$$$$-4^{(-4)(-4)(-4)...\mathrm{\infty }}=-4^{\mathrm{\infty }}=\mathrm{\infty }$$

Answered by mr W last updated on 17/Nov/19

$$-4^{-4^{-4^{-4^{...}}}}=x$$$${(-4)}^x=x\in \mathbb{C}$$$${\left(4e^{i\pi }\right)}^x=x$$$$e^{x(\ln 4+i\pi )}=x$$$${xe}^{-x(\ln 4+i\pi )}=1$$$$-x(\ln 4+i\pi )e^{-x(\ln 4+i\pi )}=-(\ln 4+i\pi )$$$$-x(\ln 4+i\pi )=W(-\ln 4-i\pi )\left(LambertWfunction\right)$$$$\Rightarrow x=-\frac{W(-\ln 4-i\pi )}{\ln 4+i\pi }$$

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