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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

slution plz...

$${slution}\:{plz}... \\ $$

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

it′s just ∞ I think  bcz ur qst =    −4^((−4)(−4)(−4)...∞)  =−4^(∞ ) = ∞

$${it}'{s}\:{just}\:\infty\:{I}\:{think} \\ $$$${bcz}\:{ur}\:{qst}\:= \\ $$$$ \\ $$$$−\mathrm{4}^{\left(−\mathrm{4}\right)\left(−\mathrm{4}\right)\left(−\mathrm{4}\right)...\infty} \:=−\mathrm{4}^{\infty\:} =\:\infty \\ $$

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

−4^(−4^(−4^(−4^(...) ) ) ) =x  (−4)^x =x ∈ C  (4e^(iπ) )^x =x  e^(x(ln 4+iπ)) =x  xe^(−x(ln 4+iπ)) =1  −x(ln 4+iπ)e^(−x(ln 4+iπ)) =−(ln 4+iπ)  −x(ln 4+iπ)=W(−ln 4−iπ)  (Lambert W function)  ⇒x=−((W(−ln 4−iπ))/(ln 4+iπ))

$$−\mathrm{4}^{−\mathrm{4}^{−\mathrm{4}^{−\mathrm{4}^{...} } } } ={x} \\ $$$$\left(−\mathrm{4}\right)^{{x}} ={x}\:\in\:\mathbb{C} \\ $$$$\left(\mathrm{4}{e}^{{i}\pi} \right)^{{x}} ={x} \\ $$$${e}^{{x}\left(\mathrm{ln}\:\mathrm{4}+{i}\pi\right)} ={x} \\ $$$${xe}^{−{x}\left(\mathrm{ln}\:\mathrm{4}+{i}\pi\right)} =\mathrm{1} \\ $$$$−{x}\left(\mathrm{ln}\:\mathrm{4}+{i}\pi\right){e}^{−{x}\left(\mathrm{ln}\:\mathrm{4}+{i}\pi\right)} =−\left(\mathrm{ln}\:\mathrm{4}+{i}\pi\right) \\ $$$$−{x}\left(\mathrm{ln}\:\mathrm{4}+{i}\pi\right)={W}\left(−\mathrm{ln}\:\mathrm{4}−{i}\pi\right)\:\:\left({Lambert}\:{W}\:{function}\right) \\ $$$$\Rightarrow{x}=−\frac{{W}\left(−\mathrm{ln}\:\mathrm{4}−{i}\pi\right)}{\mathrm{ln}\:\mathrm{4}+{i}\pi} \\ $$

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