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Question Number 139805 by mr W last updated on 01/May/21

s o l v e f o r x \in R

x^{x^{n}} = a

w i t h n, a \in R^{+}

f i n d a l s o t h e r a n g e o f a s u c h t h a t a

s o l u t i o n e x i s t s .

Commented by mr W last updated on 01/May/21

i t h o u g h t a b o u t t h i s g e n e r a l e q u a t i o n

f o r a l o n g t i m e a n d t r i e d t o f i n d

a f o r m u l a t o e x p r e s s t h e s o l u t i o n .

w e k n o w i f n = a, t h e s o l u t i o n f o r

x^{x^{a}} = a i s t h e s a m e a s f o r x^{a} = a, i . e .

x = \sqrt[a]{a} . b u t w h a t i s t h e s o l u t i o n f o r

t h e g e n e r a l c a s e w i t h n \neq a ?

Answered by mr W last updated on 02/May/21

h e r e t h e s o l u t i o n i g o t f o r t h e

g e n e r a l c a s e x^{x^{n}} = a .

x^{x^{n}} = a

x = a^{\frac{1}{x^{n}}} = e^{\frac{\ln a}{x^{n}}}

x^{n} = e^{\frac{n \ln a}{x^{n}}}

\frac{1}{x^{n}} e^{\frac{n \ln a}{x^{n}}} = 1

\frac{n \ln a}{x^{n}} e^{\frac{n \ln a}{x^{n}}} = n \ln a

\frac{n \ln a}{x^{n}} = W (n \ln a)

x^{n} = \frac{n \ln a}{W (n \ln a)}

\Rightarrow x = \sqrt[n]{\frac{n \ln a}{W (n \ln a)}}

s u c h t h a t a r e a l r o o t e x i s t s, W (n \ln a)

m u s t b e d e f i n e d . t h i s i s t h e c a s e i f

n \ln a ⩾ - \frac{1}{e}, i . e . a ⩾ e^{- \frac{1}{n e}} = \frac{1}{\sqrt[n e]{e}} .

t h e r e f o r e :

f o r a < \frac{1}{\sqrt[n e]{e}} \Rightarrow n o r o o t

f o r a = \frac{1}{\sqrt[n e]{e}} \Rightarrow o n e r o o t

f o r \frac{1}{\sqrt[n e]{e}} < a < 1 \Rightarrow t w o r o o t s

f o r a ⩾ 1 \Rightarrow o n e r o o t

e x a m p l e : x^{x^{5}} = 10

\Rightarrow x = \sqrt[5]{\frac{5 \ln 10}{W (5 \ln 10)}} \approx \sqrt[5]{\frac{5 \ln 10}{1.835928}} = 1.443664

e x a m p l e : x^{x^{3}} = \frac{9}{10}

\Rightarrow x = \sqrt[3]{\frac{3 \ln 0.9}{W (3 \ln 0.9)}} \approx {\begin{cases} \sqrt[3]{\frac{3 \ln 0.9}{- 1.656404}} = 0.575719 \\ \sqrt[3]{\frac{3 \ln 0.9}{- 0.545257}} = 0.833808 \end{cases}

Commented by Tawa11 last updated on 23/Jul/21

great

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