x-x-4-find-x-

Question Number 56214 by necx1 last updated on 12/Mar/19

x^{x} = 4

f i n d x

Answered by tanmay.chaudhury50@gmail.com last updated on 12/Mar/19

x = 2 o n l y o n e s o l u t i o n \dots

Commented by tanmay.chaudhury50@gmail.com last updated on 12/Mar/19

Commented by necx1 last updated on 12/Mar/19

s o l u t i o n p l e a s e s i r

Commented by necx1 last updated on 12/Mar/19

t h a n k s \dots . . A n y a n a l y t i c a l a p p r o a c h

Answered by mr W last updated on 13/Mar/19

g e n e r a l :

x^{x} = a w i t h a ⩾ \frac{1}{\sqrt[e]{e}} \approx 0.6922

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

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

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

\Rightarrow \frac{\ln a}{x} = W (\ln a) \leftarrow L a m b e r t f u n c t i o n

\Rightarrow x = \frac{\ln a}{W (\ln a)}

i f a = 4 :

x = \frac{\ln 4}{W (\ln 4)} = \frac{\ln 4}{0.6931} = 2

i f a = 5 :

x = \frac{\ln 5}{W (\ln 5)} = \frac{\ln 5}{0.7558} = 2.1294

Answered by MJS last updated on 12/Mar/19

x^{x} = 4

x \ln x = 2 \ln 2

so x = 2 is obvious

but is there another solution ?

x < 0

x^{x} = {(- x)}^{x} (\cos π x + i \sin π x)

\Rightarrow \sin π x = 0 \land {(- x)}^{x} \cos π x = 4

\sin π x = 0 \Rightarrow x \in Z

[{(- x)}^{x} \cos π x ∣ x = {- 1, - 2, - 3, - 4, \dots}] =

= {- 1, \frac{1}{4}, - \frac{1}{27}, \frac{1}{256}, \dots}

\Rightarrow no solution for x < 0

x = 0

0^{0} is not defined but

\lim_{x \to 0^{+}} x^{x} = 1

\Rightarrow no solution for x = 0

0 < x < 1

x^{x} has its minimum at x = \frac{1}{e} and is decreasing

in] 0; \frac{1}{e} [and increasing in] \frac{1}{e}; 1 [

\Rightarrow no solution for 0 < x < 1

x ⩾ 1

x^{x} is increasing \Rightarrow no other solution than

x = 2

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