x^x =4 find x


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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

	$x = 2 o n l y o n e s o l u t i o n . . .$
	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

	$t h a n k s . . . . . A n y a n a l y t i c a l a p p r o a c h$
	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
	$x^x =4 xln x =2ln 2 so x=2 is obvious but is there another solution? x<0 x^x =(−x)^x (cos πx +isin πx) ⇒ sin πx =0 ∧ (−x)^x cos πx =4 sin πx =0 ⇒ x∈Z [(−x)^x cos πx ∣ x={−1, −2, −3, −4, ...}]= ={−1, (1/4), −(1/(27)), (1/(256)), ...} ⇒ no solution for x<0 x=0 0^0 is not defined but lim_(x→0^+ ) x^x =1 ⇒ no solution for x=0 0<x<1 x^x has its minimum at x=(1/e) and is decreasing in ]0; (1/e)[ and increasing in ](1/e); 1[ ⇒ no solution for 0<x<1 x≥1 x^x is increasing ⇒ no other solution than x=2$
	$x^{x} = 4$ $x \ln x = 2 \ln 2$ $so x = 2 is obvious$ $but is there another solution ?$ $x < 0$ $x^{x} = {(- x)}^{x} (\cos π x + i \sin π x)$ $\Rightarrow \sin π x = 0 \land {(- x)}^{x} \cos π x = 4$ $\sin π x = 0 \Rightarrow x \in Z$ $[{(- x)}^{x} \cos π x ∣ x = {- 1, - 2, - 3, - 4, . . .}] =$ $= {- 1, \frac{1}{4}, - \frac{1}{27}, \frac{1}{256}, . . .}$ $\Rightarrow no solution for x < 0$ $x = 0$ $0^{0} is not defined but$ $\lim_{x \to 0^{+}} x^{x} = 1$ $\Rightarrow no solution for x = 0$ $0 < x < 1$ $x^{x} has its minimum at x = \frac{1}{e} and is decreasing$ $in] 0; \frac{1}{e} [and increasing in] \frac{1}{e}; 1 [$ $\Rightarrow no solution for 0 < x < 1$ $x ⩾ 1$ $x^{x} is increasing \Rightarrow no other solution than$ $x = 2$
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