(2x)^x =(1/(16))


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Question Number 71141 by Mr. K last updated on 12/Oct/19

${(2 x)}^{x} = \frac{1}{16}$
Commented by mr W last updated on 12/Oct/19

${(2 x)}^{x} = \sqrt{{(2 x)}^{2 x}} = \sqrt{t^{t}} ⩾ \sqrt{e^{- \frac{1}{e}}} = 0.832 > \frac{1}{16}$ $\Rightarrow n o r e a l s o l u t i o n f o r {(2 x)}^{x} = \frac{1}{16}$
Commented by Henri Boucatchou last updated on 12/Oct/19

$C o r r e c t S i r, n o r e a l s o l u t i o n, t h e g e o m e t r i c p r o f a l s o s h o w i t$
Commented by mr W last updated on 12/Oct/19

$b u t {(2 x)}^{x} = \frac{10}{11} h a s t w o r e a l s o l u t i o n s .$
Commented by mr W last updated on 13/Oct/19
$(2x)^x =(1/(16)) has no real solution. let′s say the question were (2x)^x =a=((10)/(11)) (2x)^x =a 2x=a^(1/x) =e^((ln a)/x) 2ln a=((ln a)/x)e^((ln a)/x) ⇒((ln a)/x)=W(2ln a) ⇒x=((ln a)/(W(2ln a))) such that real solution exists, 2ln a≥−(1/e) ⇒a≥(1/e^(1/(2e)) )=0.832 with a=((10)/(11)) which is >0.832 ⇒x=((ln ((10)/(11)))/(W(2ln ((10)/(11)))))= { ((((ln 10−ln 11)/(−0.243071))=0.392108)),((((ln 10−ln 11)/(−2.621044))=0.036363)) :}$
${(2 x)}^{x} = \frac{1}{16} h a s n o r e a l s o l u t i o n .$ ${l e t}^{'} s s a y t h e q u e s t i o n w e r e {(2 x)}^{x} = a = \frac{10}{11}$ ${(2 x)}^{x} = a$ $2 x = a^{\frac{1}{x}} = e^{\frac{\ln a}{x}}$ $2 \ln a = \frac{\ln a}{x} e^{\frac{\ln a}{x}}$ $\Rightarrow \frac{\ln a}{x} = W (2 \ln a)$ $\Rightarrow x = \frac{\ln a}{W (2 \ln a)}$ $s u c h t h a t r e a l s o l u t i o n e x i s t s,$ $2 \ln a ⩾ - \frac{1}{e} \Rightarrow a ⩾ \frac{1}{e^{\frac{1}{2 e}}} = 0.832$ $w i t h a = \frac{10}{11} w h i c h i s > 0.832$ $\Rightarrow x = \frac{\ln \frac{10}{11}}{W (2 \ln \frac{10}{11})} = {\begin{cases} \frac{\ln 10 - \ln 11}{- 0.243071} = 0.392108 \\ \frac{\ln 10 - \ln 11}{- 2.621044} = 0.036363 \end{cases}$
Commented by Mr. K last updated on 13/Oct/19

$C a n y o u s h o w t h e s o l u t i o n ?$
Commented by Mr. K last updated on 13/Oct/19

$t h a n k y o u .$
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