x^3 =3^x how do i pls use lambert to find x=3


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Question Number 151403 by 7770 last updated on 20/Aug/21

$x^{3} = 3^{x}$ $h o w d o i p l s u s e l a m b e r t t o$ $f i n d x = 3$
Commented by mr W last updated on 20/Aug/21
$x=3^(x/3) x=e^((x/3)ln 3) xe^(−(x/3)ln 3) =1 (−(x/3)ln 3)e^(−(x/3)ln 3) =−((ln 3)/3) −(x/3)ln 3=W(−((ln 3)/3)) ⇒x=−(3/(ln 3))W(−((ln 3)/3)) = { ((−(3/(ln 3))×(−1.098612289)=3)),((−(3/(ln 3))×(−0.907473042)=2.478053)) :}$
$x = 3^{\frac{x}{3}}$ $x = e^{\frac{x}{3} \ln 3}$ ${x e}^{- \frac{x}{3} \ln 3} = 1$ $(- \frac{x}{3} \ln 3) e^{- \frac{x}{3} \ln 3} = - \frac{\ln 3}{3}$ $- \frac{x}{3} \ln 3 = W (- \frac{\ln 3}{3})$ $\Rightarrow x = - \frac{3}{\ln 3} W (- \frac{\ln 3}{3})$ $= {\begin{cases} - \frac{3}{\ln 3} \times (- 1.098612289) = 3 \\ - \frac{3}{\ln 3} \times (- 0.907473042) = 2.478053 \end{cases}$
Commented by 7770 last updated on 20/Aug/21

$t h a n k y o u s i r b u t h o w d i d u g e t (- 1.098612289) p l e a s e$
Commented by mr W last updated on 21/Aug/21

$u s e a n y m e t h o d f o r s o l v i n g e q u a t i o n s$ $n u m e r i c a l l y . t h e e q u a t i o n i s$ ${x e}^{x} = a$
Commented by 7770 last updated on 21/Aug/21

$o k s i r$ $t h a n k s$
Commented by otchereabdullai@gmail.com last updated on 22/Aug/21

$thank u$
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