Solve $x^{\sqrt{x}} = 16$ $x^sqrtx = 16$ ?

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Answer 1 · 2017-04-06T09:45:02

$x = 4$ $x=4$

Making $y = \sqrt{x}$ $y = sqrt(x)$ we have

${(y^{2})}^{y} = y^{2 y} = 4^{2}$ $(y^2)^y=y^(2y)=4^2$ so $y^{y} = 4$ $y^y=4$

so $y = 2$ $y = 2$

and $x = y^{2} = 4$ $x = y^2= 4$

Now using the Lambert function $W$ $W$ we have

$y = \frac{log (4)}{W (log (4))}$ $y = log(4)/(W(log(4)))$ and also

$x = y^{2} = {(\frac{log (4)}{W (log (4))})}^{2} = 4$ $x = y^2=(log(4)/(W(log(4))))^2 = 4$

Solve x√x=16x^sqrtx = 16 ?