Solve #x^sqrtx = 16# ?

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Answer 1 · 2017-04-06T09:45:02

#x=4#

Making #y = sqrt(x)# we have

#(y^2)^y=y^(2y)=4^2# so #y^y=4#

so #y = 2#

and #x = y^2= 4#

Now using the Lambert function #W# we have

#y = log(4)/(W(log(4)))# and also

#x = y^2=(log(4)/(W(log(4))))^2 = 4#