We can solve this equation with the contribution of the so called Lambert function.
https://en.wikipedia.org/wiki/Lambert_W_function
We have
#log_e x = pmsqrt(2x) rArr x = e^(pmsqrt(2x))#
Considering first the option #x = e^sqrt(2x)#
#2x = 2 e^(sqrt(2x)) rArr sqrt(2x) = sqrt2 e^(1/2 sqrt(2x))# and
#1/2sqrt(2x) = 1/2sqrt2 e^(1/2 sqrt(2x)) rArr 1/2sqrt(2x) =1/sqrt2 e^(1/2 sqrt(2x))#
Now using the properties
#y = x e^x hArr x = W(y)# we have calling #z = 1/sqrt(2x)#
#z = 1/sqrt2 e^z rArr -ze^-z = -1/sqrt2# and then
#W(-1/sqrt2)=-z = -1/2sqrt(2x)# and finally
#x = 2 W(-1/sqrt2)^2 = -1.8070659157943643 - 2.4598345952311647i #
Considering now # x = e^-sqrt(2x)# we have correspondingly
#2x = 2 e^-sqrt(2x) rArr sqrt(2x) = sqrt2 e^(-1/2sqrt(2x))# and
#1/2 sqrt(2x) = 1/sqrt2 e^(-1/2sqrt(2x))# now calling #z = 1/2 sqrt(2x)#
follows #z e^z = 1/sqrt2 rArr z = 1/2 sqrt(2x) = W(1/sqrt2)# and finally
#x = 2 W(1/sqrt2)^2 approx 0.40608164979530725#
