Question 12234
{{{x^(1+lnx) = 7.39}}}


OK, take logs
{{{(1+lnx)lnx = ln(7.39)}}}
{{{lnx + (lnx)^2 = 2}}}

{{{(lnx)^2 + lnx -2 = 0}}}
OK, this is a quadratic, so letting y=lnx, we have {{{y^2 + y - 2 = 0}}}


This factorises to (y+2)(y-1) = 0, so that 


y+2 = 0 OR y-1 = 0

--> y = -2 or y = 1


Hence
lnx = -2 OR lnx = 1


-->
{{{x = e^(-2)}}} or x = e


jon.