SOLUTION: OK , here is the problem, solve for x. x^(1+ln x)= 7,39 Doesn´t look that complicated but I keep getting stuck. Incidentally I´ve discovered that 7,39 = e^2 and I think that the

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: OK , here is the problem, solve for x. x^(1+ln x)= 7,39 Doesn´t look that complicated but I keep getting stuck. Incidentally I´ve discovered that 7,39 = e^2 and I think that the      Log On


   



Question 12234: OK , here is the problem, solve for x.
x^(1+ln x)= 7,39
Doesn´t look that complicated but I keep getting stuck.
Incidentally I´ve discovered that 7,39 = e^2 and I think that the answer for x = square root of 7,39, but don´t know how to get there.
Thanks in advance for any assistance.

Answer by longjonsilver(2297) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E%281%2Blnx%29+=+7.39

OK, take logs
%281%2Blnx%29lnx+=+ln%287.39%29
lnx+%2B+%28lnx%29%5E2+=+2
%28lnx%29%5E2+%2B+lnx+-2+=+0
OK, this is a quadratic, so letting y=lnx, we have y%5E2+%2B+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%5E%28-2%29 or x = e

jon.