How to solve this ? 2lnx-(lnx)^2 =0

plz help me :S

2lnx - (lnx)^2 = 0

First, factor out the common lnx:

lnx(2 - lnx) = 0

Now, for the left side to be 0, either term on the left must be 0, so we have:

lnx = 0 and
2 - lnx = 0

First, lnx = 0
e^lnx = e^0
x = 1

Second, 2 - lnx = 0
lnx = 2
e^lnx = e^2
x = e^2

Thus, the two solutions are e^2 and 1.

(log.x)^2 = 2logx
so lhs is always equal to 0. so any positive value of x (which is the domain of log function) satisfies thos

2lnx-(lnx)^2 =0
let y = ln x
y^2 - 2y = 0
y(y - 2) = 0
y = 0 , 2
ln x = 0
x = e^0 = 1
ln x = 2
x = e^2
hence:
x = 1 , e^2