f= x^3 +2x^2 + x+2

Substitute with f(lnx)

f(lnx)= (lnx)^3 + 2(lnx)^2 + lnx + 2

Now assume that lnx= y

==>f(y)= y^3 + 2y^2 + y+2

Let us try and factor:

f(y)= (Y^3 + 2y^2 )+ (y+ 2)

Now factor y^2:

f(y) = y^2 ( y+2) + (y+2)

Now factor (x+2)

==> f(y) = (y+2)(y^2+1)

Then (y+2)=0 ==> y=-2

Or , y^2 + 1=0 which is impossible because this value is always >1

Then the solution is y= -2 = lnx

==> x= e^-2= 1/e^2

First, we have to write the equation, substituting x by ln x:

(lnx)^3+2(lnx)^2+(lnx)+2=0

Now, we'll solve the equation using the substitution technique. For this reason, we'll substitute ln x by t and we'll get:

t^3+2t^2+t+2 = 0

We'll group the terms and we'll factorize:

t^2(t+2) + (t+2) = 0

We'll factorize again:

(t+2)(t^2+1) = 0

We'll put each factor from the product as 0:

t+2 = 0

t = -2

So, ln x = -2

x = e^(-2)

x = 1/e^2

t^2 + 1 = 0, impossible,(lnx)^2+1>0, for any real x!

So, the only solution of the equation is x = 1/e^2.

f(x) = x^3+2x^2+x+2.

To find f(lnx)

Solution:

To solve f(lnx)= f(lny) = 0

Let us sove f(x) = 0 first.

x^3+2x^2+x+2= 0.

x^2(x+2) +1(x+2) = 0.

(x+2)(x^2+2) = 0.

x=-2 or x^2 = -1

x = -2 gives lny = -2 Or y = e^(-2)

x^2 =- 1 gives x = i or x= -i wher i = sqrt(-1).

x = i gives y = e^(pi/2*i) = cospi/2 + isinpi/2 = i

and x =-i gives y = e^(-pi/2*i) = -i