Solve the equation f(ln x)=0. f=x^3+2x^2+x+2
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:
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)
To solve f(lnx)= f(lny) = 0
Let us sove f(x) = 0 first.
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