What is x if log(x+1) (x^2-3x+2)=log(x+1) (x^2-2x), if x+1 is the base of logarithm?

We'll impose the constraints of existence of
logarithms.

The base has to be positive and it is different
from
1.

x+1>0

x>-1

x
different from 0.

Now, the arguments have to be
positive:

x^2-3x+2>0

The
expression is positive if x is in the ranges
(-inf.,1)U(2,+inf.)

x^2-2x>0

The
expression is positive if x is in the ranges
(-inf.,0)U(2,+inf.)

The common intervals of admissible
values for x are:

(-1 ; 0)U(2 ;
+infinite)

Now, we'll solve the
equation.

Since the logarithms have matching bases, we'll
apply one to one property:

x^2-3x+2 =
x^2-2x

We'll eliminate x^2 both
sides:

-3x + 2 + 2x = 0

-x + 2
= 0

-x = -2

x =
2

Since 2 doesn't belong to the intervals of admissible
values, we'll reject it.

The equation has no
solutions.