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 have to find the value of x for which log(x+1) (x^2-3x+2) = log(x+1) (x^2-2x)
As the base of the log is the same, we can equate
x^2 - 3x + 2 = x^2 - 2x
=> x = 2
But for x = 2, we get log 0 which is indeterminate.
The required value of x cannot be found.
We'll impose the constraints of existence of logarithms.
The base has to be positive and it is different from 1.
x different from 0.
Now, the arguments have to be positive:
The expression is positive if x is in the ranges (-inf.,1)U(2,+inf.)
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.