# 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?

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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+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.**