Determine x if log(x+1) (x^2-5x+6) exists.

Since the given logarithm does exist, then the base and
the argument of logarithm respect the conditions of
existence.

We'll impose the constraints of existence of
logarithms:

1) x+1 >
0

2) x + 1 different from
1.

3) x^2-5x+6 >
0

We'll solve the first
constraint:

1) x+1 >
0

We'll subtract 1 both
sides:

x > -1

2) x + 1
different from 1

We'll subtract 1 both
sides:

x different from -1+1 =
0

3) x^2-5x+6 > 0

The
expression is positive for values of x located outside the roots of the expression
x^2-5x+6 = 0.

We'll calculate the roots of the
quadratic:

x1 = [5 +
sqrt(25-24)]/2

x1 = (5+1)/2

x1
= 3

x2 = (5-1)/2

x2 =
2

The expression is positive if x is in the interval
(-infinite ; 2)U(3 ; +infinite)

From the interval
(-infinite ; 2), we'll reject the value 0.

The intervals
(-infinite ; 2)U(3 ; +infinite) - {0} will be intersected by the interval (-1 ;
+infinity).

The interval of admissible values
for the logarithm to be defined is (-1;2)U(3 ; +infinite) -
{0}.