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

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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}.

neela | High School Teacher | (Level 3) Valedictorian

Posted on

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

For log(x+1) (x^2-5x+6) to exist, the expression (x+1) (x^2-5x+6) > 0.

For this first we factorise x^2-5x+6.

x^2+5x+6 = (x+3)(x+2).

So (x+1)(x^2+5x+6) = (x+1)(x+3)(x+2) = (x+3)(x+2)(x+1) which is positive, if x > -1,

If -2< x< -1), then (x+1)(x+3)(x+2) < 0.

If -3<x<-2, then (x+1)(x+3)(x+2) > 0, as only last two factors are -ve.

If x < 3, then  (x+1)(x+3)(x+2) < 0 as all 3 factors are -ve.

So (x+1) (x^2-5x+6) = (x+1)(x+3)(x+2) > o when  x is in (-3 , -2) or x is  in (-1 , +infinity).

Therefore log(x+1) (x^2-5x+6) exists, when  x is in (-3 , -2) or x is  in (-1 , +infinity).
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