# What is x for the logarithm exist log(x+2) (15 + 2x -x^2) ?

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll impose the constraints of existence of logarithms:

1) x+2 > 0

2) x + 2 different from 1.

3) 15 + 2x -x^2 > 0

We'll solve the first constraint:

1) x+2 > 0

We'll subtract 2 both sides:

x > -2

2) x + 2 different from 1

We'll subtract 2 both sides:

x different from -2+1 = -1

3) 15 + 2x -x^2 > 0

We'll multiply by -1:

x^2 - 2x - 15 < 0

We'll calculate the roots of the quadratic:

x1 = [2 + sqrt(4+60)]/2

x1 = (2+8)/2

x1 = 5

x2 = (2-8)/2

x2 = -3

The expression is negative if x is in the interval (-3 ; 5)

From this interval, we'll reject the value -1.

The interval (-3 ; 5) - {-1} will be intersected by the interval (-2 ; +infinity).

The interval of admissible values for the logarithm to be defined is (-2;5).

neela | High School Teacher | (Level 3) Valedictorian

Posted on

To determine x for which log(x+2)(15+2x-x^2) is defined.

We know that log a exists, if a > 0. And log a is does not exist for x< 0.

Therefore log(x+2)(15+2x-x^2) exists if (x+2)(15+2x-x^2) > 0.

Or (x+2)(x^2-2x-15) < 0 ........(1), we multiplied  by (-1). So the inequality reversed.

x^2 -2x-15 = (x+3)(x-5).

Substituting in (1) , we get: (x+2)(x+3)(x-5) < 0.

Or  f(x) = (x+3)(x+2)(x-5) < 0.

Clearly for x> 5, f(x) > 0, as all factors are positive.

For   -2 < x < 5, f(x) < 0 as (x+3)(x+2) positive and x-5 < 0.

For -3<x<-2, f(x) > 0, as x+3 >0 and (x+2) <0 and (x-5) < 0.

For x< -3, f(x) < 0 as all 3 factors are negative.

Therefore log(x+2)(15+2x-x^2) exits  only if  (-2< x <5).