# Determin the values for x for the logarithm exists log(x+2)(x^2-4) .

Asked on by penarul

### 2 Answers |Add Yours

neela | High School Teacher | (Level 3) Valedictorian

Posted on

Log (x+2)(x^2-4) is real iff (x+2)(x^2-4) >0, as logarithme are not real for negative values.

So (x+2)(x^2-4) > 0.

(x+2)(x+2)(x-2) > 0.

For x > 2 all facors are > 0, so LHS is > 0

For x < -2 , LHS is negative.

So x> 2 is the only solution.

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) x^2 - 4 > 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) x^2 - 4 > 0

We'll re-write the difference of squares:

x^2 – 4 = (x - 2)(x + 2)

(x - 2)(x + 2)> 0

x – 2>0 and x + 2>0

The expression is positive  if x is in the interval (2 ; infinite)

x – 2<0 and x + 2<0

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

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

We’ve answered 319,199 questions. We can answer yours, too.
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}