Inspired by Neel Khare

Algebra Level 3

What is the domain of

log ⁡ x + 1 x + 2 ? \log \frac{x+1}{x+2} ?

Inspiration

x > − 2 x > -2 x > − 1 x > -1 x > − 1 and x < − 2 x > -1 \text{ and } x < -2 x > 0 x > 0

Calvin Lin by Calvin Lin

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2 solutions

Zee Ell
Oct 26, 2016

Since the domain of log(y) is y > 0 , therefore:

x + 1 x + 2 > 0 \frac {x+1}{x+2} > 0

1 − 1 x + 2 > 0 1 - \frac {1}{x+2} > 0

1 > 1 x + 2 1 > \frac {1}{x+2}

Now, we have two cases:

Case I : x + 2 > 0 ⟺ x > − 2 \text {Case I : } x + 2 > 0 \iff x > -2

1 > 1 x + 2 ⟺ x + 2 > 1 ⟺ x > − 1 1 > \frac {1}{x+2} \iff x + 2 > 1 \iff x > -1

As x > -1 > -2 in this case, therefore the relevant interval is x > -1 here.

Case II : x + 2 < 0 ⟺ x < − 2 \text {Case II : } x + 2 < 0 \iff x < -2

Since multiplying by a negative number changes the direction of the inequality sign to the opposite:

1 > 1 x + 2 ⟺ x + 2 < 1 ⟺ x < − 1 1 > \frac {1}{x+2} \iff x + 2 < 1 \iff x < -1

As x < -2 < -1 in this case, therefore the relevant interval is x < -2 here.

Hence, the domain of the function is:

x > − 1 and x < − 2 \boxed {x > -1 \text { and } x < -2}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Great! It's a common misconception that the domain of log ⁡ f ( x ) g ( x ) \log \frac{ f(x) } { g(x) } is either x > 0 x > 0 or f ( x ) > 0 f(x) > 0 . What we need is for the entire expression to be positive.

A much easier way to deal with these rational inequalities is to multiply throughout by the square of the denominator to ensure positivity. This way,

x + 1 x + 2 > 0 ⇔ ( x + 1 ) ( x + 2 ) > 0 \frac{x+1}{x+2} > 0 \Leftrightarrow (x+1)(x+2) > 0

and the rest follows.

Calvin Lin Staff - 4 years, 7 months ago
Md Zuhair
Oct 26, 2016

0 Helpful 0 Interesting 1 Brilliant 0 Confused

Thanks for submitting a solution. Unfortunately, this solution has numerous issues.

  1. Where did the condition that x + 1 x + 2 > 1 \frac{x+1}{x+2} > 1 come form?
  2. Is it true that if x + 1 x + 2 > 1 \frac{x+1}{x+2} >1 then x + 1 x + 2 ≥ 2 \frac{ x+1}{x+2} \geq 2 ? If so, what happens for x = 100 x = 100 ?
  3. Are you sure that x + 1 x + 2 ≥ 2 ⇔ x + 1 ≥ 2 ( x + 2 ) \frac{x+1}{x+2} \geq 2 \Leftrightarrow x+1 \geq 2 (x+2) ?
  4. Is it true that if − 3 ≥ x ⇔ x < − 2 -3 \geq x \Leftrightarrow x < -2 ?
  5. Review sat inequalities

Calvin Lin Staff - 4 years, 7 months ago

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