Question

write the domain in interval notation.

$v(x)=\log _{2}(\frac{x + 3}{x + 5})$

Explanation:

Step1: Determine the condition for the logarithm

For \(y = \log_{a}u\), \(u>0\). Here \(u=\frac{x + 3}{x + 5}\), so \(\frac{x+3}{x + 5}>0\).

Step2: Find the critical points

Set \(x+3 = 0\), then \(x=-3\); set \(x + 5=0\), then \(x=-5\).

Step3: Test intervals

• For \(x<-5\), let \(x=-6\). Then \(\frac{-6 + 3}{-6+5}=\frac{-3}{-1}=3>0\).
• For \(-5
• For \(x>-3\), let \(x=0\). Then \(\frac{0 + 3}{0+5}=\frac{3}{5}>0\).

So the domain is \((-\infty,-5)\cup(-3,\infty)\).

Answer:

$$(-\infty,-5)\cup(-3,\infty)$$