Find the domain of the function f (x) = Log (|x-7|-3) and give the solution in interval notation

(-∞, 4) ∪ (10, ∞)

Step-by-step explanation:

A logarithm y = log x can only take x-values greater than 0. It's domain is (0, ∞). If you put log (0) or log of any negative number into a calculator, it'll return an error. So when you have a logarithm like yours, f (x) = log (|x - 7| - 3), you need to make sure you take out any x value that could give you a negative result or 0.

Take the argument of the logarithm, |x - 7| - 3, and pull it aside. You can solve it one of two ways: by looking at it and plugging in values and figuring out which values give you zero or a negative number (trial and error), or by setting it equal to zero and solving for x. I'll show you the second approach:

|x - 7| - 3 = 0 ... get the absolute value bars alone

|x - 7| = 3 ... absolute value means "this distance from 0 on a number line", so what your equation means is |x - 7| is 3 units away from zero. But we don't know if it's three units to the left (negative 3) or three units to the right (positive three). So we set up both.

x - 7 = 3

x - 7 = - 3

Solve both equations, and get x = 10 for the top equation, and x = 4 for the bottom equation. This gives you the values you need to use in your interval notation. 4 and 10 both will give you 0 if you plug them into |x - 7| - 3, and 0 is outside of the domain of a logarithm. You can't have negative numbers either, and you'll see that if you try to plug in 5, 6, 7 ... all the way up to 10, you'll either get a negative result or 0. Basically, anything between 4 and 10, including 4 and 10 won't do it for you. x has to be less than 4 or greater than 10 for your original f (x) function to be defined.

Now that you've identified what ISN'T in your domain, all you have to do is set it up in interval notation. You're clear for everything from negative infinity up to 4, but at exactly 4 (if you set x = 4 and plug it in), your result is undefined. Exactly at 10, it's undefined, but anything greater than 10 works fine. So (-∞, 4) ∪ (10, ∞) is your answer.