In Exercises 75–80, find the domain of each logarithmic function....

Question

In Exercises 75–80, find the domain of each logarithmic function. f(x) = ln (x-2)²

Accepted Answer

Welcome back. I am so glad you're here. We are given the function F of X equals the natural log of X minus nine squared. And we're asked to find the domain of this function. All right, we have got a log in our function and logs do have a domain restriction what we're taking the log of. So in this case, X minus nine squared needs to be greater than zero. So now we just need to solve for X and find out what values of X are true that make this greater than zero. We recall from previous lessons on um polynomial inequalities that the first step is. We're going to find out what the boundaries of this inequality are on the X axis. So we're going to find out the boundary for X here by setting this left hand side X minus nine squared equal to zero. So we can take the square root of both sides to get rid of that. Squared. Oops started over here. There we go take the square of both sides. So on the left that's X minus nine and it equals, well, the squared of zero is zero and that's plus or minus zero, but plus or minus zero is just zero and then we can add nine to both sides and we've got X equals nine. So what did we learn? We learned that we have a boundary of nine. If we look back at our original inequality, this is a greater than, and not a greater than or equal to. So the function cannot come in contact with that nine. It's going to be an open circle at nine, so it cannot be nine. Otherwise it won't be true. Our inequality statement won't be true. The next step we're going to do is we're going to pick some test points on both sides of our boundary, both sides of nine to see if our test points are true. Basically is the function heading toward negative infinity or positive infinity or neither one or both. So for a test point that's less than nine heading toward negative infinity, let's choose zero because zero is usually pretty easy to fill in. So instead of X minus nine, it's now zero minus nine squared. And we're seeing if it's true, if that's greater than 00 minus nine is negative nine and negative nine squared is positive 81. Yes, that is greater than zero. So our function can head in this direction to the left through our test point toward negative infinity. Now let's pick a test point on the right hand side heading toward positive infinity and I'm just going to choose 10. So we plug a 10 in for X 10 minus nine squared. We'll see if that's true. If it's greater than 0, 10 minus nine is one. So we've got one squared, greater than 01 squared is one and yes, one is greater than, than zero. This is true. So our function can head to the right toward positive infinity that is included in the domain. So basically X can be any number other than nine. So how do we express that in interval notation? Well, it can, starting on the left, it can start all the way down from negative infinity and keep going all the way up until we hit nine. And then we put a parentheses here to say it can't be nine. And that's in union with, we're going to start with another parentheses and then it's going to start just after nine and go all the way up to positive infinity. And we close that with a parentheses as well to say, we'll never quite obtain infinity. We look at our answer choices and this matches with answer choice. A well done. We'll catch you on the next one.

In Exercises 75–80, find the domain of each logarithmic function. f(x) = ln (x-2)²

Key Concepts

Logarithmic Functions

Domain of a Function

Inequalities

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