How do I determine the range of an ln function?

Question

The specific question I'm looking at is ln(9-x^2). How do I determine the range?

Answer 1

Since the argument for the Ln function must be greater than zero (ie ln 0 does not exist), all you need to do is to set the argument to zero to find out the bounds.

In other words, set 9 - x^2 = 0, you will get x = -3, 3.

Pick a value great or smaller than of the solutions, and see whether the resulting function is greater than zero.

For example, since we got -3 and 3 as the bound, we can pick a value in between or a value not in between to test whether the greater than zero condition is satisfied.

For x = -4, x = 4 => 9 - 16 = -7 (Negative solution)
For x = 0 => 9 - 0 = 9 (Postive solution)


Therefore, the range is in between -3 and 3, which can be express mathematically as -3 < x < 3

(Note: the point x = -3 and x = 3 are not included! Because both of them will give ln(0), which does not exist!)

Answer 2

Range Of Ln Function

Answer 3

We generally say that negative numbers have no logarithms (which isn't quite true if we extend things to the complex numbers. But I digress ☺)

As a result of this, you have to make sure that 9-x² > 0 which means that -3 < x < 3. Over this domain, the largest value that 9-x² can assume is 9 (when x is 0) and, in the limit, it can approach 0 (as x approachces 3 or -3) and ln(0) = -∞ so that the range of this function is
-∞ < ln(9-x²) ≤ ln(9)

HTH ☺

Doug

Answer 4

First the domain. 9-x^2>0 so -3

Answer 5

Well if you look at the graph at ln, you will see its general shape. Now, you use the derivative to find the points where the slope is infinity and where the slope becomes 0 and you got your x and y constraints. I think...