MATH 155 SUPPLEMENTAL NOTES 1

NATURAL LOGARITHMS
THE DOMAIN OF THE LOGARITHMIC FUNCTION

Consider the function y = ln x. The domain of this function is the interval (0, ¥). The first thing that you should remember about any logarithmic function is that the argument of the function cannot be a negative number or zero. Therefore, your domain will be all values of x that makes the argument greater than zero.

EXAMPLE 1: Give the domain for the function y = ln (x - 1).

SOLUTION: Set x - 1 > 0 ® x > 1. Therefore, the domain for this function is (1, ¥).

EXAMPLE 2: Give the domain for the function y = ln (x ² - 4).

SOLUTION: Set x ² - 4 = 0 ® x = ± 2.

Since there are two values of x that make x ² - 4 = 0, we must look at the three intervals that these two points create and determine which intervals will be positive.

(- ¥ , -2)	(-2, 2)	(2, ¥ )
x ² - 4 > 0	x ² - 4< 0	x ² - 4 > 0

The domain for this function are the interval for which x ² - 4 > 0. Therefore, the domain for this function is (- ¥, -2) È (2, ¥).

EXAMPLE 3: Give the domain for the function y = ln [(x + 2)/(x + 4)].

SOLUTION: First of all, x + 4 ¹ 0 ® x ¹ -4. Second of all, x + 2 > 0 ® x > -2. Is the interval (-2, ¥) the domain for this function? NO! It may be part of the domain. The argument for this logarithmic function is a rational expression. Recall that a positive divided by a positive is a positive, but so does a negative divide a negative. Therefore, we will have to determine if there is any other interval that will make the argument positive. To do this, I will use a sign pattern for the argument.

	(-¥ , -4)	(-4, -2)	(-2, ¥ )
x + 2	negative	negative	positive
x + 4	negative	positive	positive
	positive	negative	positive

Looking at this chart, the domain for this function is (-¥, -4) È (-2, ¥).

PROPERTIES OF LOGARITHMS

There are seven properties of logarithms that allow us to simplify logarithmic expressions. You should become familiar with these properties because the will be used to simplify expressions later on.

For any numbers a > 0 and x > 0
1.	PRODUCT RULE:	ln(ax) = ln a + ln x
2.	QUOTIENT RULE:	ln(a /x) = ln a - ln x
3.	RECIPROCAL RULE:	ln(1 /x) = - ln x
4.	POWER RULE:	ln x ⁿ = n ln x
5.	ln 1 = 0
6.	ln e ^x = x
7.	e ^{ln x} = x

Simplify the following logarithmic expression as a sum and difference of logarithms.

EXAMPLE 4: Simplify: ln [(x ² - 4)(x + 3)]

SOLUTION: ln [(x ² - 4)(x + 3)] = ln [(x + 2)(x - 2)(x + 3)] = ln (x + 2) + ln (x - 2) + ln (x + 3)

EXAMPLE 5:

SOLUTION:

EXAMPLE 6:

SOLUTION:

These properties will be important when we start to take derivative of logarithmic and other functions.

Therefore know them, use them, and let them be your friends!! J
THE DERIVATIVE OF THE NATURAL LOG

Consider the graphs of the following two functions: y = 1/x and y = ln x for x > 0.

Notice that y = ln x is always an increasing function and y = 1/x is always positive. Therefore y = 1/x could be the derivative of y = ln x. So let us define the ln x the following way:

Notice the following traits of the natural logarithm.

(#3 is the definition of property 5)

From all of this, let us define the following for the derivative of the natural logarithm.

This result comes from the Fundamental Theorem of Calculus.

EXAMPLE 7:

SOLUTION:

We will have to use the chain rule to find this derivative. Therefore, let us review how to do the chain rule. If y = g (u) and u = f (x), then y' (x) = g' (u)* f '(x).

Let u = x + 2, then y = ln u.

EXAMPLE 8:

SOLUTION:

Let u = x ² - 4, then y = ln u.

EXAMPLE 9:

SOLUTION:

First of all, let us simplify y = ln [x ³(x + 2)].

y = ln [x ³(x + 2)] ® y = ln x ³ + ln (x + 2) ® y = 3ln x + ln (x + 2)

Now, to find the derivative. (Notice that the derivative of ln (x + 2) will be a chain rule.)

EXAMPLE 10:

SOLUTION:

Using the chain rule, let u = ln (x + 2), therefore y = u ⁴.

EXAMPLE 11:

SOLUTION:

To find this derivative, we will have to use the quotient rule and the product rule along with the derivative of the natural logarithm. Here is a review of the quotient rule and product rule from calculus I.

Quotient Rule

Product Rule

EXAMPLE 12:

SOLUTION:

THE INTEGRAL OF 1/ X

FACT:

EXAMPLE 13:

Integrate the function y = tan x.

SOLUTION:

Recall that the tan x = (sin x)/(cos x), therefore replace the tan x with this and integrate using a u-substitution.

Why are the two solutions, -ln |cos x| + C and ln |sec x| + C, equal to each other?

FACT:

FACT:

EXAMPLE 14:

SOLUTION:

Let u = x ² + 4, therefore, du = 2x dx.

EXAMPLE 15:

SOLUTION:

Let u = 3 + cos x, then du = - sin x dx or -du = sin x dx.

EXAMPLE 16:

SOLUTION:

Let u = 3x + 15, then du = 3dx.

Work through these examples. If you have any problems with any of them, feel free to come and receive help. We will be building onto this topic, so it is best to understand this topic now, and not later.

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