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• Elementary FunctionsPart 3, Exponential Functions & LogarithmsLecture 3.3a, Logarithms: Basic Properties

Dr. Ken W. Smith

Sam Houston State University

2013

Smith (SHSU) Elementary Functions 2013 1 / 29

The logarithm as an inverse function

In this section we concentrate on understanding the logarithm function. Ifthe logarithm is understood as the inverse of the exponential function,then the properties of logarithms will naturally follow from ourunderstanding of exponents.

The meaning of the logarithm.The logarithmic function g(x) = logb(x) is the inverse of the exponentialfunction f(x) = bx.

The meaning of y = logb(x) is by = x.

The expressionby = x

is the exponential form for the logarithm y = logb(x).

The positive constant b is called the base (of the logarithm.)

Smith (SHSU) Elementary Functions 2013 2 / 29

Logarithms

Some worked exercises.Write each of the following logarithms in exponential form and then usethat exponential form to solve for x.

1 log2(8) = xSolution. The exponential form is 2x = 8. Since 23 = 8 the answer isx = 3 .

2 log2(247) = x

Solution. The exponential form is 2x = 247. So x = 47 .

3 log2(12) = x

Solution. The exponential form is 2x = 12 . Since 21 =

1

is x = 1 .

Smith (SHSU) Elementary Functions 2013 3 / 29

Logarithms

4 log2(18) = x

Solution. The exponential form is 2x = 18 . Since 23 = 18 the answer

is x = 3 .5 log2(

32) = x

Solution. The exponential form is 2x = 32 = 21/3. So x = 1/3 .

Notice how we use the exponential form in each problem!

Smith (SHSU) Elementary Functions 2013 4 / 29

• The graph of a logarithm function

The graph of y = 2x was drawn in an earlier lecture (see below.)

The graph of the inverse function y = log2 x is obtained by reflecting thegraph of y = 2x across the line y = x.

Smith (SHSU) Elementary Functions 2013 5 / 29

The graph of a logarithm function

The graph of y = log2 x :

Smith (SHSU) Elementary Functions 2013 6 / 29

The graph of a logarithm function

If we draw them together, we have the picture below.

Smith (SHSU) Elementary Functions 2013 7 / 29

The graph of a logarithm function

The graph of the exponential function y = 2x:

The graph of the logarithmic function y = log2 x:Smith (SHSU) Elementary Functions 2013 8 / 29

• Logarithms

We agreed earlier that the exponential function f(x) = bx has domain(,) and range (0,).

Since g(x) = logb x is the inverse function of f(x) the domain of the logfunction will be the range of the exponential function, and vice versa.

So the domain of logb x is (0,) and the range is (,).

The most useful base for logarithms is e. We will abbreviate loge(x) byln(x) and speak of the natural logarithm.

Sometimes, for historical reasons, we may use base 10. It is customary tospeak then of the common logarithm and abbreviate log10(x) by log(x),dropping the subscript. However (warning!), in higher mathematics andengineering applications, log(x) usually means base e and is equivalent toln(x).

In these notes we will use log(x) to mean log10(x).

One more abbreviation in computer science, because computers storedata in binary (in bits of zeroes and ones), one uses base 2. Someabbreviate log2(x) as lb(x) and speak of the binary logarithm.

Smith (SHSU) Elementary Functions 2013 9 / 29

Logarithms

In summary, here are our abbreviations:

1 lnx means the logarithm base e,

2 log x means the logarithm base 10 and

3 lb x means the logarithm base 2.

Smith (SHSU) Elementary Functions 2013 10 / 29

Logarithms

A few more worked exercises.Write each of the following logarithms in exponential form and then usethat exponential form to solve for x.

1 log(1000) = xSolution. The exponential form is 10x = 1000. Since 103 = 1000 theanswer is x = 3 .

2 ln(1

e3) = x

Solution. The exponential form is ex = e3 so the answer is 3 .

3 lb(12) = x

Solution. The exponential form is 2x =12. Since 21/2 =

2 then

21/2 =12

and so the answer is x = 1/2 .

Smith (SHSU) Elementary Functions 2013 11 / 29

Properties of exponential functions in terms of logarithms

The logarithm function plucks the exponent from an expression. For thisreason, the properties of exponents translate into properties of logarithms.For example, we know that when we multiply two terms with a commonbase, we add the exponents:

(bx)(by) = bx+y (1)

Suppose we call the first term M := bx and the second term N := by.Then one may ask the question, What is the exponent on b in theproduct MN?The answer is We add the exponents appearing in M and N . In otherwords (if we learn to translate logb as the exponent on b that...), wecan restate this exponent property as when we multiply numbers we addtheir exponents. This is the product property for logarithms:

logb(MN) = logbM + logbN (2)

Smith (SHSU) Elementary Functions 2013 12 / 29

http://en.wikipedia.org/wiki/Binary_logarithm

• Logarithms

What happens when we divide two terms with a common base?

bx

by= bxy (3)

When we do division, we subtract exponents. So, in the language oflogarithms, we have the quotient property, the exponent in a quotient isthe difference of the two exponents:

logb(M

N) = logbM logbN (4)

Smith (SHSU) Elementary Functions 2013 13 / 29

Logarithms

A third important property of exponents: when we raise a term like bx to apower, we multiply exponents.

(bx)c = bxc (5)

In our logarithm language (thinking of M as bx) we have the exponentproperty

logb(Mc) = c logbM (6)

Each of these three properties is merely a restatement, in the language oflogarithms, of a property of exponents.

Smith (SHSU) Elementary Functions 2013 14 / 29

Logarithms

We review the three basic logarithm rules we have developed so far.

Product Property of Logarithms:

logb(MN) = logbM + logbN

Quotient Property of Logarithms:

logb(M

N) = logbM logbN

Exponent Property of Logarithms:

logb(Mc) = c logbM

Each of these properties is a restatement, in the language of logarithms, ofa property of exponents.

Smith (SHSU) Elementary Functions 2013 15 / 29

Exponential Functions

In the next presentation, we develop several more properties of logarithms.

(END)

Smith (SHSU) Elementary Functions 2013 16 / 29

• Elementary FunctionsPart 3, Exponential Functions & Logarithms

Lecture 3.3b, Logarithms: Basic Properties, Continued

Dr. Ken W. Smith

Sam Houston State University

2013

Smith (SHSU) Elementary Functions 2013 17 / 29

Logarithms

We review the three basic logarithm rules we have developed so far.

Product Property of Logarithms:

logb(MN) = logbM + logbN

Quotient Property of Logarithms:

logb(M

N) = logbM logbN

Exponent Property of Logarithms:

logb(Mc) = c logbM

Each of these three properties is merely a restatement of a property ofexponents.

Smith (SHSU) Elementary Functions 2013 18 / 29

Changing the base

Suppose we want to change the base of our logarithm. This often occurswhen we want to use a good base like e on a problem which began witha different base.Suppose we want to work with base c but our problem began with base b:

y = logb x.

Rewrite this in exponential form:

by = x.

Now take the log of both sides of the equation. If we want to work in basec then let us apply logc() to both sides of our equation.

logc(by) = logc(x).

Now we use the exponent property pulling the exponent y outside thelogarithm:

y logc(b) = logc(x).

Solve for y:

y =logc x

logc b.Smith (SHSU) Elementary Functions 2013 19 / 29

Changing the base

We began withy = logb x.

We rewrote this as

y =logc x

logc b.

So y, which was originally equal to logb x is now

logb x =logc x

logc b

Lets call this the change of base equation or change of baseproperty.

One way to remember this is to note that on the left side of the equal sign(logb x), b is lower than x.Then on the right side of the equal sign ( log xlog b ), b is still lower than x!

Smith (SHSU) Elementary Functions 2013 20 / 29

• Logarithms

Example. Suppose we want to compute log2(17) but our calculator onlyallows us to use the natural logarithm ln. Then, by the change of baseequation we can write

log2(17) =ln 17

ln 2 4.087463.

Smith (SHSU) Elementary Functions 2013 21 / 29

More on the logarithm as an inverse function

We began this lecture by defining g(x) = logb(x) as the inverse function off(x) = bx. Since these functions are inverses, we know then that

(f g)(x) = (g f)(x) = x. (7)

Let us examine this in more detail.

Note that (g f)(x) = g(f(x)) = g(bx) = logb(bx). Since the log functionand the exponential function are inverse functions, this must be equal tojust x and we have

logb(bx) = x.

This equation is really fairly easy to understand. If we translate logb(x)as the exponent on b that give x then we should translate logb(b

x) asthe exponent on b which gives bx.Obviously this should be x since x is the exponent one places on b to getbx. (If that doesnt make sense, read through it one more time slowly....)

Smith (SHSU) Elementary Functions 2013 22 / 29

More on the logarithm as an inverse function

Since (f g)(x) = x we also havex = (f g)(x) = f(g(x)) = f(logb x) = blogb x. So

blogb x = x.

This is almost as easy to understand as the