2

Exponential and Logarithmic Functions

One of the overall themes in high-school mathematics is the concept of function. The idea was introduced in math 20, then we looked at the simplest class of functions: polynomials. These included linear, quadratic, cubic, and quartic functions. In math 30 we add to our knowledge of functions by studying exponential and logarithmic functions (in this chapter), and trigonometric functions in subsequent chapters. These functions are all known as transcendental functions, which means they transcend our ability to write them in polynomial form. In other words if we try to write any of these functions in polynomial form, we get a polynomial of infinite degree. They have many applications in physics and engineering, and we use them all the time in calculus.

2.1

Geometric Sequences and Series

We begin our study of exponential functions by going back to a topic that was touched on in math 10: geometric sequences. The following are all examples of geometric sequences: • 5, 15, 45, 135, 405, 1215 • 64, 32, 16, 8, 4, 2, 1, .5, .25 √ √ √ • 2, 2, 2 2, 4, 4 2 • 2, −6, 18, −54, 162, . . . In all of these examples, we start with a certain number, and then multiply each time by the same value to produce the next number. The value that we multiply by each time is called the common ratio. In the above examples important the common ratios are term √ 3, 1/2, 2, −3 We want to deal with geometric sequences in an algebraic way, so we need to have a completely algebraic description of them. The notation we use in high-school is: • the first term is denoted a • the common ratio is denoted r

1

Using these symbols, we can now write the general geometric sequence as a, ar, ar2 , ar3 , . . . , arn−1 , . . . There is one unfortunate thing about this notation; you can see that the number of the exponent is one less than the term number. So for example mistake in the fourth term the exponent on r is only 3. This is because in the first alert term there is no r, or the exponent on r is zero. This is a common source of error, because many students just assume the exponent on r should be equal to the position of the term in the sequence; try to avoid this mistake. Another symbol that is common in high-school is t to denote a term. For example t3 means the third term in a sequence, t102 means the 102nd term, and tn means the nth term, or what is called the general term. Using all of our symbols, we can now say that for any geometric sequence: tn = arn−1 Examples: • find the 7th term in the sequence: −3, −6, −18, . . .. We see that a = −3, and r = 2 so that: tn = arn−1 t7 = (−3)(2)(7−1) t7 = (−3)(2)6 t7 = (−3)(64) t7 = −192 • the first term in a geometric sequence is 240, and the ninth term is 0.9375. Find the common ratio. Using our symbols, we have: tn = arn−1 t9 = ar(9−1) 0.9375 = (240)(r)8 0.00390625 = r8 1/2 = r Note that to go from the second last to the last line above, we took the eighth root of both sides. 2

Now we turn to the topic of adding up geometric sequences: 2 + 8 + 32 = 42 This is a nice simple case, where a = 2, r = 4, and the number of terms (n) is 3. This is an example of a geometric series which is defined as the sum important of the terms of a sequence. We certainly do not want to have to add up definition series in the ordinary way, even using a calculator, since it would be tedious and error prone to add a list of hundreds or even thousands of numbers. Therefore we look for a technique (or formula). You might remember that we used the trick of pairing terms when adding arithmetic sequences. This does not work with geometric series (you should try it to see why), but another technique does work: we multiply the series by r, then subtract the original series. Sound complicated? It’s not. First an easy numerical example: Find the sum of S = 1 + 2 + 4 + 8 + 16 We see that r = 2, so we multiply both sides of the equation above by 2: S = 1 + 2 + 4 + 8 + 16 2S =

2 + 4 + 8 + 16 + 32

(2S − S) = 32 − 1 S = 33 We aligned the second series to the right in order to make subtraction easier; we also subtracted going up in order to avoid negatives. Now let’s do it in general (we assume n terms): Sn = a + ar + ar2 + · · · + arn−1 rSn =

ar + ar2 + · · · + arn−1 + arn

rSn − Sn = arn − a Sn (r − 1) = a(rn − 1) a(rn − 1) Sn = r−1

Actually we use two versions of this formula:

3

Sn =

a(rn − 1) r−1

and

Sn =

a(1 − rn ) 1−r

The one on the right is used when r > 1, and the one on the left is used when r < 1. It’s ok if you forget this: you’ll get the right answer either way, but if you use the correct version you avoid negatives. (Actually I’m not sure if that’s very important since all of you use calculators all the time; anyway there are two versions of the formula).

Example: find the sum of the first 8 terms of the geometric sequence 3 + 12 + 48 + · · · Clearly a = 3, r = 4, and n = 8: a(rn − 1) r−1 3[(4)8−1 − 1] S8 = 4−1 3(47 − 1) S8 = 3 S8 = 47 − 1

Sn =

S8 = 16383

2.2

Sigma Notation

In this section we will learn about a mathematical symbol that is use to write out series in a more compact form. This notation uses the symbol Σ, which is a capital S in the Greek alphabet. This letter was chosen because S is the first letter in the word sum, and sigma notation is a short way of writing sums, or series. The best way to learn this notation is to examine some examples: 1.

4 X

i2 = 12 + 22 + 32 + 42

i=1

2.

3 X

2k + 1 = 1 + 3 + 5 + 7

k=0

4

3.

5 X

2j = 1 + 2 + 4 + 8 + 16 + 32

j=0

4.

∞ X

2i = 2 + 4 + 6 + · · ·

i=1

The letters i, j, k are called dummy variables; it does not matter which letters X we use, the meaning of the notation is the same. The number below the sign tells you what value the dummy starts at, and the number on top tells you the final value. Sometimes you need to look at a sigma Pb notation and just know how many terms there are in the sum. For i=a , the number of terms is b − a + 1. It is a very common error in high-school to think mistake that the number of terms is b − a. We now give a few more examples of alert sigma notation, beginning with a restatement of the formula for the sum of a geometric series: n−1 X

ari =

i=0

1. evaluate

7 X

a(rn − 1) r−1

0.125(2)i We can see that this is a geometric series with

i=0

a = 0.125 (just put 0 in for i to see this), and r = 2. There are 7 − 0 + 1 = 8 terms, so n = 8 and the formula gives S8 =

[0.125(27 − 1)] 2−1

or S8 =

127 8

2. write the following sum in sigma notation: 4+12+36+108+324+972 This is a geometric series, with a = 4, r = 3, and n = 6. So each term in this series is of the form 4(3)k , where k starts at 0 and ends at 5 (one less than the number of terms). Thus the sigma notation for this series is: 5 X 4(3)k k=0

5

You might be surprised that the exponent was k, and not k − 1, since when we learned the general term for geometric sequences the exponent was n − 1. The reason for this is that when we were learning the general term for geometric sequences, the variable (n in that case) started at 1, whereas in the example we just did, the dummy variable started at zero. This is a very important point because it means that there is not just one way to describe key point a series in sigma notation; there are many ways. We can start the dummy variable at whatever we want. For example, all of the following are equal: 5 X

k

4(3) =

k=0

6 X

k−1

4(3)

=

k=1

22 X

4(3)k−17

k=17

The choice for the initial value of the dummy variable is usually one of convenience; it is usually easiest to start it at 1.

2.3

From Sequences to Exponential Functions

Let’s back up a bit, and look at the graphs of two sequences. The first will be tn = 3(2)n−1 . The number of each term will be on the horizontal axis, and the value of each term will be on the vertical axis: tn = 3(2)n−1 60 c

50 40 30 term value

c

20 10

c

0 -10 -1

0

1

c

c

term 2 number 3

4

5

6

This is a function, because it passes the vertical line test, but right now it is only defined for the positive integers, because when we count terms (or anything) we start at 1, and use the positive integers. But if we draw a smooth curve through these points, we get a function defined on the real numbers (i.e. decimals etc.). Here it is:

6

tn = 3(2)n−1 60 c

50 40 30 term value

c

20 10

c

0 -10 -1

0

1

c

c

term 2 number 3

4

5

6

What is this function? Well all we have done is allow the sequence formula to be used for all real numbers, so we just replace n (which refers to integers) with x to get y = 3(2)x−1 Actually we prefer to write this a bit differently (putting in a more standard form) by doing the following: y = 3(2)x−1 y = 3(2)x (2)−1 y = 3(1/2)(2)x y = (3/2)(2)x This equation now has the typical form of an exponential function, which is any function that can be written in the form y = a(c)x , c > 0 In our example a was 3/2, and c was 2. We will see in a little while why we must insist that c > 0.

Let’s look at a second example. This time we will graph a sequence whose terms get smaller. We will use tn = 128(.5)n−1 . 7

tn = (128)(1/2)n−1 140

c

120 100 80

c

term value 60 40

c c

20 0 -1

0

1

term 2 number 3

4

c

c

5

6

Once more we have a function defined only for the positive integers, because that is the meaning of the variable n, but we can get a function for all real numbers by using any real value for the variable. The graph looks like this: tn = 128(.5)n−1 140

c

120 100 80 term value 60

c

40

c c

20 0 -1

0

1

term 2 number 3

4

c

c

5

6

While we could write this as y = 128(1/2)x−1 , we again prefer to rearrange it as before (you should be able to do this yourself) to get y = 64(1/2)x so that once more we have a typical exponential function y = a(c)x , c > 0 with a = 64, and c = 1/2.

8

So let us summarize: An exponential function is any function that can be written in the form y = a(c)x , c > 0 . If we were to allow c to be negative we would get a crazy graph that jumped back and forth from negative to positive values, and that is not defined for many values of x. (specifically, it would not be defined for any x that in reduced fraction form, looked like k/2n, where k, and n are both integers.) Here is the typical exponential function, when c > 1:

10

8

6

4

2

0 -4

-3

-2

-1

0

1

2

3

4

This is often called a growth graph, since it models such things as growth of money in a bank account, or growth of population. Here is the typical case when 0 < c < 1:

9

10

8

6

4

2

0 -4

-3

-2

-1

0

1

2

3

4

This is called a decay graph, since it can model the amount of radioactive material when it decays (among other things). The characteristics of both forms above are: memorize 1. domain: x< 2. range: y > 0 3. no x-intercepts; y-intercept: a 4. asymptotes: the x-axis

2.4

Logarithms

This chapter is about exponential functions and logarithmic functions, and we haven’t yet explain what logarithms are, so we take a break from exponential things for a while to take care of this. Let’s begin with examples: 1. since 102 = 100, we say that 2 is the logarithm of 100 (to the base 10) 2. since 103 = 100, we say that 3 is the logarithm of 1000 (to the base 10) 3. since 10.3010 = 2, we say that .3010 is the logarithm of 2 (to the base 10) 10

4. since 10−1 = 0.10, we say that −1 is the logarithm of 0.10 (to the base 10) We can generalize this by saying: If 10k = m, then we say that k is the logarithm of m, base 10. The symbols for this are: if 10k = m, then log10 m = k. In simple English, k is the logarithm of m (base 10) if k is the exponent we need to put on 10, in order to produce m. So far we have confined ourselves to using 10 as the base, but we can use any base. So, for example, since 23 = 8, we can say that the logarithm of 8 (base 2) is 3. In symbols we write: log2 8 = 3

2.5

The Laws of Logarithms

There are three very important laws of logarithms; these laws are used both to simplify complicated expressions, and also to solve equations algebraically, that could not be solved in any other way. The laws will look strange at first, but you should always keep in mind that they are just the familiar exponent laws in another form.

log(ab) = log a + log b

(1)

log(a/b) = log a − log b

(2)

n

n log a = log a

(3)

We will now take each law, one at a time, and give some examples and then a proof. The intent throughout will be to emphasize the connection between these laws and the more familiar laws of exponents. In all of this work we will be working in base 10, but we could easily use any other base. important point 1. on your calculator you can verify that log(1000) = log 10 + log 100 to prove this law, let’s begin with this: 10m × 10n = 10m+n

11

Now if we let a = 10m , and b = 10n ,then ab = 10m+n Now from the very definition of logarithms, you should be able to see that: log a = m, log b = n, and log (ab) = m + n, therefore it is clear that log (ab) = log a + log b 2. we know that 10m ÷ 10n = 10m−n If, as before, a = 10m , b = 10n and a/b = 10m−n , then from the definition of logarithms you can see that log(a/b) = log a − log b 3. we prove the third law by repeated application of the first law. e.g. 2 log n = log n + log n = log(n × n) = log n2 3 log n = (log n + log n) + log n = log n2 + log n = log(n2 × n) = log n3 all I did here was apply the first law again and again. It would be easy to extend this to n terms. (in fact I would ask this on a chapter test) Even though our proof of this law requires that n be an integer, the law works even for non integers. e.g. π log a = log aπ To be very sure that you understand what we have done here, I restate the 3 laws of logarithms, with their corresponding exponent laws: 1. log(ab) = log a + log b means the same as 10m × 10n = 10m+n 2. log(a/b) = log a − log b means the same as 10m ÷ 10n = 10m−n

12

3. n log a = log an means the same as (10m )n = 10mn Now even though all of these laws are important, I want to draw your attention to the third law in particular. We use this law to isolate variables in the exponent. Till now, this was not possible. Example: Solve the equation 7 = 2x Before having the third law, we would have solved this by guess and check, or graphing, but now we proceed like this: 7 = 2x log 7 = log 2x log 7 = (x) log 2 log 7 = x log 2 And while we’re at it, let’s notice something else: we should know from the definition of logarithms, that if 7 = 2x then x = log2 7 Therefore we have proved the important result that log2 7 =

log 7 log 2

Or in general , we just proved that

key point

loga b =

13

log b log a

2.6

The Logarithm Function

Now that we have some idea what logarithms are, we can consider the function y = log x Note that we are still confining ourselves to base 10 only for convenience, and we could replace the 10 by any other (positive) number. If we graph y = log x, we get the following:

1 0.5 0 -1

-0.5

0

0.5

1

1.5

2

2.5

3

-0.5 -1 -1.5 -2 -2.5 -3

You should realize right away that this function is only defined for x > 0. The reason is that, for example, log10 (−5) = k implies that 10k = −5. Since this is impossible, we cannot take the logarithm of any num- we can in ber that is not positive. You might have also noticed that this graph the complex looks like a reflection in the line y = x of the graph of y = 10x . To make this numbers clear, we illustrate both graphs, along with the mirror line:

14

10x and log x 3 2 1 0 -3

-2

-1

0

1

2

3

-1 -2 -3

To see why this is so, recall that to find the inverse of a given function, we • switch x and y • isolate y So let’s try this with the function y = 10x : y = 10x x = 10y log x = log(10y ) log x = y log(10) log x = y To go from the second last line to the last line, we made use of the fact that log10 10 = 1. Thus we have proved that y = 10x and y log10 x are inverses of each other. In general we see that y = ax and loga x are inverses of each other. We conclude this section by noting that there are actually two looks for log graphs, depending on whether the base is larger, or smaller than 1. (compare to the two looks for exponential graphs) 15

For y loga = x, a > 1, the graph looks like this: 3 2 1 0 -1

-0.5

0

0.5

1

1.5

2

2.5

3

2.5

3

-1 -2 -3

and for y = loga = x, 0 < a < 1, the graph looks like this: 3 2 1 0 -1

-0.5

0

0.5

1

1.5

2

-1 -2 -3

In either case, the characteristics are: (just switch x stuff and y stuff) 16

1. domain: x > 0 2. range: y< 3. no y-intercepts; x-intercept: 1 4. asymptotes: the y-axis

2.7

Exponential and Logarithmic Equations

Many equations in science and higher mathematics are in exponential form. To solve such equations we must know both the laws of exponents and those of logarithms. 2.7.1

Exponential Equations

For the purposes of math 30, we can consider two sorts of exponential equations: those in which both sides of the equation can be put to the same base, and those in which this is not possible. Let’s look at some examples of the first type: 1. 92x−1 = 27x−2 (32 )2x−1 = (33 )x−2 34x−2 = 33x−6 4x − 2 = 3x − 6 x = −4 2. (1/8)3x+1 = 322x−1 (2−3 )3x+1 = (25 )2x−1 2−9x−3 = 210x−5 −9x − 3 = 10x − 5 −19x = = 2 −2 x = 19

17

This is by far the easier of the two types; you can usually solve this kind without a calculator. In the second type you must make use of the third law of logarithms in order to get the variable out of the exponent: log an = n log a Examples: 1. 3x+2 = 5x log 3x+2 = log 5x (x + 2) log 3 = (x) log 5 (x) log 3 + 2 log 3 = (x) log 5 2 log 3 = (x) log 5 − (x) log 3 log 9 = x(log 5 − log 3) log 9 x = log 5 − log 3 x = 4.301 2. 2(3)x+1 = 62x−1 log(2(3)x+1 ) = log(62x−1 ) log 2 + (x + 1) log 3 = (2x − 1) log 6 log 2 + log 3 + log 6 = 2x log 6 − x log 3 log 36 = x log 62 − x log 3 log 36 = x log 36 − x log 3 log 36 = x(log 36 − log 3) log 36 = x(log 12) log 36 x = log 12 x = 1.44 It is easy to see that this type of equation is tedious, but it is definitely not hard; after we get the variable out of the exponent, we are really dealing with a simple linear equation that looks hard because of all the log symbols around. As I pointed out in class, there are ways of making substitutions that make the algebra easier. 18

2.7.2

Logarithmic Equations

In order to solve logarithmic equations you must be conversant with the laws of logarithms and you must also know how to change a statement from exponential form and back. The key fact here is: loga b = c ⇒ ac = b Students constantly forget this; if you have trouble remembering it, just make up an easy example: since 102 = 100 then log10 100 = 2 The first sort of logarithmic equation we will look at is the type which simply involves changing to exponential form: 1. logx 3 = 2 x2 = 3 √ x = 3

2. log3 x = 2 32 = x 9 = x 3. log5 7 = x log 5 = x log 7

Next, we look at the type which has more than one log symbol on a side of the equation. The first thing to do in this type is to use the laws of logarithms to simplify:

19

1. log5 (x − 5) + log5 x = log5 6 log5 (x2 − 5x) = log5 6 x2 − 5x = 6 x2 − 5x − 6 = 0 (x − 6)(x + 1) = 0 x = 6, −1 x = 6

(−1 is extraneous)

2. log2 (2m + 4) − log2 (m − 1)  2m + 4  log2 m−1 2m + 4 m−1 2m + 4

= 3 = 3 = 23 = 8m − 8

12 = 6m 2 = 2 There is actually very little here besides the laws of logarithms and some basic algebra.

20