Chapter 5: exponential and logarithmic function

The logarithm of a number is the exponent to which another fixed value, the base, must be

raised to produce that number. Logarithms are exponents. They were once used t simplify

calculations, but the advent of calculators and computers caused calculation with

logarithms to be used less and less.

Section 5-1: Integral Exponents Basic Laws of Exponents

Each of the above rules should be familiar to you from algebra I. Here are some sample problems with their solutions. 1) Watch the difference between these two: a) (-3)-2 b) -(3)-2 The first one is squaring a negative number and the second is squaring a positive number and then making the whole result negative. a) = 1/(-3)2 = 1/9 b) 1/-(3)2 = -1/9

c) 7 . 2-3 = d) (7 . 2)-3 = The first one raises the power then multiplies, while the second one multiplies first then raises the power. c) = 7/8 d) = 14-3 = 1/143 = 1/2744

In the above example our first step is to work inside the grouping symbols and get a common denominator. Then add the two fractions. Only when you have a single fraction, is it permitted to invert the fraction.

In the above example we again simplify inside the grouping symbols and get a common denominator. Once we have a single fraction in step 3 we can invert the fraction. Notice the factoring in the last step!

In this example, the numerator and denominator have the same base. We can apply the division rule by subtracting the exponents. Then simplify. Remember, no negative exponents should be left in the answer.

An important type of rule can now be stated using exponents. It is a growth or decay problem. We can mathematically model this function by using the following: A(t) = Ao(1 + r)t

Where Ao is the initial amount at time t = 0 r is the rate (as a decimal) t is the time A(t) is the amount after the time t. If r > 0, then it is an exponential growth. If -1 < r < 0, then it decays exponentially.

1) Suppose a bike costs $100 now and it increases at a rate of 5% per year. What will be the cost of the bike in 4 years? Solution: Ao = 100, r = 5% = .05, t = 4 A(4) = 100(1 + .05)4 = 100(1.05)4 = 121.55 The bike will cost $121.55 in 4 years. (Rounded to the nearest penny)

2) Suppose a car is worth $15,000 new. What will it be worth in 3 years if it decreases at a rate of 12% per year? Solution: This is a decrease problem with Ao = 15000, r = -.12, t = 3 A(3) = 15000(1-.12)3 = 15000(.88)3 = 10222.08 The car will be worth $10222.08 in 3 years.

One of the easier ways to do this problem is to multiply both numerator and denominator by the positive power of the biggest negative exponent. In this case we multiplied by 44. This greatly simplifies the problem. Section 5-2: Rational Exponents All rules presented in the previous section were defined for integers only. All of the properties in the last section can also be extended to include rational exponents according to the following definitions:

Examples

1)

2)

3)

4)

5)

6) 7) (1001/2 - 361/2)2 = (10 - 6)2 = 42 = 16 8) x1/2(x3/2 + 2x1/2) = x2 + 2x

9)

We can use these rules to solve for x when x is the exponent. This method will only work if the bases are the same. Check back in section 5-1 for the appropriate rule!! Examples 1) 16x = 25 We can write both sides in base two. 24x = 25 Now use the fact that the bases are the same, the exponents are = 4x = 5 And solve for x!!

x = 5/4 To check it, take the 5/4 root of 16 = 25!! 2) 271-x = (1/9)3-x You need to make both bases the same. How about 3!! 33(1-x) = 3-2(3-x) Notice the power on the right side is negative. 3(1 - x) = -2(3 - x) Because the bases are =, the exponents must be = 3 - 3x = -6 + 2x Solve for x. 3 = -6 + 5x 9 = 5x 9/5 = x

The growth and decay formula can also be used with rational numbers. Consider the following: 1) The cost of a computer has been increasing at 7% per year. If it costs $1500 now, find the cost: a) 2 years and 6 months from now b) 3 years and 3 months ago. Solutions: a) Ao = 1500, r = .07 and t = 2.5 A(2.5) = 1500(1 + .07)2.5 = 1500(1.07)2.5 = 1776.44 b) Ao = 1500, r = .07, and t = -3.25 A(-3.25) = 1500(1.07)-3.25 = 1203.91 Section 5-3: Exponential Functions Any function in the form f(x) = abx, where a > 0, b > 0 and b not equal to 1 is called an exponential function with base b. Let's take a look at a couple of simple exponential graphs. f(x) = 2x

X f(x)

3 8

2 4

1 2

0 1

-1 1/2

-2 1/4

-3 1/8

Notice the domain is all real numbers and the range is y > 0. As x gets larger (right), y gets very large. As x gets smaller(left), y approaches zero asymptotically. Notice also that the graph crosses the y-axis at (0, 1). The above is the general shape of an exponential with b > 1. This is an example of exponential growth.

Now let's look at the graph of f(x) = (1/2)x

x f(x)

3 1/8

2 1/4

1 1/2

0 1

-1 2

-2 4

-3 8

Observe that this graph is the reflection about the y-axis of the first graph. The domain is still all real numbers and the range is y > 0. The y-intercept is (0, 1). This is the general form of an exponential graph if 0 < b < 1. It is an example of anexponential decay.

Look at the following graphs that illustrate the general properties of exponentials.

Do you see the similarities of each graph? How about this one?

Many of the functions associated with exponential growth or decay are functions of time. We have already had one form: A(t) = Ao(1 + r)t A second form looks like: A(t) = Aobt/k where k = time needed to multiply Ao by b

Rule of 72 If a quantity is growing at r% per year then the doubling time is approximately 72/r years.

For example, if a quantity grows at 10% per year, then it will take 72/10 or 7.2 years to double in value. In other words, it will take you 7.2 years to double your money if you put it into an account that pays 10% interest. At the current bank rate or 2%, it will take you 72/2 or 36 years to double your money!! Boy, jump all over that investment!!

Sample Problems 1) Suppose you invest money so that it grows at A(t) = 1000(2)t/8 a) How much money did you invest? b) How long will it take to double your money? Solutions: a) The original amount in the formula is $1000. b) This means what time will it take to get $2000. 2000 = 1000(2)t/8 2 = 2t/8 1 = t/8 8 = t

It will take 8 years to double your money!! 2) Suppose that t hours from now the population of a bacteria colony is given by: P(t) = 150(100)t/10 a) What is the initial population? b) How long does it take for the population to be multiplied by 100)20/10 = 150(100)2 = 1,500,000 3 100? c) What is the population at t = 20? Solutions: a) It is 150 from the original equation. b) It takes 10 hours. That's the definition of the exponential function. c) P(20) = 150() The half life of a substance is 5 days. We have 4 kg present now. a) Write a formula for this decay problem. b) How much is left after 10 days? 15 days? 20 days? Solutions: a) A(t) = 4(1/2)t/5 b) A(10) = 4(1/2)10/5 = 4(1/2)2 = 1 kg. A(15) = 4(1/2)15/5 = 4(1/2)3 = 1/2 kg. A(20) = 4(1/2)20/5 = 4(1/2)4 = 1/4 kg. 4) The value of a car is given by the equation V(t) = 6000(.82)t a) What is the annual rate of depreciation? b) What is the current value? c) What will be the value in three years? Solutions: a) It is 1 - .82 or .18 = 18% b) The current value is given in the formula, $6000. c) V(3) = 6000(.82)3 = 3308.21 Which is $3308.21 Section 5-4 The number e and the function ex Try the quiz at the bottom of the page! go to quiz

Definition of the irrational number e

Without getting into a discussion of limits right now, we can get an idea of what's happening by taking increasingly larger values of n. We will talk about limits later on in the year. Study the following table of values and use your calculator to double check the results:

n (1 + 1/n)n

10 2.593742

100 2.704814

1000 2.716924

10,000 2.718146

100,000 2.718268

1,000,000 2.718280

If you study this chart, you see that the number e approaches a value of 2.718 . . . The function ex is called the natural exponential function. The graph of ex and e-x are graphed below:

Notice, they fit the pattern of the previous section. The number e appears in many applications of physics and statistics. We will take a close look at the number e and how it relates to compound interest.

Compound Interest Formula

Where: A(t) = amount after time t. Ao = Initial amount r = rate in decimal n = number of times compounded in a year. t = time in years Thus, if the interest was paid semiannually, n = 2. Paid quarterly would make n = 4, Paid monthly, n = 12, etc.

Sample Problems 1) Find the value of a $1 if it is invested for 1 year at 10% interest compounded quarterly. Solution: Initial amount is $1 with r = .10, n = 4 and t = 1. A(1) = 1(1 + .10/4)4 = 1.1038 This means that at the end of a year, each dollar invested in worth 1.1038 or slightly more than $1.10. The effect of compounding adds another .0038 % to the interest rate. Thus the effective annual yield is 10.38%. 2) You invest $5000 in an account paying 6% compounded quarterly for three years. How much will be in the account at the end of the time period? Solution: Initial amount is $5000, with r = .06, n = 4 and t = 3 A(3) = 5000(1 + .06/4)12 = $5978.09. This account pays $978.09 in interest over the three years. 3) What is the effective annual yield on $1 invested for one year at 15% interest compounded monthly? Solution: Initial amount $1, with r = .15, n = 12 and t = 1 A(1) = 1(1 + .15/12)12 = 1.1608. The effective annual yield is 16.08%. This is a relatively big increase because of the number of times compounded in the year.

The above problems all had one thing in common. The number of times compounded was a finite number. We can also havecontinuous compounding. That is, compounding basically every second on the second. This would be rather cumbersome to calculate because the compounding is extremely large. We can use a similar formula if the compounding is continuous. P(t) = Poert

Notice the appearance of the number e. If you look closely at the compound interest formula, you will see imbedded the definition of the number e. Only use this formula if you are sure the compounding is continuous.

Problems 1) $500 is invested in an account paying 8% interest compounded continuously. They leave it in the account for 3 years. How much interest is accumulated? Solution: Initial amount $500, with r = .08 and t = 3. P(3) = 500(e.08(3)) = 635.62. This means the interest is $135.62. 2) A population of insects rapidly increases so that the population after t days from now is given by A(t) = 5000e.02t. Answer the following questions: a) What is the initial population? b) How many will there be after a week? c) How many will there be after a month? (30 days) Solutions: a) The initial population is 5000 from the formula. b) A(7) = 5000e.14 = 5751 c) A(30) = 5000e.6 = 9111 Section 5-5: Logarithmic Functions Common Logarithm Demo: Log Funtion Applet log x = a if and only if 10a = x The important thing to remember is the log represents the exponent. In the case of common logs, the base is always base 10. Study the following examples. 1) log 100 = 2 because 102 = 100. 2) log 1000 = 3 because 103 = 1000. 3) log 1 = 0 because 100 = 1. 4) log .1 = -1 because 10-1 = .1 5) log .01 = -2 because 10-2 = .01 The log function is the inverse function of the exponential function and as such their graphs are reflections about the y = x line. Here is the graph of the common log and the inverse.

Some important facts you need to understand from the log graph. The domain of the log is x > 0. The range is all real numbers. The zero is at x = 1. You can only find the log of positive numbers. Logs of numbers less than one are negative and logs of numbers greater than one are positive.

We can find the log of other bases by using the following formula similar to the common log definition. logb x = n if and only if x = bn. Here are some examples: 1) log2 8 = 3 because 23 = 8 2) log3 81 = 4 because 34 = 81. 3) log4 1/16 = -2 because 4-2 = 1/16 4) log8 1 = 0 because 80 = 1

One of the most important log function is called the natural log which has the number e as the base. When e is used as a base we use the following notation: ln x = a if and only if ea = x Most natural logs need to be calculated on your calculator. The graph of the natural log is shown below:

Solving Simple Log Equations 1) Log x = 3 Solution: To solve an equation of this type, rewrite the equation in exponential form. x = 103 = 1000 2) Log |x| = 2 Solution: To solve an equation of this type, again rewrite the equation in exponential form and solve for x. |x| = 102 = 100 x = 100 or -100 3) Log (x2 + 19) = 2 Solution: Again, rewrite as an exponential equation and solve for x. x2 + 19 = 102 x2 + 19 = 100 x2 = 81 x = 9 or -9 4) Log x = .3 Again, rewrite exponentially. x = 10.3 Use your calculator and round to hundredths. x = 2.00 5) Ln x = -1.2 Solution: Same as above. x = e-1.2 x = .30

Section 5-6: Laws of Logarithms

1) Logb MN = Logb M + Logb N 2) Logb M/N = Logb M - Logb N 3) Logb M = Logb N if and only if M = N

4) Logb Mk = k Logb M

5) Logb b = 1 6) Logb 1 = 0 7) Logb bk = k 8) bLogb x = x

Sample problems Write each log in expanded form. 1) Log5 xy2 = Solution: Log5 x + Log5 y2 = Log5 x + 2 Log5 y 2) Log7 (xy/z2) = Solution: Log7 x + Log7 y - 2 Log7 z

3) Express each as a single log. 1) Log x + Log y - Log z = Solution: Log (xy)/z 2) 2 Ln x + 3 Ln y = Solution: Ln x2y3 3) (1/2) Ln x - (1/3) Ln y =

Solution:

Writing logs as single logs can be helpful in solving many log equations. 1) Log2 (x + 1) + Log2 3 = 4

Solution: First combine the logs as a single log. Log2 3(x + 1) = 4 Now rewrite as an exponential equation. 3(x + 1) = 24 Now solve for x. 3x + 3 = 16 3x = 13 x = 13/3 Since this doesn't make the number inside the log zero or negative, the answer is acceptable. 2) Log (x + 3) + Log x = 1 Solution: Again, combine the logs as a single log. Log x(x + 3) = 1 Rewrite as an exponential. x(x + 3) = 10 Solve for x. x2 + 3x = 10 x2 + 3x - 10 = 0 (x + 5)(x - 2) = 0 x = -5 or x = 2 We have to throw out 5. Why? Because it makes (x + 3) negative and we can't take the log of a negative number. So the only answer is x = 2. 3) Ln (x - 4) + Ln x = Ln 21 Solution: Notice, this time we have a log on both sides. If we write the left side as a single log, we can use the rule that if the logs are equal, the quantity inside must be equal. Ln x(x - 4) = Ln 21 Since the logs are equal, what is inside must be equal. x(x - 4) = 21 Solve for x. x2 - 4x = 21 x2 - 4x - 21 = 0 (x - 7)(x + 3) = 0 x = 7 or x = -3 Again, we need to throw out one of the answers because it makes both quantities negative. Throw out -3 and keep 7. Thus, the answer is x = 7.

Simplify each log 1) ln e5 Solution: This is rule number 7. The answer is 5!

2) Log 10-3 Solution: This is again rule #7. The answer: -3 (This answers the question: what power do you raise 10 to get 10 to the third?

3) eln 7 Solution: This is rule #8. The answer is 7. 4) e2ln 5 Solution: We can use rule #8 as soon as we simplify the problem. Rewrite as: eln

25 = 25 The 25 came from 52. 5) 10Log 6 Solution: Rule #8 again. Answer: 6 6) 102 + log 5 Solution: We need to simplify before we can apply one of the rules. Rewrite as: (102)(10log 5) Adding exponents means you are multiplying the bases. = 100(5) Use rule #8 again. = 500

Section 5-7: Change of Base Formula

An exponential equation is an equation that contains a variable in the exponent. We solved problems of this type in a previous chapter by putting the problem into the same base. Unfortunately, it is not always possible to do this. Take for example, the equation 2x = 17. We cannot put this equation in the same base. So, how do we solve the problem? We use thechange of base formula!! We can change any base to a different base any time we want. The most used bases are obviously base 10 and base e because they are the only bases that appear on your calculator!! Change of base formula Logb x = Loga x/Loga b Pick a new base and the formula says it is equal to the log of the number in the new base divided by the log of the old base in the new base.

Examples 1) Log2 37 = Solution: Change to base 10 and use your calculator. = Log 37/log 2 Now use your calculator and round to hundredths. = 5.21 This seems reasonable, as the log2 32 = 5 and log2 64 = 6. 2) Log7 99 = Solution: Change to either base 10 or base e. Both will give you the same answer. Try it both ways and see.

= Log 99/Log 7 or Ln 99/ Ln 7 Use your calculator on both of the above and prove to yourself that you get the same answer. Both ways give you 2.36.

Solving Exponential Equations using change of base Now, let's go back up and try the original equation: 2x = 17 To put these in the same base, take the log of both sides. Either in base 10 or base e. Hint. Use base e only if the problem contains e. Log 2x = Log 17 Using the log rules, we can write as: x Log 2 = Log 17 Now isolate for x and use your calculator. x = Log 17/Log 2 x = 4.09 To check your answer, type in 24.09 and see what you get! The answer will come out slightly larger than 17 do to rounding.

Sample Problems 1) e3x = 23 Solution: Use natural log this time. Ln e3x = Ln 23 3x Ln e = Ln 23 3x = Ln 23 ( Ln e = 1) x = (Ln 23)/3 x = 1.05 2) How long does it take $100 to become $1000 if invested at 10% compounded quarterly? Solution: Ao = 100, A(t) = 1000, r = .1, n = 4 1000 = 100(1 + .1/4)4t 10 = 1.0254t Use the change of base formula Log 10 = Log 1.0254t 1 = 4t Log 1.025 (Log 10 = 1) 1/(4Log 1.025) = t t = 23.31 It will take 23.3 years to have $1000 from the $100 investment.