Chapter 2 Exponentials and Logarithms

Transcription

1 Chapter Eponentials and Logarithms The eponential function is one of the most important functions in the field of mathematics. It is widely used in a variety of applications such as compounded interest, population growth, and carbon dating. This chapter is a brief introduction to the eponential function and its inverse function, the logarithmic function. We will look at various properties of these two functions, learn to solve equations involving them, and then finally use them to model real world situations..1 Eponential Functions An eponential function is a function of the form f( ) = b where b > 0 and b 1. Some properties of eponential functions are summarized in Table.1. Figure.1 shows the two different types of curves that are generated by an eponential function. Table.1 Properties of Eponential Functions Let f( ) = b where b > 0 and b If b > 1 then f( ) is an eponential growth function.. If 0< b < 1then f( ) is an eponential decay function. 3. The domain of f( ) is (, ). 4. The range of f( ) is (0, ). 5. The y-intercept is (0, 1). Figure.1 Graphs of eponential growth and decay functions.

2 EXAMPLE.1 Determine whether each of the following are eponential growth or decay functions. Then graph each function and verify the domain, range and y-intercept of the function are as stated in Table.1. a. 3 f( ) = b. 1 f( ) = 4 Solution a. This is an eponential growth function because b = 1.5 and that is greater than 1. Figure. Graph of Eample.1 a. By inspection of the graph in Figure., the domain is (, ) and the range is ( 0, ). f() 5 f() b. This is an eponential decay function because b = 0.5 and that is less than 1. Figure.3, ( ) and the range is ( ) Graph of Eample.1 b. By inspection of the graph in Figure.3, the domain is 0,. f() 18 f() The function f( ) = is an eponential function because the independent variable,, is part of the eponent. If we were given a table of values, or a data set, how could we determine if it modeled an eponential function? One method is to calculate the ratio of successive f( ) values when the values ACOW Group Page //003

3 are equally spaced. If the ratios are all the same (or close to the same) the table of values would model an eponential function. The value of the successive ratios is the base of the eponential function. EXAMPLE. Determine if the table of values given below represents an eponential function. If so, find the base of the function f() Solution Compute the successive ratios, as shown below. f(3) f(4) f(5) f(6) = =.5, =.5, =.5, = =.5 f() 6.5 f(3) f(4) f(5) Since all ratios are approimately equal to.5 we conclude that the table of values does represent an eponential growth function with a base of.5. Eponential functions have a variety of applications in the business world and it is often necessary to solve these equations for different variables. In order to solve these equations we must know the laws and definitions associated with eponential functions that are listed in Table.. Table. Properties and Laws of eponents ***a,b 0, positive?? Let a and b be positive numbers and let m and n be real numbers. Then, 1. 0 a = 1. a m 1 =, where a 0 m a m/ n n m n 3. a a ( a) m = = 4. m n m n a a = a + 5. a b m n m n = a 6. ( ) m m m ab = ab 7. ( a ) m n a mn = 8. m a a = b b m m Eample.3 se the laws and definitions of eponents from Table. to solve the following equations for. ACOW Group Page 3 //003

4 a. d = 5 b. 9 = = c. e = = 0 7 Solution a. The epressions on both sides of the equation have a base of 5. Since these epressions and the bases are equal we can conclude that their eponents must also be equal. So we set up an equation stating this fact and solve it for. + 3= 3 = b. Here the bases of each term are not equal to each other. Thus, our first step is to search for a common base. We know that 64= 8 so we can rewrite 64 as 8. se the laws from Table. to find ( ) 5 Now solve for by setting the eponents equal, 8 = 8 = 8 = 8 ( 5) 4 10 = = = c. Since the bases on each term are different we must rewrite 7 as eponents, 3 and apply laws and definitions of ( ) 3 = 3 3 = 3 3 = 0 3 Now solve for, 3 = = 6 7 = 1 1 =. 7 d. This problem differs from the previous three because there is an in each term that accompanies the eponential function. Our strategy in solving this problem will still involve getting the same bases, but now we must factor the epression and solve using zero product property1. Begin with the left side of the equation and writing 9 as 3, 1 That is, in a series of products, if any one of the terms are zero, then the entire product is zero. ACOW Group Page 4 //003

5 Now work with the right side of the equation, Net set them equal to each other, ( ) 9 = 3 = 3 3 = = = 3 3 ( 3 ) 3 = 3 Rearrange to have all the variables on the left and factor. 3 3 = 0 Now by applying the zero product property we obtain: = 0, ( ) 3 1 = =, and 1 = 0. Since an eponential function can never equal zero, 3 0, the only solutions are = 0,1. e. Since each eponential function is accompanied by an, we will use the strategy from the previous problem. That is, we will get the bases the same and then factor. Let u = 3 for ease in factoring and find ( ) = 7 ( ) u u ( u )( u ) = 1 + 7= 9 3 = 0 The first term has the solution u = 9 or 3 = 9 = 3, so =. The second term has the solution u = 3 or 1 3 = 3= 3, so = 1. Thus, the two solutions to the given equation are = 1,. A check of all these answers through substitution of the solution value into the original equation is encouraged.. Applications of Eponential Functions One of the most common types of eponential functions in the banking industry is compound interest, which can be easily derived from the simple interest formula. ACOW Group Page 5 //003

6 Simple Interest: If P dollars is deposited into an account that earns an annual interest of r% (epressed as a decimal), then the amount of interest accumulated, I, after t years is given by I = Prt Suppose you accumulated $1,000 in cash from your high school graduation. If you deposit this money into an account that earns 5% simple interest, then at the end of 4 years you would have earned I = (1,000)(0.05)(4) = $00. So you would have $1,000 + $00 = $1,00 in the account at the end of 4 years. In general, the amount in an account that earns simple interest for t years is A= P+ I A= P+ Prt ( 1 ) A= P + rt Another type of interest commonly used in the banking industry is compound interest. This type of interest is often computed more than once a year and the account earns interest on the interest computed the previous compounding period. This is what makes compound interest more appealing to investors than simple interest. For eample, if you take the $1,000 accumulated from your high school graduation and deposit it into an account that earns 5% compounded quarterly, we can compute the amount in the account at the end of 1 year. To solve this we use A= P(1 + rt) with t = 1 4 because the interest is figured quarterly ( 1 4 of a year). Quarter A = P(1 + rt) Amount at the end of the Quarter A = 1, ( ) = 1,000(1.015) $1, st ( ) 1 4 nd A = 1,01.5( 1.015) = 1,000(1.015)(1.015) = 1,000(1.015) $1, rd 3 A = 1,05.16(1.015) = 1,000(1.015) (1.015) = 1,000(1.015) $1, th 3 4 A = 1,037.97(1.015) = 1,000(1.015) (1.015) = 1,000(1.015) $1, If we look at the epression for A, near the end of the formula we can see a pattern that is developing for compound interest. The eponent on is the same as the compounding period. Thus for the 16th 16 compounding period (4 years) the amount in the account would be 1,000(1.015) = 1, If we compare this number to the amount in the account that earned simple interest for 4 years, we see that the compounded account earned $19.89 more. This difference is greater for larger initial deposits and when the money is left in the accounts for longer time periods. ACOW Group Page 6 //003

7 It is without a doubt that the computation above was long and tedious. We would not want to use this method of computing for larger numbers, therefore we generalize the computation and derive a formula for compound interest. Compound Interest Formula: If P dollars is deposited into an account earning r% annual interest (epressed as a decimal) and is compounded n times a year, then the amount in the account, A, after t years is given by r A= P 1+ n nt Interest can be compounded a number of different ways. Below are the n values for the different compounding periods. Table.3 Compounding Periods Frequency, n Annually 1 Semi-annually Quarterly 4 Monthly 1 Weekly 5 Daily 365 Eample.4 Find the amount in an account after 10 years if $,500 is compounded monthly at 8%. Solution Here P =,500, t = 10, n= 1, r= Thus, the amount, A, would be (1)(10) 0.08 A =, $5, Or, we can us the compound interest calculator applet to find the same result as shown in Figure.4. Figure.3 Compound Interest Applet for Eample.4 ACOW Group Page 7 //003

8 Eample.5 How much money should be deposited into an account that earns 6.5% compounded semi-annually if, after 7 years, the account must be worth $10,000? Solution Here A= 10,000, r = 0.065, n =, t = 7. Plugging in these values into the formula, we have ()(7) ,000= P 1 + = P(1.035) 14. Now solve for P, 10,000 P = $6, (1.035) Figure.4 Compound Interest Applet for Eample.5 Suppose you invest $1 in an account that earns 100% interest for 1 year. How would the amount of money at the end of the year change if the interest was compounded more and more often? The compound interest formula with these values would be 1 A = 11 + n In Table.4 we see how more frequent compounding (larger n) gives you a larger and larger A, up to a point. n Table.4 Amount of Money vs. Compounding Frequency n ,000 10, ,000 1,000,000 A We see that as n grows larger, A becomes approimately.718. This number (called e) is frequently used in the field of mathematics and is used in the formula for continuously compounded interest. ACOW Group Page 8 //003

9 Continuously Compounded Interest: If P dollars is deposited into an account earning r% annual interest (epressed as a decimal) and is compounded continuously, then the amount in the account, A, after t years is A= Pe rt Eample.6 Find the amount in an account after 10 years if $,500 is deposited into an account that earns 8% compounded continuously. Solution Here P =,500, r = 0.08, and t = 10. When we substitute into the continuous compound interest formula we get: A= (0.08)(10),500e 5, In the module there is a continuous compounding calculator that will give the same result, shown below, Figure.5 Continuous Compounding Applet for Eample.6 So there will be approimately $5, in the account after 10 years. Comparing this answer to the one obtained in Eample.4 shows that continuously compounded interest earns $14.75 more than compounding quarterly..3 Logarithmic Functions We can solve eponential equations when the bases are the same, but how would we solve an equation like = 10? To solve this we need to use the inverse of an eponential function, the logarithmic function. Definition of Logarithm: For all b > 0, b 1, and y > 0, y = b if and only if log b y = Note: log b y = is read log to the base b of y equals. Two commonly used logarithmic functions are the common logarithm ( log 10 ) and the natural logarithm ( log e ). Typically we write log for log 10 and ln for log e. Eample.7 Rewrite k 6 = 36 and e = 4 as logarithmic equations. ACOW Group Page 9 //003

10 Solution 6 = 36 is in the eponential form, y = b, with y = 36, b = 6 and =. Match this to k the logarithmic form, log b y =, to find log6 36 =. Matching terms for the eponential form e = 4 gives y = 4, b= e and = k. Put these values into the logarithmic form gives log 4 = ln4 = k. Eample.8 Rewrite logb 9 = and lnk = as eponential equations. Solution logb 9 = is in the logarithmic form, log b y =, with b= b, y = 9 and =. Match this to the eponential form and find b = 9. Rewrite lnk = as loge k = and find that b= e, y = k and =. Put this into the eponential form and find e = k. e Figure.6 below shows the graph of y = log and its corresponding table of values. Notice that the table of values is constructed by using the eponential form of y = log. Figure.6 Graph of y = log. 3 = y y Table.5 Logarithmic Functions Let f( ) = log b, where b > 0 and b 1. Then, 1. The domain of f( ) is (0, ).. The range of f( ) is (, ). 3. The -intercept is (1, 0). Eample.9 Graph f( ) = ln and verify its domain and range. Solution The graph is shown in Figure.7. The domain is (0, ) and the range is (, ). ACOW Group Page 10 //003

11 Figure.7 Graph of y = ln. 3 y To solve equations involving logarithms, we will need to know the Laws of Logarithms listed in Table.6. Table.6 Properties of Logarithms If M, N, and b are all positive numbers where b 1 and c is any real number, then: 1. log MN = log M + log N 5. log 1 0 b b b M. logb logb M logb N N = 6. log b = c b b = c c 3. log M = clog M 7. b 4. logb b = 1 b log b M b = M Eample.10 Given logb 4= and logb 3= , use the definition and laws of logarithms to evaluate the following epressions: a. log 1 b b. log 9 b c. 16 logb 3 Solution log 1 = log 43 = log 4+ log 3= = a. ( ) b b b b b. c. logb9 = logb3 = logb3= (0.5645) = logb = logb = logb 4 logb3 = logb4 logb3 = (0.714) = Eample.11 se the definition and laws of logarithms to solve the following equations for. a. ln( + ) = 4 b. log(log ) = 1 ACOW Group Page 11 //003

12 Solution a. First rewrite the logarithmic equation as an eponential equation then solve for : Substitute this into the original equation to check: e e 4 4 = + =. 4 4 ln( ) ln( e ) ln( e ) 4lne 4 + = + = = =. b. Since this is an embedded logarithm we must rewrite as an eponential equation twice, and then solve for : Check this solution in the original equation, 1 10 = log = = log log = log( log 10 ) = log( 10log[ 10] ) = log( 10 1) = 1 Eample.1 Solve the following eponential equations for. e = 4 b. e = 9 c. 5 a. ( ) log(+ 1) 10 = ln Solution a. Divide both sides of the equation by, 5 e = Since the variable is in the eponent, rewrite as a logarithmic equation and then solve for : 5= ln = ln+ 5 ln+ 5 = b. Rewrite as a logarithmic equation, ln9= Solve for by squaring both sides, ACOW Group Page 1 //003

13 = ( ln9) c. se logarithmic properties from Table.6 on the left side of the equation and then solve for, + 1= ln = ln 1 ln 1 = Modeling As we have seen in earlier sections, we can model real world data with eponential and logarithmic functions. Eample.13 The data below represents the amount of goods (in millions of dollars) the nited States imported from Hungary from 1993 to 000. Find the best fitting eponential model for this data set and use the model to predict the amount of imported goods in the year 010. Year Amount of goods imported (in millions of dollars) Source: Solution We will make the independent variable,, the import year where the value of represents the number of years after 1990 (i.e. 4 represents 1994). The amount of goods imported will be the dependent variable, y. sing the regression calculator available for these modules we input the data, click on plot data, and then choose the eponential regression button. In doing so we obtain y = ( 1.33) with R = as shown in Figure.8. ACOW Group Page 13 //003

14 Figure.8 Modeling Applet for Eample.13 To predict the amount of imported goods in the year 010 we input = 0 into the regression calculator and find y = 1, million dollars. Eample.14 The table below represents the amount of goods (in millions of dollars) the nited States imported from Indonesia from 1993 to 000. Find the best fitting logarithmic model for this data set and use the model to predict the amount of imported goods in the year 005. Year Amount of goods imported (in millions of dollars) Source: Solution sing the same set up as in Eample.13 we find the logarithmic regression model to be y = (ln ) with R = , as shown below. ACOW Group Page 14 //003

15 Figure.9 Modeling Applet for Eample.14 To predict the amount of imported goods in the year 005 we input = 15 into the regression calculator and find y = 11, million dollars. Eample.15 The table below represents the average price (in dollars) ranchers in the nited States received per head of cattle from 1996 to 001. Find the best fitting model for this data set and use the model to predict the average price per head in Year Average price paid per head of cattle (in dollars) Source: hog/cattle/sheep Solution Let be the number of years after 1990 and y be the dollar amount paid per head of cattle. sing the regression calculator we get the following equations and R values. ACOW Group Page 15 //003

16 Regression Model Equation R Eponential y = (1.08) Logarithmic y = (ln ) Linear y = Quadratic Cubic y = y 0.61 Power y ( ) 3 = = By comparing the R values we see that the eponential, quadratic, and cubic models have the highest value. Since the cubic value is not significantly higher than the quadratic we can eliminate the cubic model because the difference in the R value does not justify the more complicated model. Thus, we must decide between the eponential and quadratic model by studying their overall shape as seen in Figure.10. Figure.10 Modeling Applet for Eample.15 Both graphs etend to positive infinity with the same general shape, but from 0 to 5 the graphs have slightly different characteristics. It appears that the eponential model has the same general shape as the data from [0, 5] whereas the quadratic model tends to deviate from the data s shape. It is for this reason that the eponential model would best fit this data set. Now using this model to predict the price per head of cattle in 1993 we need to find y when = 3. sing the regression calculator we find y $399. ACOW Group Page 16 //003

17 Sample Quiz Question.1 Determine whether the following table represents an eponential function by calculating successive f() ratios. If it represents an eponential function, give the value of the base and state whether it is a growth or decay function f() Question. Solve 4 15 = 36 for. Question.3 How much money should Dave deposit into an account that earns 5.5% annual interest compounded monthly if he wants to have $30,000 in the account 18 years from now? Question.4 After some research, you found two investment options for your $18,000 graduation gift. Bank One offers 8% compounded semiannually and Bank One-Plus offers 7.5% compounded monthly. a. Which bank would be the better investment if you left the money in the account for 1 year? b. How many years would it take for Bank One to become the better investment? Question.5 Find the amount in an account after 7 years if $3,000 is compounded continuously at 5.5%. Question.6 Solve ln(ln ) = 0 for. Question.7 The number of DVD players supplied to an electronics store is given by the equation 0.004p S( p) = 150e where p is the price in dollars. If the store is supplied 15 DVD players, at what price should they sell them? Question.8 Given logb 5=.319, logb 3= , and logb 7 =.8074, find 1 logb 5. Question.9 The following table represents the assets (in billions of dollars) of FDIC insured Commercial Banks (y) in the nited States from 1987 to 001 (). Find the best fitting eponential and logarithmic model, along with their R^ values, and then determine which of the two would be the better model y,913 3,056 3,07 3,361 3,377 3,438 3,569 3, y 4,171 4,397 4,771 5,181 5,469 5,983 6,360 Source: ACOW Group Page 17 //003

18 Question.10 The following table represents the assets (in billions of dollars) of FDIC insured Savings Institutions (y) in the nited States from 1987 to 001 (). Find the best fitting model for this data and state why it is the best model y 1,441 1,556 1,51 1,317 1,161 1,078 1, y 1,017 1,03 1,09 1,045 1,16 1,179 1,75 Source: ACOW Group Page 18 //003