1 An Exponential Function with base b is a function of the form: fx) = b x, where b > 0, b is a real number. We know the meaning of b r if r is a rational number. What if r is irrational? What we do is we approximate the value of b r by using rational approximate for r. For example, to approximate 5 π, we may approximate it as 5 3., 5 3., 5 3.5, In advance mathematics one can define the value of 5 π to be the limit of these approximations. For now, just realize that such approximation can be used to approximate the value of b r for any positive number b and any real number r. Since b x can be defined for all real numbers r, the domain of an exponential function is all real numbers. The range of b x is all real numbers greater than 0. Note the difference between an exponential function and a power function. x is a power function. In this function, the exponent is the constant. x is an exponential function. In this function, the base is the constant but the exponent is the variable input). An exponential function is always positive. And if in addition 0 < b <, f is a decreasing function. That is fx) decreases as x increases. fx) = ) x
2 If b >, then f is an increasing function. I.e. fx) increases in value as x increases. fx) = x Note from the property of exponents that x = x = ) x In other words, the exponential functions fx) = b x and gx) = b) x are reflections of each other with respect to the y axis. Simple Interest When you borrow money from a bank, you have to pay interest based on a particular interest rate. The amount you borrow is called the principal. If the interest is being calculated based on the principal only, we call this simple interest. The following formula gives the relationship of these quantities: Let P = principal, the amount borrowed. Let r = the simple annual interest rate. r the rate at which interest will be paid if calculated after year. r is usually given as a percentage, but when used in calculation, r will be written as a decimal. Let t = the amount of time, in terms of years, passed when interest is being
3 calculated. Let I = interest. Then we have: I = P rt. Example: You borrowed $3500 for 5 months, for an interest rate of 6%. How much interest will you have to pay? Ans: P = $ 3500, r = 6% = t = 5 = 5 =.5 I = = 6.5. You will have to pay $6.5 in interest. If you borrowed P in principal, on an annual simple interest rate of r, then after t years, how much total, A, will you have to pay? In interest problems like this, the total amount at the end of the period, A, is the future value, and P the present value. According to the simple interest formula, you will have to pay I = P rt in interest. Of course, you will still have to pay off the original amouunt, the principle P, so the total A you will have to pay is: A = P + P rt = P + rt) where A is the total you have to pay or the future value). Exmaple: You deposited $8000 into an account that pays 3.5% annual simple interest. After 8 months, how much should you have in your account? Ans: P = $8000, r = 3.5% = 0.035, t = 8 months =.5 years. A = ) = 80 You will have total of $80 in your account. $0 is the interest. Compound Interest: If you deposit money into a bank and earns an interest), after your interest is being paid, if you do not take out the money, the next time you earn interest, you will earn interest on the original amount, P, and the interest from the first period, too. When you earn interest on the interest that you earned from a previous period, you are being paid compound interest. In calculating compound interest, since you will receive interest on interest that is being paid from the pervious period, how often you are being paid interest makes a difference. Example: You invested $000 into a compound interest account that pays % rate, compounded quarterly. How much will you have at the end of two years? Ans: By interest being compounded quarterly, this means that the interest you
4 earn is being calculated and paid) every 3 months 3 months is of a year). Using the same terminology, if we say interest is compounded monthly, that means interest will be calculated and paid each month; compounded daily means interest will be calculated and paid each day...etc. At the end of the first three months, you will be paid interest. The time elasped since you last received interest is 3 months, or t = year. A = P + rt) = )) = 0 At the end of 6 months, you will )) be paid interest again, this time, you will already have A = in your account. This is your new principal. We use this number in the formula to calculate the new balance: )) [ )] [ A = = )] 3.60 At the end of 9 months, you will be paid interest again, this time using yet a new principal: [ )] [ )] [ )] 3 A = = Looking at the pattern, we see that for each additional compounding period, the power of the expression + rt), increased by one. At the end of two years, interest would be have been compounded 8 times, the total amount we will have is: [ )] 8 A = We have the following formula: If a principal amount P is earning an annual interest rate of r, and interest is compounded m times a year, then the total at the end of t years will be given by: A = P + r ) mt m Example: $0, 000 is being invested into an account that pays r = 8% nominal interest. How much will be in the account after 5 years if interest is compounded a. annualy? b. semi-annually? c. quarterly? d. monthly? e. daily? Ans: a. If interest is compounded annually, then r m = 0.08 = In 5 years interest will be compounded 5 times, so mt = 5, using the formula we have:
5 A = 0, ) b. If interest is compounded semi-annually, then r m = r = 0.08 = 0.0. In 5 years interest will be compounded 0 times two times a year times 5 years), so mt = 0, using the formula again gives: A = 0, ) c. If interest is compounded quarterly, then r m = r = 0.08 = 0.0. In 5 years interest will be compounded 0 times four times a year times 5 years), so mt = 0, we have: A = 0, ) d. If interest is compounded monthly, then r m = r = In 5 years interest will be compounded 60 times twelve times a year times 5 years), so mt = 60, we have: A = 0, ) e. If interest is compounded daily, then r m = r 365 = In years interest will be compounded 85 times 365 times a year times 5 years), so mt = 85, we have: A = 0, ) We see that the more often interest is compounded, the more money we get at the end. What if we compound interest more often, say compounded every hour, or every minute, or every second? Will we get infinite amount of money, or will the amount we get approach a particular number? Looking at the formula A = P + r ) mt m If we let x = m r, then m = xr, r m = m/r =, then the formula becomes: x A = P + x) xrt [ = P + ) x ] rt x As we compound more often, m keeps on increasing while r remains the same, so x also keeps on increasing.
6 If we look at the expression, + x) x, and try some large values of x to see what is the value of the expression, we give a table of the value of the expression for some large values of x: x 0 00, 000 0, , 000, 000, ) x x + 0) ) ) ) ) ) It appears that the value of the expression + x) x approaches a particular number as x becomes incresingly large. This is indeed the case. It can be shown that the limit of this expression, when the value of x becomes arbitrarily large we say that the value of x increases without bound, or that x approaches infinity), the value of the expression: + x) x approaches a particular irrational) real number. This number is very important in mathematics and science, and is represented by the letter e and has an approximate value of e The exponential function using e as its base is called the natural expenential function. In other words, the natural exponential function is the function fx) = e x Using e to replace the above expression, we get the formula: A = P e rt This is the formula used when interest is compounded continuously. In reality, we can not really compound interest continuously. The formula gives an upperbound on how much interest can be earned regardless of how often interest is being compounded.
7 Example: $0000 is being invested into an account that pays r = 8% nominal interest. How much will be in the account after 5 years if interest is compounded continuously? Ans: P = 0000, r = 0.08, t = 5 We use the formula for continuous compound interest: A = 0000e 0.085) In other words, given the above condition, you can never earn more than $ no matter how often interest is compounded. Logarithmic Functions: An exponential function is a one-to-one function. It has in inverse. However, the expression for the inverse of an exponential function cannot be solved by any algebraic means, therefore we do not have an algebraic expression for it. We just define such a function and study its property knowing that it is the inverse of b x Definition: gx) = log b x read log base b of x ) is the inverse function of fx) = b x log b is the name of the function. x is the argument input) to log b, and the value output) is log b x. Since the range of an exponential function is all positive real numbers, the domain of a log function is all positive real numbers. Because of the fact that log b x is the inverse of b x, we have this by definition: log b b x ) = x for all x b log b x = x for all x > 0 The following two equations are equivalent: y = b x log b y = x The logarithm with base e, which is the inverse function of e x, is given a special name, the natural logarithm, and written ln. I.e. log e x = ln x
9 Example: Evaluate: log 6 Ans: Set log 6 = y, this is the same as asking: y = 6, what is y? From observation we see that y =, so log 6 = Example: log 5 5 =? Ans: Since 5 3 = 5, log 5 5 = 3 Example: log x = 5, what is x? Ans: 5 = x, so x = 3 log x 00 =, what is x? This is the same as saying, x = 00, so x = 0
10 Because of the fact that logs are inverse functions of the exponential functions, they have many properties that are similar to that of the exponential functions, and can be easily proved using the definition: Properties of Logarithm For any real number r, any base b > 0, a > 0, any x > 0, y > 0, we have: log b xy) = log b x + log b y ) x log b = log y b x log b y log b x r = r log b x Example: ln0) = ln 5) = ln) + ln5) Example: log 3 3xy) = log 3 3) + log 3 x) + log 3 y) = + log 3 x) + log 3 y) ) Example: ln = ln) ln5) 5 Example: log x 5 ) = 5 log x) Example: log b x 3 y ) = log b )+log b x 3 )+log b y ) = log b )+3 log b x)+ log b y) Change of Base Formula: log a x = log b x log b a Example: log 5) = log 0 5 log 0 Example: log 5 3) = ln3) ln5) The change of base formula says that we can easily change from log of one base to another, so the choice of log of which base to use usually for convenience only. We like to use natural log for most of our studies because that is the most conveninent one in math and science. Also make sure not to misuse the properties of logs, for example, the property of logarithm does not give us this: logx + y) = log x + log y Here s some example of how to incorrectly use the property of log: log b x + 5) = log b x) + log b 5) log b x 3) = log b x ) log b 3) log b x) 3 = 3 log b x)
11 To solve an equation involving exponential function, one would need to use logarithm, and to solve an equation involving logarithm, one uses exponential functions. E.g. Solve 3 x = 0 Ans: We take the log of both side log 3 3 x = log 3 0 Since log and exponential functions are inverse, the left hand side is just x, we have x = log 3 0 One may approach the same problem by taking the ln of both sides and use properties of logarithm: 3 x = 0 ln3 x ) = ln 0 x ln 3 = ln 0 ln 0 x = ln 3
12 E.g. Solve log x + ) = 3 Ans: We turn this into an equation involving exponential function: log x + ) = 3 3 = x + 6 = x + x = 6 E.g. Solve log x + ) + log x + ) = Ans: We use property of log to combine the left hand side: log x + ) + log x + ) = log x + )x + )) = log x + 5x + ) = = x + 5x + = x + 5x + x + 5x = 0 x = 0 or x = 5 Notice that x = 5 is an extraneous solution, so our only solution is x = 0
13 Application to compound interest: Example: You invested $8000 into an account that pays 5% nominal rate. How long before you will have $000 if interest is compounded: a. quarterly? b. monthly? Ans: We are solving for time t in the formula: A = P + r ) mt m a. We are told that the nominal rate, r, is r = 5% = If compounded quarterly, then m =. You started with P = You want to end up with A = 000, so we have the equation: 000 = ) t This is an exponential equation which can be solved using logarithm: = ) t 3 = ) t ) [ 3 ln = ln ) ] t ln ) 3 t = = t ln ln ) 3 ln ) ) ) ln 3 t = ln ) It takes about 8 years and two months for the $8000 to become $000 Example: You have some money to invest into an account that pays 6% nominal interest rate compounded monthly. How long would it take you to double your investment? Ans: It does not really matter the amount you started with. The time it takes you to double your investment is dependent only on the interest rate and the ways interest is compounded. If you started with initial investment P, you want to end up with future value of A = P. r = 0.06, m =, we want to solve for t:
14 P = P ) t = ) t [ ln = ln ) ] t ln = t ln ) ln t = ln ) It takes about and a half years to double your investment. Example: You want to deposit some money into an account that pays 3% nominal interest rate compounded monthly. If you want to be able to withdraw $50000 to buy a car in 5 years, how much must you deposit now? Ans: We are trying to find P, given A = 50000, r = 0.05, t = 5, m = : = P ) 5) P = ) ) Exponential Growth and Decay: There are many situations where we may use an exponential function as a model. For example, the growth of a population may be modeled as an exponential growth, the amound of radioactive material remain after a certain period of time may be modeled as an exponential decay. An exponential growth or decay is defined by the equation: y = y 0 e kt In this equation, t is time, y 0 and k are constants. y 0 is the initial value of the population, or amount of radioactive material present). If k > 0, the equation represents an exponential growth, and k is the growth constant. If k < 0, then the equation represents an exponential decay, and k is the decay constant. Example: A biologist starts with 000 bacteria in a petri dish. After three hours, the population has increased to, 600. Assuming that the population of the bacteria increases exponentially, provide an equation that express the number of
15 bateria as a function of time. Ans: Since we assume that the rate of growth of the bacteria is exponential, we use the equation y = y 0 e kt, where y is the number of bacteria at time t. At time t = 0, the initial population size is y = y 0 e k0) =, 000 y 0 =, 000 This is the initial population, y 0. After t = hours, the population y has grown to, 600, we then set up the equation:, 600 = 000e k) e k = ek = 3 0 k = ln.3) ln.3) k = 0.3 The equation we can use to model this population growth is: y = 000e 0.3t How long does it take this bacteria culture to double in size from the original population size? Ans: To double in size, the bateria will have to increase to 000, we set y = 000 to solve the equation: 000 = 000e 0.3t e 0.3t = 0.3t = ln) t = ln) It takes approximately 5.8 hours for the population to double in size. This number, the amount of time it takes for a population to double in size in an exponential growth, is called the doubling time. This time is independent of the amount of the initial population. If you started with 50 bacteria, it will take 5.8 hours to double to 00 bacteria. If you started with 5000 bacteria, it will take the same 5.8 hours to double to 0,000 bacteria. Example: 50 grams of certain type of radioactive material is placed inside a detector. After 0 days, it is observed that only 7 grams of the radioactive material is left. Assuming that the radioactive material decays exponentially, find an equation that describes the amount of material left as a function of time t in days). Ans: We let y = the amount of radioactive material present at any time t. At t = 0 we have 50 grams, so we have: y = y 0 e k0) = 50 y 0 = 50 The initial amount of radioactive material is 50 grams, as we know. After t = 0 days, y = 7 grams left, we have:
16 7 = 50e k0) e 0k = k = ln ) 7 k = ln ) The function that can be used to model this radioactive decay is: y = 50e t How much time t does it take this radioactive material to decrease to half of what we started? Ans: We started with 50 grams, to decrease to half is to end up with only 5 grams, we set up the equation: 5 = 50e t e t = t = ln0.5) t = ln0.5) It takes approximately 8 days for the radioactive material to decay to half of what it has originally. This number, the amount of time it takes a radioactive material to decay to half of its original amount, is called the half-life of a radioactive material, and is independent of the amount of radioactive material we started with. Example: After the Chernobyl nuclear meltdown, it is estimated that the concentration of cesium-37, a radioactive material with a half-life of 33 years, was 0 times the normal level. How long will it take the concentration level of cesium-37 to return to normal level. Ans: We first come up with the equation that models the decay of cesium-37. Since its half-life is 33 years, it means it takes t = 33 years for the amount to reduce to half of the original content, so we set up the equation: y 0 = y 0 e k33) = e33k 33k = ln0.5) k = ln0.5) The equation we can use to mode the decay of cesium-37 is: y = y 0 e 0.0t, where t is expressed in years, and y 0 is however much of the material we started with. Since the amount we started with, y 0, is 0 times the normal level, we need the amount to drop down to only one-tenth of y 0 to be normal, so we set up this equation to solve for t: 0 y 0 = y 0 e 0.0t 0 = e 0.0t 0.0t = ln0.) t = ln0.) It will take about 0 years for the radiation to drop back down to normal level.