Exponential Functions, Logarithms, and e

Transcription

1 chapter 3 Starry Night, painted by Vincent Van Gogh in 889. The brightness of a star as seen from Earth is measured using a logarithmic scale. Eponential Functions, Logarithms, and e This chapter focuses on eponents and logarithms, along with applications of these crucial concepts. Each positive number b leads to an eponential function b. The inverse of this function is the logarithm base b. Thus log b y = means b = y. We will see that the important algebraic properties of logarithms follow directly from the algebraic properties of eponents. We will use eponents and logarithms to model radioactive decay, earthquake intensity, sound intensity, and star brightness. We will also see how functions with eponential growth describe population growth and compound interest. Our approach to the magical number e and the natural logarithm lead to several important approimations that show the special properties of e. These approimations demonstrate why the natural logarithm deserves its name. This chapter concludes by relooking at eponential growth through the lens of our new knowledge of e. We will see how e is used to model continuously compounded interest and continuous growth rates. 225

2 226 chapter 3 Eponential Functions, Logarithms, and e 3. Logarithms as Inverses of Eponential Functions learning objectives By the end of this section you should be able to use eponential functions; evaluate logarithms in simple cases; use logarithms to find inverses of functions involving b ; compute the number of digits in a positive integer from its common logarithm. Eponential Functions We defined the meaning of a rational eponent in Section 2.3, but an epression such as 7 2 has not yet been defined. Nevertheless, the following eample should make sense to you as the only reasonable (rational?) way to think of an irrational eponent. eample Find an approimation of 7 2. Because 2 is approimately.44, we epect that 7 2 should be approimately 7.44 (which has been defined, because.44 is a rational number). A calculator shows that If we use a better approimation of 2, then we should get a better approimation of 7 2. For eample, is a better rational approimation of 2 than.44. A calculator shows that , which turns out to be correct for the first five digits after the decimal point in the decimal epansion of 7 2. We could continue this process by taking rational approimations as close as we wish to 2, thus getting approimations as accurate as we wish to 7 2. We want to define functions f such as f () = 2 whose domain is the set of all real numbers. Thus we must define the meaning of irrational eponents. The eample above gives the idea for defining the meaning of an irrational eponent: Irrational eponent Suppose b > 0 and is an irrational number. Then b is the number that is approimated by numbers of the form b r as r takes on rational values that approimate. The definition of b above does not have the level of rigor epected of a mathematical definition, but the idea should be clear from the eample above. A rigorous approach to this question would take us beyond material appropriate for a precalculus course. Thus we will rely on our intuitive sense of the loose definition given above.

3 section 3. Logarithms as Inverses of Eponential Functions 227 The bo below summarizes the key algebraic properties of eponents. These are the same properties we saw earlier, but now we have etended the meaning of eponents to a larger class of numbers. 32 Algebraic properties of eponents Suppose a and b are positive numbers and and y are real numbers. Then b b y = b +y, (b ) y = b y, a b = (ab), b 0 =, b = b, a a y = a y, a ( a ). b = b Now that we have defined b for every positive number b and every real number, we can define a function f by f () = b. These functions (one for each number b) are sufficiently important to have their own name: 6 Eponential function Suppose b is a positive number, with b. Then the eponential function with base b is the function f defined by f () = b. 8 For eample, taking b = 2, we have the eponential function f with base 2 defined by f () = 2. The domain of this function is the set of real numbers, and the range is the set of positive numbers. The potential base b = is ecluded from the definition of an eponential function because we do not want to make eceptions for this base. For eample, it is convenient (and true) to say that the range of every eponential function is the set of positive numbers. But the function f defined by f () = has the property that f () = for every real number ; thus the range of this function is the set {} rather than the set of positive numbers. To eclude this kind of eception, we do not call this function f an eponential function. Be careful to distinguish the function 2 from the function 2. The graphs of these functions have different shapes. The function g defined by g() = 2 is not an eponential function. For an eponential function such as the function f defined by f () = 2, the variable appears in the eponent The graph of the eponential function 2 on the interval [ 5, 5]. Here the same scale is used on both aes to emphasize the rapid growth of this function The graph of 2 on the interval [ 3, 3]. Unlike the graph of 2, the graph of 2 is symmetric about the vertical ais.

4 228 chapter 3 Eponential Functions, Logarithms, and e Each time increases by, the value of 2 doubles; this happens because 2 + = 2 2. Logarithms Base 2 Consider the eponential function f defined by f () = 2. The table here gives the value of 2 for some choices of. We now define a new function, called the logarithm base 2, that is the inverse of the eponential function 2. Logarithm base 2 Suppose y is positive number. The logarithm base 2 of y, denoted log 2 y, is defined to be the number such that 2 = y. Short version: log 2 y = means 2 = y. For eample, log 2 8 = 3 because 2 3 = 8. Similarly, log = 32. = 5 because eample 2 Find a number t such that 2 /(t 8) = 5. y log 2 y y The graph of log 2 (red) on the interval [, 8] is 8 obtained by flipping the graph of 2 (blue) on the interval [ 3, 3] across the line y =. The equation above is equivalent to the equation log 2 5 = t 8. Solving this equation for t gives t = log The definition of log 2 y as the number such that 2 = y means that if f is the function defined by f () = 2, then the inverse function of f is given by the formula f (y) = log 2 y. Thus the table here giving some values of log 2 y is obtained by interchanging the two columns of the earlier table giving the values of 2, as always happens with a function and its inverse. Epressions such as log 2 0 and log 2 ( ) make no sense because there does not eist a number such that 2 = 0, nor does there eist a number such that 2 =. The figure here shows part of the graph of log 2. Because the function log 2 is the inverse of the function 2, flipping the graph of 2 across the line y = gives the graph of log 2. Note that log 2 is a function; thus log 2 (y) might be a better notation than log 2 y. In an epression such as log 2 5 log 2 5, we cannot cancel log 2 in the numerator and denominator, just as we cannot cancel a function f in the numerator and denominator of f (5) f (5). Similarly, the epression above is not equal to log 2 3, just as f (5) f (5) is usually not equal to f (3).

5 section 3. Logarithms as Inverses of Eponential Functions 229 Logarithms with Any Base We now take up the topic of defining logarithms with bases other than 2. No new ideas are needed for this more general situation we simply replace 2 by a positive number b. Here is the formal definition: Logarithm Suppose b and y are positive numbers, with b. The logarithm base b of y, denoted log b y, is defined to be the number such that b = y. The base b = is ecluded because = for every real number. Short version: log b y = means b = y. (a) Evaluate log (b) Evaluate log eample 3 (c) Evaluate log 3 8. (a) Because 0 3 = 000, we have log = 3. (b) Because 7 2 = 49, we have log 7 49 = 2. (c) Because 3 4 = 8, we have log 3 8 = 4. Two important identities follow immediately from the definition: The logarithm of and the logarithm of the base If b is a positive number with b, then log b = 0; log b b =. The first identity holds because b 0 = ; the second holds because b = b. The definition of log b y as the number such that b = y has the following consequence: Logarithm as an inverse function Suppose b is a positive number with b and f is the eponential function defined by f () = b. Then the inverse function of f is given by the formula f (y) = log b y. Logs have many uses, and the word log has more than one meaning.

6 230 chapter 3 Eponential Functions, Logarithms, and e If y 0, then log b y is not defined. Because a function and its inverse interchange domains and ranges, the domain of the function f defined by f (y) = log b y is the set of positive numbers, and the range of this function is the set of real numbers. eample 4 Suppose f is the function defined by f () = Find a formula for f. To find a formula for f (y), we solve the equation = y for. Dividing by 3, we have 5 7 = y 3. Thus 7 = log y 5 3, which implies that y = 7 + log 5 3. Hence f y (y) = 7 + log 5 3. Most applications of logarithms involve bases bigger than. Because the function log b is the inverse of the function b, flipping the graph of b across the line y = gives the graph of log b. If b > then log b is an increasing function (because b is an increasing function). As we will see in the net section, the shape of the graph of log b is similar to the shape of the graph of log 2 obtained earlier. The definition of logarithm implies the two equations displayed below. Be sure that you are comfortable with these equations and understand why they hold. Note that if f is defined by f () = b, then f (y) = log b y. The equations below could then be written in the form (f f )(y) = y and (f f )() =, which are equations that always hold for a function and its inverse. Inverse properties of logarithms If b and y are positive numbers, with b, and is a real number, then b log b y = y; log b b =. Common Logarithms and the Number of Digits In applications of logarithms, the most commonly used values for the base are 0, 2, and the number e (which we will discuss in Section 3.5). The use of a logarithm with base 0 is so frequent that it gets a special name: Common logarithm The logarithm base 0 is called the common logarithm. To simplify notation, sometimes logarithms base 0 are written without the base. If no base is displayed, then the base is assumed to be 0. In other words, log y = log 0 y. John Napier, the Scottish mathematician who invented logarithms around 64. Thus, for eample, log 0000 = 4 (because 0 4 = 0000) and log 00 = 2 (because 0 2 = 00 ). If your calculator has a button labeled log, then it will compute the logarithm base 0, which is often just called the logarithm.

7 section 3. Logarithms as Inverses of Eponential Functions 23 Note that 0 is a two-digit number, 0 2 is a three-digit number, 0 3 is a four-digit number, and more generally, 0 n is an n-digit number. Thus the integers with n digits are the integers in the interval [0 n, 0 n ). Because log 0 n = n and log 0 n = n, this implies that an n-digit positive integer has a logarithm in the interval [n, n). Digits and logarithms The logarithm of an n-digit positive integer is in the interval [n, n). Alternative statement: If M is a positive integer with n digits, then n log M < n. The conclusion above is often useful in making estimates. For eample, without using a calculator we can see that the number , which has nine digits, has a logarithm between 8 and 9 (the actual value is about 8.09). The net eample shows how to use the conclusion above to determine the number of digits in a number from its logarithm. Suppose M is a positive integer such that log M 73.. How many digits does M have? Because 73. is in the interval [73, 74), we can conclude that M is a 74-digit number. eercises For Eercises 6, evaluate the indicated quantities assuming that f and g are the functions defined by (f g)( ) 2 (g f )(0) 3 (f g)(0) f () = 2 and g() = (g f )( 3 2 ) 5 (f f )( 2 ) 6 (f f )( 3 5 ) For Eercises 7 8, find a formula for f g given the indicated functions f and g. 7 f () = 5 2, g() = 8 8 f () = 7 2, g() = 3 For Eercises 9 24, evaluate the indicated epression. Do not use a calculator for these eercises. 9 log log log log log log log log log log log 000 eample 5 Always round up the logarithm of a number to determine the number of digits. Here log M 73. is rounded up to show that M has 74 digits. 20 log log log log log Find a number y such that log 2 y = Find a number t such that log 2 t = Find a number y such that log 2 y = Find a number t such that log 2 t = 9. For Eercises 29 36, find a number b such that the indicated equality holds. 29 log b 64 = 30 log b 64 = 2 3 log b 64 = 3 32 log b 64 = 6 33 log b 64 = 2 34 log b 64 = 8 35 log b 64 = log b 64 = 6 5

8 232 chapter 3 Eponential Functions, Logarithms, and e For Eercises 37 48, find all numbers such that the indicated equation holds. 37 log = 2 38 log = 3 39 log = 2 40 log = 3 4 log 3 (5 + ) = 2 42 log 4 (3 + ) = = = = = = = 8 For Eercises 49 66, find a formula for the inverse function f of the indicated function f. For Eercises 67 74, find a formula for (f g)() assuming that f and g are the indicated functions. 67 f () = log 6 and g() = f () = log 5 and g() = f () = 6 3 and g() = log 6 70 f () = and g() = log 5 7 f () = log 4 and g() = 0 72 f () = log 2 7 and g() = 0 73 f () = log 4 and g() = f () = log 2 7 and g() = Find a number n such that log 3 (log 5 n) =. 76 Find a number n such that log 3 (log 2 n) = Find a number m such that log 7 (log 8 m) = f () = 3 50 f () = f () = f () = f () = f () = f () = f () = f () = problems 58 f () = f () = log 8 60 f () = log 3 6 f () = log 4 (3 + ) 62 f () = log 7 (2 9) 63 f () = 5+3 log 6 (2+) 64 f () = 8+9 log 2 (4 7) 65 f () = log 3 66 f () = log Find a number m such that log 5 (log 6 m) = Suppose N is a positive integer such that log N How many digits does N have? 80 Suppose k is a positive integer such that log k How many digits does k have? Some problems require considerably more thought than the eercises. 8 Show that (3 2 ) 2 = Give an eample of three irrational numbers, y, and z such that ( y ) z is a rational number. 83 Is the function f defined by f () = 2 for every real number an even function, an odd function, or neither? 84 Suppose f () = 8 and g() = 2. Eplain why the graph of g can be obtained by horizontally stretching the graph of f by a factor of Suppose f () = 2. Eplain why shifting the graph of f left 3 units produces the same graph as vertically stretching the graph of f by a factor of Eplain why there does not eist a polynomial p such that p() = 2 for every real number. [Hint: Consider the behavior of p() and 2 for near.] 87 Eplain why there does not eist a rational function r such that r () = 2 for every real number. [Hint: Consider the behavior of r () and 2 for near ±.] 88 Eplain why log 5 5 = Eplain why log 3 00 is between 4 and Eplain why log 40 3 is between 4 and 3. 9 Show that log 2 3 is an irrational number. [Hint: Use proof by contradiction: Assume log 2 3 is equal to a rational number m ; write out what this n means, and think about even and odd numbers.] 92 Show that log 2 is irrational. 93 Eplain why logarithms with a negative base are not defined. 94 Give the coordinates of three distinct points on the graph of the function f defined by f () = log Give the coordinates of three distinct points on the graph of the function g defined by g(b) = log b Suppose g(b) = log b 5, where the domain of g is the interval (, ). Is g an increasing function or a decreasing function?

9 section 3. Logarithms as Inverses of Eponential Functions 233 worked-out s to Odd-Numbered Eercises Do not read these worked-out s before attempting to do the eercises yourself. Otherwise you may mimic the techniques shown here without understanding the ideas. Best way to learn: Carefully read the section of the tetbook, then do all the odd-numbered eercises and check your answers here. If you get stuck on an eercise, then look at the worked-out here. For Eercises 6, evaluate the indicated quantities assuming that f and g are the functions defined by (f g)( ) 3 (f g)(0) f () = 2 and g() = (f g)( ) = f ( g( ) ) = f (0) = 2 0 = (f g)(0) = f ( g(0) ) = f ( 2 ) = 2/2.44 log 2 28 If we let = log 2, then is the number such that = 2. Because 28 = 2 7 = 2 7, we see that = 7. Thus log 2 28 = 7. 3 log 4 2 Because 2 = 4 /2, we have log 4 2 = 2. 5 log 4 8 Because 8 = 2 4 = 4 /2 4 = 4 3/2, we have log 4 8 = (f f )( 2 ) (f f )( 2 ) = f ( f ( 2 )) = f (2 /2 ) f (.442) = For Eercises 7 8, find a formula for f g given the indicated functions f and g. 7 f () = 5 2, g() = 8 (f g)() = f ( g() ) = f ( 8 ) 2 = 8 5( ) = 5 6 = log 0000 log 0000 = log 0 4 = 4 9 log 000 log 000 = log 000 /2 = log (0 3 ) /2 = log 0 3/2 = log log = log 2 (2 3 ) 3. = log For Eercises 9 24, evaluate the indicated epression. Do not use a calculator for these eercises. 9 log log 6 32 = 9.3 such that If we let = log 2 64, then is the number 64 = 2. Because 64 = 2 6, we see that = 6. Thus log 2 64 = 6.

10 234 chapter 3 Eponential Functions, Logarithms, and e log 6 32 = log = log 6 (2 4 ) 5/4 = log 6 6 5/4 35 log b 64 = 3 2 The equation log b 64 = 3 implies that 2 b 3/2 = 64. = Find a number y such that log 2 y = 7. The equation log 2 y = 7 implies that y = 2 7 = Find a number y such that log 2 y = 5. Raising both sides of this equation to the 2/3 power, we get b = 64 2/3 = (2 6 ) 2/3 = 2 4 = 6. The equation log 2 y = 5 implies that y = 2 5 = 32. For Eercises 37 48, find all numbers such that the indicated equation holds. For Eercises 29 36, find a number b such that the indicated equality holds. 29 log b 64 = Thus b = log b 64 = 3 The equation log b 64 = implies that b = 64. The equation log b 64 = 3 implies that b 3 = 64. Because 4 3 = 64, this implies that b = log b 64 = 2 37 log = 2 The equation log = 2 is equivalent to the equation = 0 2 = 00. Thus the two values of satisfying this equation are = 00 and = log = 2 The equation log = 2 means that log = 2 or log = 2, which means that = 0 2 = 00 or = 0 2 = log 3 (5 + ) = 2 The equation log 3 (5 + ) = 2 implies that 5 + = 3 2 = 9. Thus 5 = 8, which implies that = 8 5. Thus The equation log b 64 = 2 implies that b 2 = 64. b = 64 /2 = (2 6 ) /2 = 2 6/2 = 2 /2 = = The equation 3 = 0 2 implies that log 3 2 = log 3. Thus =, which is approimately = 0.8 Multiplying both sides of the equation above by 0 + 2, we get 0 + = Solving this equation for 0 gives 0 = 3, which means that = log

11 section 3. Logarithms as Inverses of Eponential Functions = 2 Note that 0 2 = (0 ) 2. This suggests that we let y = 0. Then the equation above can be rewritten as y 2 + y 2 = 0. The s to this equation (which can be found either by using the quadratic formula or by factoring) are y = 4 and y = 3. Thus 0 = 4 or 0 = 3. However, there is no real number such that 0 = 4 (because 0 is positive for every real number ), and thus we must have 0 = 3. Thus = log For Eercises 49 66, find a formula for the inverse function f of the indicated function f. 49 f () = 3 By definition of the logarithm, the inverse of f is the function f defined by 5 f () = 2 5 f (y) = log 3 y. To find a formula for f (y), we solve the equation 2 5 = y for. This equation means that 5 = log 2 y. Thus = 5 + log 2 y. Hence 53 f () = f (y) = 5 + log 2 y. To find a formula for f (y), we solve the equation = y for. Subtract 7 from both sides, getting 6 = y 7. This equation means that = log 6 (y 7). Hence 55 f () = 4 5 f (y) = log 6 (y 7). To find a formula for f (y), we solve the equation 4 5 = y for. Divide both sides by 4, getting 5 = y 4. This equation means that = log y 5 4. Hence f y (y) = log f () = To find a formula for f (y), we solve the equation = y for. Subtract from both sides, then divide both sides by 2, getting 9 = y 2. This equation means that = log 9 y 2. Hence 59 f () = log 8 f (y) = log 9 y 2. By the definition of the logarithm, the inverse of f is the function f defined by 6 f () = log 4 (3 + ) f (y) = 8 y. To find a formula for f (y), we solve the equation log 4 (3 + ) = y for. This equation means that 3 + = 4 y. Solving for, we get = 4y 3. Hence 63 f () = log 6 (2 + ) f (y) = 4y. 3 To find a formula for f (y), we solve the equation log 6 (2 + ) = y for. Subtracting 5 from both sides and then dividing by 3 gives log 6 (2 + ) = y 5 3. This equation means that 2 + = 6 (y 5)/3. Solving for, we get = 6(y 5)/3 2. Hence 65 f () = log 3 f (y) = 6(y 5)/3. 2 To find a formula for f (y), we solve the equation log 3 = y for. This equation means that y = 3. Raising both sides to the power y, we get = 3 /y. Hence f (y) = 3 /y. For Eercises 67 74, find a formula for (f g)() assuming that f and g are the indicated functions. 67 f () = log 6 and g() = 6 3 (f g)() = f ( g() ) = f (6 3 ) = log = 3

12 236 chapter 3 Eponential Functions, Logarithms, and e 69 f () = 6 3 and g() = log 6 (f g)() = f ( g() ) = f (log 6 ) 7 f () = log 4 and g() = 0 = 6 3 log 6 = (6 log 6 ) 3 = 3 (f g)() = f ( g() ) = f (0 ) 73 f () = log 4 and g() = 00 = log 0 4 = log 4 log 0 = log 4 (f g)() = f ( g() ) = f (00 ) = log 00 4 = log 4 log 00 = log 4 log 0 2 = log Find a number n such that log 3 (log 5 n) =. The equation log 3 (log 5 n) = implies that log 5 n = 3, which implies that n = 5 3 = Find a number m such that log 7 (log 8 m) = 2. that The equation log 7 (log 8 m) = 2 implies log 8 m = 7 2 = 49. The equation above now implies that m = Suppose N is a positive integer such that log N How many digits does N have? Because 35.4 is in the interval [35, 36), we can conclude that N is a 36-digit number The graphs of log 2 (blue), log 3 (red), log 4 (green), log 5 (orange), and log 6 (purple) on the interval [0., 00].

13 section 3.2 Applications of the Power Rule for Logarithms Applications of the Power Rule for Logarithms learning objectives By the end of this section you should be able to apply the formula for the logarithm of a power; model radioactive decay using half-life; apply the change-of-base formula for logarithms. Logarithm of a Power Logarithms convert powers to products. To see this, suppose b and y are positive numbers, with b, and t is a real number. Let = log b y. Then the definition of logarithm base b implies b = y. Raise both sides of this equation to the power t and use the identity (b ) t = b t to get b t = y t. Again using the definition of logarithm base b, the equation above implies log b (y t ) = t = t log b y. Thus we have the following formula for the logarithm of a power: An epression without parentheses of the form log b y t should be interpreted to mean log b (y t ), not (log b y) t. Logarithm of a power If b and y are positive numbers, with b, and t is a real number, then log b (y t ) = t log b y. The net eample shows a nice application of the formula above. How many digits does have? eample We can answer this question by evaluating the common logarithm of Your calculator cannot Using the formula for the logarithm of a power and a calculator, we see that log( ) = 5000 log evaluate Thus the formula for the logarithm of a power is needed even though a calculator is being Thus has 2386 digits. used.

14 238 chapter 3 Eponential Functions, Logarithms, and e Before calculators and computers eisted, books of common logarithm tables were frequently used to compute powers of numbers. As an eample of how this worked, consider the problem of evaluating The key to performing this calculation is the formula log( ) = 3.7 log.7. Books of logarithms have now mostly disappeared. However, your calculator is using the formula log b (y t ) = t log b y when you ask it to evaluate an epression such as Let s assume we have a book that gives the logarithms of the numbers from to 0 in increments of 0.00, meaning that the book gives the logarithms of.00,.002,.003, and so on. The idea is first to compute the right side of the equation above. To do that, we would look in the book of logarithms, getting log Multiplying the last number by 3.7, we would conclude that the right side of the equation above is approimately Thus, according to the equation above, log( ) Hence we can evaluate by finding a number whose logarithm equals To do this, we would look through our book of logarithms and find that the closest match is provided by the entry showing that log Thus Logarithms are rarely used today directly by humans for computations such as evaluating However, logarithms are used by your calculator for such computations. Logarithms also have important uses in calculus and several other branches of mathematics. Furthermore, logarithms have several practical uses, as we will soon see. Radioactive Decay and Half-Life Marie Curie, the only person ever to win Nobel Prizes in both physics (903) and chemistry (9), did pioneering work on radioactive decay. Scientists have observed that starting with a large sample of radon atoms, after 92 hours one-half of the radon atoms will decay into polonium. After another 92 hours, one-half of the remaining radon atoms will also decay into polonium. In other words, after 84 hours, only one-fourth of the original radon atoms will be left. After another 92 hours, one-half of those remaining one-fourth of the original atoms will decay into polonium, leaving only one-eighth of the original radon atoms after 276 hours. After t hours, the number of radon atoms will be reduced by half t/92 times. Thus after t hours, the number of radon atoms left will equal the original number of radon atoms divided by 2 t/92. Here t need not be an integer multiple of 92. For eample, after five hours the original number of radon atoms will be divided by 2 5/92. Because 0.963, 25/92 this means that after five hours a sample of radon will contain 96.3% of the original number of radon atoms. Because half the atoms in any sample of radon will decay to polonium in 92 hours, we say that radon has a half-life of 92 hours. Some radon atoms eist for less than 92 hours; some radon atoms eist for much longer than 92 hours.

15 section 3.2 Applications of the Power Rule for Logarithms 239 The half-life of any radioactive isotope is the length of time it takes for half the atoms in a large sample of the isotope to decay. The table here gives the approimate half-life for several radioactive isotopes (the isotope number shown after the name of each element gives the total number of protons and neutrons in each atom of the isotope). isotope neon-8 nitrogen-3 radon-222 polonium-20 cesium-37 carbon-4 plutonium-239 half-life 2 seconds 0 minutes 92 hours 38 days 30 years 5730 years 24,0 years Half-life of some radioactive isotopes. Some of the isotopes in this table are human creations that do not eist in nature. For eample, the nitrogen on Earth is almost entirely nitrogen-4 (7 protons and 7 neutrons), which is not radioactive and does not decay. The nitrogen-3 listed here has 7 protons and 6 neutrons; it can be created in a laboratory, but it is radioactive and half of it will decay within 0 minutes. If a radioactive isotope has a half-life of h time units (which might be seconds, minutes, hours, days, or years), then after t time units the number of atoms of this isotope is reduced by half t/h times. Thus after t time units, the remaining number of atoms of the isotope will equal the original number of atoms divided by 2 t/h. Because 2 t/h = 2 t/h, we have the following result: Radioactive decay If a radioactive isotope has half-life h, then the function modeling the number of atoms in a sample of this isotope is a(t) = a 0 2 t/h, where a 0 is the number of atoms of the isotope in the sample at time 0. The radioactive decay of carbon-4 has led to a clever way of determining the age of fossils, wood, and other remnants of plants and animals. Carbon-2, by far the most common form of carbon on Earth, is not radioactive and does not decay. Radioactive carbon-4 is produced regularly as cosmic rays hit the upper atmosphere. Radioactive carbon-4 then drifts down, getting into plants through photosynthesis and then into animals that eat plants and then into animals that eat animals that eat plants, and so on. Carbon-4 accounts for about 0 0 percent of the carbon atoms in living plants and animals. When a plant or animal dies, it stops absorbing new carbon because it is no longer engaging in photosynthesis or eating. Thus no new carbon-4 is absorbed. The radioactive carbon-4 in the plant or animal decays, with half of it gone after 5730 years, as shown in the table above. By measuring the amount of carbon-4 as a percentage of the total amount of carbon in the remains of a plant or animal, we can then determine how long ago it died. The 960 Nobel Prize in Chemistry was awarded to Willard Libby for his invention of this carbon-4 dating method.

16 240 chapter 3 Eponential Functions, Logarithms, and e eample 2 Suppose a cat skeleton found in an old well has a ratio of carbon-4 to carbon-2 that is 6% of the corresponding ratio for living organisms. Approimately how long ago did the cat die? If t denotes the number of years since the cat died, then 0.6 = 2 t/5730. To solve this equation for t, take the logarithm of both sides, getting Solve this equation for t, getting log 0.6 = t log t = 5730 log 0.6 log Because we started with only two-digit accuracy (6%), we should not produce such an eact-looking estimate. Thus we might estimate that the skeleton is about 400 years old. The author s first cat. Change of Base Your calculator can probably evaluate logarithms for only two bases. One of these is probably the logarithm base 0 (the common logarithm, probably labeled log on your calculator). The other is probably the logarithm base e (this is the natural logarithm that we will discuss in Section 3.5; it is probably labeled ln on your calculator). If you want to use a calculator to evaluate something like log , then you probably need a formula for converting logarithms from one base to another. To derive this formula, suppose a, b, and y are positive numbers, with a and b. Let = log b y. Then the definition of logarithm base b implies b = y. Take the logarithm base a of both sides of the equation above, getting Thus log a b = log a y. = log a y log a b. Replacing in the equation above with its value log b y gives the following formula for converting logarithms from one base to another: Change of base for logarithms If a, b, and y are positive numbers, with a and b, then log b y = log a y log a b.

17 section 3.2 Applications of the Power Rule for Logarithms 24 A special case of this formula, suitable for use with calculators, is to take a = 0, thus using common logarithms and getting the following formula: Change of base with common logarithms If b and y are positive numbers, with b, then log b y = log y log b. Caution: In WolframAlpha and other advanced software, log with no base specified means logarithm base e. Thus use log_0 6.8 to evaluate the common logarithm of 6.8 in WolframAlpha; the underscore _ denotes a subscript. Evaluate log eample 3 Use a calculator with b = 2 and y = 73.9 in the formula above, getting log = log 73.9 log The change-of-base formula for logarithms implies that the graph of the logarithm using any base can be obtained by vertically stretching the graph of the logarithm using any other base (assuming both bases are bigger than ), as shown in the following eample. Sketch the graphs of log 2 and log on the interval [ 8, 8]. What is the relationship between these two graphs? eample 4 The change-of-base formula implies that log 2 = (log )/(log 2). Because /(log 2) 3.32, this means that the graph of log 2 is obtained from the graph of log by stretching vertically by a factor of approimately The graphs of log 2 (red) and log (blue) on the interval [ 8, 8]. 2 3 More generally, for fied positive numbers a and b, neither of which is, the change-of-base formula log a = (log a b) log b implies that the graph of log a can be obtained from the graph of log b by stretching vertically by a factor of log a b.

18 242 chapter 3 Eponential Functions, Logarithms, and e eercises The net two eercises emphasize that log( y ) does not equal (log ) y. For = 5 and y = 2, evaluate each of the following: (a) log( y ) (b) (log ) y 2 For = 2 and y = 3, evaluate each of the following: (a) log( y ) (b) (log ) y 3 Suppose y is such that log 2 y = Evaluate log 2 (y 00 ). 4 Suppose is such that log 6 = Evaluate log 6 ( 0 ). For Eercises 5 8, find all numbers such that the indicated equation holds. 5 3 = = = = 9 9 Suppose m is a positive integer such that log m 3.2. How many digits does m 3 have? 0 Suppose M is a positive integer such that log M How many digits does M 4 have? How many digits does have? 2 How many digits does have? 3 Find an integer k such that 8 k has 357 digits. 4 Find an integer n such that 22 n has 222 digits. 5 Find an integer m such that m 234 has 99 digits. 6 Find an integer N such that N 432 has 604 digits. 7 Find the smallest integer n such that 7 n > Find the smallest integer k such that 9 k > Find the smallest integer M such that 5 /M < Find the smallest integer m such that 8 /m < Suppose log 8 (log 7 m) = 5. How many digits does m have? 22 Suppose log 5 (log 9 m) = 6. How many digits does m have? A prime number is an integer greater than that has no divisors other than itself and. 23 At the time this book was written, the third largest known prime number was How many digits does this prime number have? 24 At the time this book was written, the second largest known prime number was How many digits does this prime number have? 25 About how many hours will it take for a sample of radon-222 to have only one-eighth as much radon- 222 as the original sample? 26 About how many minutes will it take for a sample of nitrogen-3 to have only one sity-fourth as much nitrogen-3 as the original sample? 27 About how many years will it take for a sample of cesium-37 to have only two-thirds as much cesium- 37 as the original sample? 28 About how many years will it take for a sample of plutonium-239 to have only % as much plutonium- 239 as the original sample? 29 Suppose a radioactive isotope is such that one-fifth of the atoms in a sample decay after three years. Find the half-life of this isotope. 30 Suppose a radioactive isotope is such that fivesiths of the atoms in a sample decay after four days. Find the half-life of this isotope. 3 Suppose the ratio of carbon-4 to carbon-2 in a mummified cat is 64% of the corresponding ratio for living organisms. About how long ago did the cat die? 32 Suppose the ratio of carbon-4 to carbon-2 in a fossilized wooden tool is 20% of the corresponding ratio for living organisms. About how old is the wooden tool? For Eercises 33 40, evaluate the indicated quantities. Your calculator probably cannot evaluate logarithms using any of the bases in these eercises, so you will need to use an appropriate change-of-base formula. 33 log log log log log log log log

19 section 3.2 Applications of the Power Rule for Logarithms 243 problems 4 Eplain why there does not eist an integer m such that 67 m has 9236 digits. 42 Do a web search to find the largest currently known prime number. Then calculate the number of digits in this number. [The discovery of a new largest known prime number usually gets some newspaper coverage, including a statement of the number of digits. Thus you can probably find on the web the number of digits in the largest currently known prime number; you are asked here to do the calculation to verify that the reported number of digits is correct.] 43 Suppose f () = log and g() = log( 4 ), with the domain of both f and g being the set of positive numbers. Eplain why the graph of g can be obtained by vertically stretching the graph of f by a factor of Eplain why log b 27 3 = log b for every positive number b. 45 Suppose and b are positive numbers with b. Show that if 27, then log b 3 log b Find a positive number such that log b 4 = log b 4 for every positive number b. 47 Eplain why epressing a large positive integer in binary notation (base 2) should take approimately 3.3 times as many digits as epressing the same positive integer in standard decimal notation (base 0). [For eample, this problem predicts that five trillion, which requires the 3 digits 5,000,000,000,000 to epress in decimal notation, should require approimately digits (which equals 42.9 digits) to epress in binary notation. Epressing five trillion in binary notation actually requires 43 digits.] 48 Suppose a and b are positive numbers, with a and b. Show that log a b = log b a. worked-out s to Odd-Numbered Eercises The net two eercises emphasize that log( y ) does not equal (log ) y. For = 5 and y = 2, evaluate each of the following: (a) log( y ) (b) (log ) y (a) log(5 2 ) = log (b) (log 5) 2 ( ) Suppose y is such that log 2 y = Evaluate log 2 (y 00 ). log 2 (y 00 ) = 00 log 2 y 5 3 = 8 Take the common logarithm of both sides, getting log(3 ) = log 8, which can be rewritten as log 3 = log 8. Thus 7 6 = 2 = log 8 log Take the common logarithm of both sides, getting log(6 ) = log 2, which can be rewritten as log 6 = log 2. Thus = log 2 log 6. = = 767 Hence = ( ) 2 log log 6 For Eercises 5 8, find all numbers such that the indicated equation holds. 9 Suppose m is a positive integer such that log m 3.2. How many digits does m 3 have? Note that

20 244 chapter 3 Eponential Functions, Logarithms, and e log(m 3 ) = 3 log m = Because 39.6 is in the interval [39, 40), we can conclude that m 3 is a 40-digit number. How many digits does have? Using the formula for the logarithm of a power and a calculator, we have log( ) = 4000 log Thus has 338 digits. 3 Find an integer k such that 8 k has 357 digits. We want to find an integer k such that 356 log(8 k ) < 357. Using the formula for the logarithm of a power, we can rewrite the inequalities above as Dividing by log 8 gives 356 k log 8 < log 8 k < 357 log 8. Using a calculator, we see that 357 log log and Thus the only possible choice is to take k = 284. Again using a calculator, we see that log(8 284 ) = 284 log Thus indeed has 357 digits. 5 Find an integer m such that m 234 has 99 digits. We want to find an integer m such that 990 log(m 234 ) < 99. Using the formula for the logarithm of a power, we can rewrite the inequalities above as Dividing by 234 gives Thus log m < log m < /234 m < 0 99/234. Using a calculator, we see that 0 990/ and 0 99/ Thus the only possible choice is to take m = 4. Again using a calculator, we see that log(4 234 ) = 234 log Find the smallest integer n such that 7 n > Suppose 7 n > Taking the common logarithm of both sides, we have which can be rewritten as This implies that log(7 n ) > log(0 00 ), n log 7 > 00. n > 00 log The smallest integer that is bigger than 8.33 is 9. Thus we take n = 9. 9 Find the smallest integer M such that 5 /M <.0. Suppose 5 /M <.0. Taking the common logarithm of both sides, we have which can be rewritten as This implies that log(5 /M ) < log.0, log 5 M < log.0. M > log 5 log The smallest integer that is bigger than 6.7 is 62. Thus we take M = Suppose log 8 (log 7 m) = 5. How many digits does m have? that The equation log 8 (log 7 m) = 5 implies log 7 m = 8 5 = The equation above now implies that m = To compute the number of digits that m has, note that log m = log( ) = log Thus m has digits. A prime number is an integer greater than that has no divisors other than itself and. Thus indeed has 99 digits.

21 section 3.2 Applications of the Power Rule for Logarithms At the time this book was written, the third largest known prime number was How many digits does this prime number have? To calculate the number of digits in , we need to evaluate log( ). However, is too large to evaluate directly on a calculator, and no formula eists for the logarithm of the difference of two numbers. The trick here is to note that and have the same number of digits, as we will now see. Although it is possible for a number and the number minus to have a different number of digits (for eample, 00 and 99 do not have the same number of digits), this happens only if the larger of the two numbers consists of followed by a bunch of 0 s and the smaller of the two numbers consists of all 9 s. Here are three different ways to see that this situation does not apply to and (pick whichever eplanation seems easiest to you): (a) cannot end in a 0 because all positive integer powers of 2 end in either 2, 4, 6, or 8; (b) cannot end in a 0 because then it would be divisible by 5, but is divisible only by integer powers of 2; (c) cannot consist of all 9 s because then it would be divisible by 9, which is not possible for a prime number. Now that we know that and have the same number of digits, we can calculate the number of digits by taking the logarithm of and using the formula for the logarithm of a power. We have log( ) = log Thus has, 85, 272 digits; hence also has, 85, 272 digits. 25 About how many hours will it take for a sample of radon-222 to have only one-eighth as much radon- 222 as the original sample? The half-life of radon-222 is about 92 hours, as shown in the chart in this section. To reduce the number of radon-222 atoms to one-eighth the original number, we need 3 half-lives (because 2 3 = 8). Thus it will take 276 hours (because 92 3 = 276) to have only one-eighth as much radon-222 as the original sample. 27 About how many years will it take for a sample of cesium-37 to have only two-thirds as much cesium- 37 as the original sample? The half-life of cesium-37 is about 30 years, as shown in the chart in this section. Thus if we start with a atoms of cesium-37 at time 0, then after t years there will be a 2 t/30 atoms left. We want this to equal 2 a. Thus we must 3 solve the equation a 2 t/30 = 2 3 a. To solve this equation for t, divide both sides by a and then take the logarithm of both sides, getting t 30 log 2 = log 2 3. Now multiply both sides by, replace log 2 3 by log 3, and then solve for t, getting 2 t = 30 log 3 2 log Thus two-thirds of the original sample will be left after approimately 7.5 years. 29 Suppose a radioactive isotope is such that one-fifth of the atoms in a sample decay after three years. Find the half-life of this isotope. Let h denote the half-life of this isotope, measured in years. If we start with a sample of a atoms of this isotope, then after 3 years there will be a 2 3/h atoms left. We want this to equal 4 a. Thus we must 5 solve the equation a 2 3/h = 4 5 a. To solve this equation for h, divide both sides by a and then take the logarithm of both sides, getting 3 h log 2 = log 4 5. Now multiply both sides by, replace log 4 5 by log 5, and then solve for h, getting 4 h = 3 log 2 log Thus the half-life of this isotope is approimately 9.3 years. 3 Suppose the ratio of carbon-4 to carbon-2 in a mummified cat is 64% of the corresponding ratio for living organisms. About how long ago did the cat die? The half-life of carbon-4 is 5730 years. If we start with a sample of a atoms of carbon-4, then after t years there will be a 2 t/5730

22 246 chapter 3 Eponential Functions, Logarithms, and e atoms left. We want to find t such that this equals 0.64a. Thus we must solve the equation a 2 t/5730 = 0.64a. To solve this equation for t, divide both sides by a and then take the logarithm of both sides, getting t 5730 Now solve for t, getting log 2 = log log 0.64 t = log 2 Thus the cat died about 3689 years ago. Carbon-4 cannot be measured with etreme accuracy. Thus it is better to estimate that the cat died about 3700 years ago (because a number such as 3689 conveys more accuracy than will be present in such measurements). 39 log log = log 7. log For Eercises 33 40, evaluate the indicated quantities. Your calculator probably cannot evaluate logarithms using any of the bases in these eercises, so you will need to use an appropriate change-of-base formula. 33 log 2 3 log 2 3 = 35 log log 3 log log = 37 log log = log 9.72 log log 0.23 log The author s second cat proofreading the manuscript.

23 section 3.3 Applications of the Product and Quotient Rules for Logarithms Applications of the Product and Quotient Rules for Logarithms learning objectives By the end of this section you should be able to apply the formula for the logarithm of a product; apply the formula for the logarithm of a quotient; model earthquake intensity with the logarithmic Richter magnitude scale; model sound intensity with the logarithmic decibel scale; model star brightness with the logarithmic apparent magnitude scale. Logarithm of a Product Logarithms convert products to sums. To see this, suppose b,, and y are positive numbers, with b. Let u = log b and v = log b y. Then the definition of logarithm base b implies b u = and b v = y. Multiplying together these equations and using the identity b u b v = b u+v gives b u+v = y. Again using the definition of logarithm base b, the equation above implies log b (y) = u + v = log b + log b y. Thus we have the following formula for the logarithm of a product: Logarithm of a product If b,, and y are positive numbers, with b, then log b (y) = log b + log b y. In general, log b (+y) does not equal log b + log b y. There is no nice formula for log b ( + y). Use the information that log to evaluate log eample log = log(0 4 6) = log(0 4 ) + log 6 = 4 + log

24 248 chapter 3 Eponential Functions, Logarithms, and e Logarithm of a Quotient Logarithms convert quotients to differences. To see this, suppose b,, and y are positive numbers, with b. Let u = log b and v = log b y. Then the definition of logarithm base b implies b u = and b v = y. Divide the first equation by the second equation and use the identity bu b v = bu v to get b u v = y. Again using the definition of logarithm base b, the equation above implies log b y = u v = log b log b y. Thus we have the following formula for the logarithm of a quotient: Here and in the bo below, we assume that b,, and y are positive numbers, with b. Logarithm of a quotient log b y = log b log b y. eample 2 Suppose log 4 = 8.9 and log 4 y = 2.2. Evaluate log 4 4 y. Using the quotient and product rules, we have log 4 4 y = log 4(4) log 4 y = log log 4 log 4 y = = 7.7. As a special case of the formula for the logarithm of a quotient, take = in the formula above for the logarithm of a quotient, getting log b y = log b log b y. Recalling that log b = 0, we get the following result: Logarithm of a multiplicative inverse log b y = log b y.

25 section 3.3 Applications of the Product and Quotient Rules for Logarithms 249 Earthquakes and the Richter Scale The intensity of an earthquake is measured by the size of the seismic waves generated by the earthquake. These numbers vary across such a huge scale that earthquakes are usually reported using the Richter magnitude scale, which is a logarithmic scale using common logarithms (base 0). Richter magnitude scale An earthquake with seismic waves of size S has Richter magnitude log S S 0, where S 0 is the size of the seismic waves corresponding to what has been declared to be an earthquake with Richter magnitude 0. A few points will help clarify this definition: The value of S 0 was set in 935 by the American seismologist Charles Richter as approimately the size of the smallest seismic waves that could be measured at that time. The size of the seismic wave is roughly proportional to the amount of ground shaking. The unit used to measure S and S 0 does not matter because any change in the scale of this unit disappears in the ratio S S 0. An increase of earthquake intensity by a factor of 0 corresponds to an increase of in Richter magnitude, as can be seen from the equation log 0S S 0 = log 0 + log S S 0 = + log S S 0. The world s most intense recorded earthquake struck Chile in 960 with Richter magnitude 9.5. The most intense recorded earthquake in the United States struck Alaska in 964 with Richter magnitude 9.2. Approimately how many times more intense was the 960 earthquake in Chile than the 964 earthquake in Alaska? eample 3 Let S C denote the size of the seismic waves from the 960 earthquake in Chile and let S A denote the size of the seismic waves from the 964 earthquake in Alaska. Thus 9.5 = log S C and 9.2 = log S A. S 0 S 0 Subtracting the second equation from the first equation, we get Thus 0.3 = log S C log S ( A SC / SA ) = log = log S C. S 0 S 0 S 0 S 0 S A S C S A = In other words, the 960 earthquake in Chile was approimately twice as intense as the 964 earthquake in Alaska. As this eample shows, even small differences in the Richter magnitude can correspond to large differences in intensity.