UNIT 10 Eponentials and Logarithms Bacteria like Staph aureus are ver common. Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
In the right environment, the rate of growth of the number of bacteria in a petri dish ma continuousl increase. You can describe this kind of growth with an eponential function. Big Ideas A function is a correspondence between two sets, the domain and the range, that assigns to each member of the domain eactl one member of the range. Functions can model man events in the phsical world. In a famil of functions, once ou know the graph of the parent function, ou can use transformations to graph related functions. If ou can create a mathematical model for a situation, ou can use the model to solve other problems that ou might not be able to solve otherwise. Unit Topics Foundations for Unit 10 Eponential Epressions and Equations Graphing Eponential Functions Applications: Growth and Deca Logarithms Using Logs to Solve Eponential Equations Solving Logarithmic Equations Graphing Logarithmic Functions Applications: Logarithms EXPONENTIALS AND LOGARITHMS 357 Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Foundations for Unit 10 Before ou learn about eponential and logarithmic epressions, equations, and functions, take time to ensure ou know how to do the following: Use eponential properties to simplif eponential epressions. Simplif products of powers and powers of products. Simplif quotients of powers and powers of quotients. Zero and Negative Eponent Properties ZERO AND NEGATIVE EXPONENTS Let a be a nonzero real number and let m be an integer. Propert Statement Zero Eponent a 0 = 1 Negative Eponent a m = a 1 Eample 1 Evaluate each epression. A 0 0 = undefined B ( 3.71) 0 = 1 C π 0 = 1 Eample Simplif each epression. Use positive eponents onl. A = = _ 3 1 B = C ( 3 ) 3 = ( 3 ) 3 = 33 3 = _ 7 Problem Set A Simplif each epression. Use positive eponents onl. 1. 7 0. 0 3.. ( 5 ) 5. ab 0. () 0 z 3 35 UNIT 10 EXPONENTIALS AND LOGARITHMS Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Unit 10 Foundations Using Product Properties of Eponents Use the product properties of eponents when ou multipl powers with the same base or raise a product to a power. PRODUCT PROPERTIES OF EXPONENTS Let a and b be real numbers. Let m and n be integers. Propert Product of Powers Power of a Product Statement a m a n = a m + n (ab) m = a m b m Eample 3 Simplif each epression. Use positive eponents onl. A 3 = 1 + = 1 B (rs) = r s = 1r s C ( ) ( ) ( ) 3 = ( ) + ( ) + 3 = ( ) 1 = Problem Set B Simplif each epression. Use positive eponents onl. 7. () 3. 3 9 3 1 9. (z) 10. ( 5 3 ) ( 5 3 ) 7 11. ( pq) 0 ( pq) ( pq) 3 1. 5 3 13. (3a b) 5 1. ( ) 9 ( ) Using Quotient Properties of Eponents Use the quotient properties of eponents to divide powers with the same base or raise a quotient to a power. QUOTIENT PROPERTIES OF EXPONENTS Let a and b be nonzero real numbers. Let m and n be integers. Propert Statement Quotient of Powers _ am = a a n m n Power of a Quotient ( a b ) m = _ am b m FOUNDATIONS FOR UNIT 10 359 Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Unit 10 Foundations Eample Simplif each epression. Use positive eponents onl. A _ 1 = 9 1 9 = 3 = 1 B ( 3 ) = 3 = 1 5 C _ ()3 _ (3) = 3 3 3 3 = _ 1 3 3 = _ Problem Set C 1 1 1 = 1 Simplif each epression. Use positive eponents onl. 15. _ 7 15 7 1 1. ( 9 ) 17. _ a a 1. 19. 0. 3 (ab) (ab) 1 5 1. (3pq) 5 ( pq) 3. ( 3 ) 3 ( 3 ) Using the Power of a Power Propert of Eponents To raise a power to another power, simpl multipl the eponents. POWER OF A POWER PROPERTY OF EXPONENTS Let a be a real number. Let m and n be integers. (a m ) n = a mn Eample 5 Simplif each epression. Use positive eponents onl. A ( ) 3 = 3 = = B ( a 3 b ) = a ( 3)( ) b ()( ) = a b = a b C ( _ 33 5 ) _ ( 3 = 3 ) ( = 3 3 ) = ( 5 ) 5 ( ) 9 3 _ Problem Set D = 9 5 5 Simplif each epression. Use positive eponents onl. 3. ( 3 3 ). ( f g 3 ) 5 5. ( 1 z ) _. ( a 3b ) 3 5 7. _ ( z 3 ) 0 (z) 5. ( s t ) 3 ( st ) 30 UNIT 10 EXPONENTIALS AND LOGARITHMS Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Eponential Epressions and Equations An root of a number can be written as a fractional power of the number. Rational Eponents and Radical Form PROPERTY OF RATIONAL EXPONENTS For an positive integers n and m, where n 0, and an b, m b n = ( n b ) m = n _ b m, ecept when b < 0 and n is even. REMEMBER A rational eponent is an eponent that can be written as a fraction. Eample 1 Epress each of the following in radical form. 5 A 10 B ( 3) 3 3 C ( ) 5 _ 10 = ( 10 ) 5 _ 3 ( 3) = 3 ( 3) ( 3 ) = _ = 10 5 = 3 9 = 3 = THINK ABOUT IT If there were no parentheses in Eample 1B, the result would be different: 3 3 = 3 3 = 3 9. Eample Epress each of the following in rational eponent form. _ A 3 B 5 ( ) 3 C ( 3 7 ) 3 = 3 _ 5 ( ) 3 5 = ( ) 3 ( 3 7 ) = ( 7 3 ) 3 = 7 3 = 7 Using Properties of Eponents to Simplif Epressions The properties of eponents can be used to simplif epressions containing rational eponents. EXPONENTIAL EXPRESSIONS AND EQUATIONS 31 Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Eample 3 A 3 Simplif. Assume all variables are positive. Here are two strategies ou can use to simplif this epression. Method 1 Combine eponents first. 3 = ( + 3 ) = = = 1 B 3 1 = 3 Quotient of Powers Propert 3 3 = 3 Subtract: 3 3 = 3. = ( 3 ) Propert of Rational Eponents = The cube root of is. = Simplif. ( C 1 a b ) ( 1 a b ) = (a b ) _ 1 = 1 = D _ ( r r z r Method Simplif each factor first. 3 = ( ) 3 = 3 = = 1 = a b Power of a Power Propert = ab 1 Simplif. = a Propert of Rational Eponents b = _ a Simplif. b 5z ) r _ ( r r z r ( 5z ) r r z r ) = r ( 5z r ) Power of a Quotient Propert = r r z r _ Power of a Product Propert 5 z r = r r z r Propert of Rational Eponents r 5 z _ = r r z r r 5 _ = r r z 3r 5 Quotient of Powers Propert Simplif. THINK ABOUT IT When ou solve problems with fractional eponents, values are usuall easier to work with if ou find the root before raising to the power: 3 = ( ) 3 = 3 = has smaller values than 3 = 3 = _ =. THINK ABOUT IT If,, or z in Eample 3D is less than zero and n is even, the propert of rational eponents is not valid. If = 3, for eample, ( ) = _ ( 3) = 3. 3 UNIT 10 EXPONENTIALS AND LOGARITHMS Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Using Rational Eponents to Epress Radical Epressions in Simplified Radical Form Eample Epress each of the following in simplified radical form with the smallest inde possible. _ 50 5 A 10 _ 10 50 5 _ 5 10 = 50 Rewrite in rational eponent form. = 50 Simplif. = (5 ) Factor. = 5 Power of a Product Propert _ = 5 Propert of Rational Eponents = 5 Simplif. B a b a b = ( a b ) Rewrite in rational eponent form. C ( 1 ( 1 = ( a b ) Simplif. = ( ab ) Product Propert of Eponents _ = ab Propert of Rational Eponents 1 ) 1 ) = ( 1 _ ) 1 = = ( ) 1 1 Rewrite in rational eponent form. = ( ) 1 1 ( ) 1 1 ( ) 1 1 Product Propert of Eponents = ( _ 1 1 )( _ 1 1 )( _ 1 1 ) Power of a Power Propert = ( 3 )( 3 )( 3 ) Simplif. = ( ) 3 Product Propert of Eponents = 3 Propert of Rational Eponents Solving Eponential Equations An eponential equation is an equation with variable epressions as eponents. PROPERTY OF EQUALITY FOR EXPONENTIAL EQUATIONS If a is a positive number other than 1, then a = a if and onl if =. The propert of equalit for eponential equations can be used to solve some tpes of eponential equations. EXPONENTIAL EXPRESSIONS AND EQUATIONS 33 Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Eample 5 Solve, then check. A = = ( ) = 3 Rewrite each term so that the bases are the same. = 3 Power of a Power Propert = 3 Propert of Equalit for Eponential Equations = 3 Division Propert of Equalit Check 3 3 _ = B 5 = 5 + 5 = 5 + = + = Solve for. Propert of Equalit for Eponential Equations Check 5 5 ( ) + 5 5 + 5 = 5 C = = = Rewrite each term so that the bases are the same. = = ± Solve for. Propert of Equalit for Eponential Equations Check Substitute for : Substitute for : ( ) ( ) = = So the solution set is { ± }. TIP Alwas check for etraneous solutions when ou solve eponential equations. REMEMBER Carefull follow the order of operations: ( ) = ( ) ( ) = = 3 UNIT 10 EXPONENTIALS AND LOGARITHMS Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Problem Set Epress each of the following in simplified radical form. 1. 3 3. ( 3 ). ( ) 5. ( ) Epress each of the following in rational eponent form. _ 5 5. 10 3 7. ( 3 ) 3 _ 3. ( 3). 1 Simplif b using properties of eponents. 9. 3 7 3 7 13. ( ) 10. 5 5 3 1. ( 3 5 ) 11. 1. _ 10 10 5 5 3 5 5 15. ( 73t 9t z 1t 3t 1. 3 1pq p 3 r Simplif the radical epressions. Use the smallest inde possible. _ 1 17. 0 3 0. a 3 b 3 1. n 9 3 1. 7 7 z 15 _ 19. 1 z 1. f g h Solve and check. 3. 3 = 1 7. 3 =. 9 n = 1. 5 = 15 5. d 1 = d + 1 * 9. Challenge 5 = 0. 7 1 = ( 9 ) * 30. Challenge (3a 1) 3 = ) 3 EXPONENTIAL EXPRESSIONS AND EQUATIONS 35 Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Graphing Eponential Functions An eponential function has an equation of the form f() = ab h + k, where a 0, b > 0, and b 1. The function f () = b is the parent function for the famil of eponential functions. Eponential Functions of the Form: f () = b When b > 1, the function f () = b is called an eponential growth function. General shape: When 0 < b < 1, the function f () = b is called an eponential deca function. General shape: = 0 End behavior: As increases in value, f () increases in value. As decreases in value, f () decreases in value, approaching the -ais. The following properties are common to both tpes of eponential functions: Asmptote: Since the graph gets close to the ais, but never meets or crosses it, the line = 0 (the -ais) is the asmptote of the graph. Domain: the set of all real numbers Range: all real numbers greater than 0 -intercept: 1 = 0 End behavior: As decreases in value, f () increases in value. As increases in value, f () decreases in value, approaching the -ais. 3 UNIT 10 EXPONENTIALS AND LOGARITHMS Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Using a Table of Values to Graph an Eponential Function Eample 1 Use a table of values to graph. A f () = In the function f () =, b =. Since b > 1, this is an eponential growth function. Step 1 Make a table of values. Use both positive and negative values of. Since this is an eponential growth function of the form f () = b, ou know that the -intercept is 1 and that the point (1, ) lies on the graph. 3 1 0 1 3 THINK ABOUT IT Point (1, b) will alwas lie on the graph of an eponential function of the form f () = b since an number raised to the power 1 equals that number: b 1 = b. f () 1 Step Plot the points from the table. Then connect the points in a smooth curve. Use the general shape of an eponential growth function as a guide. Since is alwas a positive number, the range of the function is f () > 0. The graph of the function gets close to the -ais, but does not meet or cross it, so the line = 0 is an asmptote of the function. B f () = (0.5) ( ) 3, 1 f () = ( 1, 1 ) (, 1 ) In the function f () = (0.5), b = 0.5. Since 0 < b < 1, this is an eponential deca function. Step 1 Make a table of values. Since this is an eponential deca function of the form f () = b, ou know that the -intercept is 1 and that the point (1, 0.5) lies on the graph. 10 (, ) (3, ) (1, ) (0, 1) = 0 3 1 0 1 3 f () 1 0.5 0.5 0.15 Step Plot the points from the table. Then connect the points in a smooth curve. Use the general shape of an eponential deca function as a guide. Since (0.5) is alwas a positive number, the range of the function is f () > 0. The -ais is an asmptote of the function. Notice that the graph of f () = (0.5) is a reflection across the -ais of the graph of f () =. ( 3, ) 10 f () = (0.5) (, ) (1, 0.5) (, 0.5) ( 1, ) (3, 0.15) = 0 (0, 1) GRAPHING EXPONENTIAL FUNCTIONS 37 Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Eponential Functions Graph Famil: f () = a b h + k Functions of the form f () = a b h + k are in the famil of eponential functions whose parent function is f () = b. The graph of f () = a b h + k intersects the -ais at (0, a b h + k) and has asmptote = k. The domain of the function is the set of real numbers, and the range, for b > 0, is the set of all real numbers greater than k. 1 1 10 10 f () = 3 = 0 f () = 3 + = f () = 3 = 0 f () = 1 3 f () = 3 = 0 f() = 3 Graph of f () = b, where b > 0. A change in the parameter h shifts the graph left or right h units. A change in the parameter k shifts the graph up or down k units. If a < 0, reflect the graph across the -ais. If a > 1, stretch verticall b a factor of a. If 0 < a < 1, shrink verticall b a factor of a. Graphing a Function in the Famil f() = a b h + k Eample Graph f () = 3 1 + 3. Then find its asmptote, domain, and range. Method 1 Make a table of values. To sketch the graph, use the ordered pairs and the knowledge that the function is an eponential growth function. 3 1 + 3 f () REMEMBER The values a, h, and k all have the same effect on the graph that the do in transforming the parent function p() = into the function f () = a( h) + k. 3 ( 1) + 3 = 3 3 + 3 = 3 + 3 3 3 1 3 ( 1 1) + 3 = 3 + 3 = 3 + 3 3 3 10 f () = 3 ċ 1 (, 9) 0 3 (0 1) + 3 = 3 1 + 3 = 3 + 3 1 3 (1 1) + 3 = 3 0 + 3 = 3 1 + 3 3 ( 1) + 3 = 3 1 + 3 = 3 + 3 9 ( ) 1, 3 3 (, 3 3 ) (1, ) ( 0, 1 ) = 3 3 UNIT 10 EXPONENTIALS AND LOGARITHMS Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Method Transform the graph of the parent function. In the function f () = 3 1 + 3, a = 3, b =, h = 1, and k = 3. The green graph is f () =. Since a > 0, the function f () = 3 1 + 3 does not get reflected across the -ais. Since a > 1, the parent function is stretched verticall b a factor of 3. The blue graph is the graph of the function f () = 3. The graph of f () = 3 is then translated horizontall b 1 unit (h) and verticall b 3 units (k). The black line shows the graph of f () = 3 1 + 3. Domain: the set of real numbers Range: the set of real numbers greater than 3 Asmptote: = 3 1 10 f () = 3 ċ 1 + 3 f () = f () = 3 ċ = 3 Using the Graph of a Function to Find an Equation for the Function Eample 3 Find the equation of the function shown in the graph, given the parent function f () = ( ). The graph intersects the -ais at the point (0, ) and has asmptote = 5, so k = 5. Find the unknown parameter a b substituting the point (0, ). f () = a ( ) 5 = a ( ) 0 5 = a 1 5 3 = a Check f () = 3 5 1 3 5 = _ 13 10 1 3 = 5 So f () = 3 ( ) 5 is an accurate equation for the function. 9 1 15 f () ( ) 1, 1 (0, ) GRAPHING EXPONENTIAL FUNCTIONS 39 Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Problem Set Use a table of values to graph each eponential function. 1. f () = 3. g() = 3. f () = ( 3 ). p(t) = 0.5 t 1 5. b() =. f (t) = 3 t + 7. g() = 0.3. r(t) = 11.9 t 9. h() = 5 1 Graph each function. Then find its domain and range, and the equation of its asmptote. 10. f () = 3 + 1 1. f () = + 3 3 1. f () = 0.5 (0.) + 5 11. f () = + 1 + 15. f () = 0.7 (0.1) + 13. f () = ( _ 10 1 ) 1 + Describe the effect each parameter has on the graph, given that the parent function is f () =. 1. f () = + 7 1. f () = 17. f () = 3 19. f () = 3 + 9 Find an equation of the function shown in the graph, given the parent function. 0. f () = + 5 7 1. f () = 5 10. f () =. f () = 3 = 0 ( 1, 1) (0, ) 10 10 ( 1, 1) (0, 3) = 0 3. f () = 5. f () = 0.5 10 10 ( 1, ) (0, ) ( 1, ) = 0 = 0 (0, 3) 370 UNIT 10 EXPONENTIALS AND LOGARITHMS Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
. f () = ( ) 9. f () = (0.) 10 (, 5) (0, ) = 1 10 = (1,.) ( 0, 7) 10 1 1 7. f () = 3 * 30. Challenge f () = ( ) (1, 5) (0, 1) = 1 10 1 = 3 (, 5) 10 1 1 (1, 15). f () = ( ) 31. f () = ( 3 ) 10 = 3 ( 0, ) ( 1, 7) 10 1 1 10 ( 1, 5) 10 (0, 1) = GRAPHING EXPONENTIAL FUNCTIONS 371 Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Applications: Growth and Deca Eponential equations can be used to solve man real-world problems. Mone in a bank and populations of people or bacteria often grow, while radioactive substances graduall lose their radioactivit. Eponential formulas can help ou solve man problems involving growth or deca. Solving Growth Problems EXPONENTIAL GROWTH FORMULA If a quantit is growing eponentiall from initial amount b for time t and with growth rate r, then the total amount is = b(1 + r ) t. Eample 1 Suki starts a bacterial culture in her Biolog class. She records her observations for the first several hours. How man bacteria should Suki epect to see after 5 hours, 10 hours, 15 hours, and hours? Hour 0 1 Bacteria (in thousands) 10 0 0 The initial amount is 10,000 bacteria, so b = 10. The bacteria are epected to grow at a rate of 100% each hour, so r = 1. Use the eponential growth formula, = b(1 + r) t. = b(1 + r) t THINK ABOUT IT At some point in time the bacteria will stop growing because of the limited amount of food in a petri dish. = 10(1 + 1) t = 10 t Hour 0 1 5 10 15 Bacteria (in thousands) 10 0 0 30 10,0 37,0 17,77,10 37 UNIT 10 EXPONENTIALS AND LOGARITHMS Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Eample In 000, the population of a countr was 10,3,7. The population is epected to grow at a rate of 0.% each ear. A Predict the population of the countr in 015. B In what ear will the population first eceed 1 million? A The initial population is 10,3,7, so b = 10,3,7. The growth rate is 0.%, so r = 0.00. The time period is ears, and the difference between 015 and 000 is 15 ears, so t = 15. Use the eponential growth formula. THINK ABOUT IT Populations can also decrease at a constant rate. In such a case, ou would use an eponential deca function. = b(1 + r) t = 10,3,7(1 + 0.00) 15 = 10,3,7(1.00) 15 = 10,3,7(1.9551) 11,53,059 If the growth rate of 0.% remains constant, the population in 015 will be approimatel 11,53,059. B Since the population is predicted to be 11,53,059 in 015, ou can predict that it will eceed 1 million sometime after 015. Plug in some values of t that are greater than 15 to test if the population becomes greater than 1 million. It seems reasonable to begin b testing t = 0. = 10,3,7(1 + 0.00) 0 = 10,3,7(1.00) 0 = 10,3,7(1.17703) 1,00, The population is just over 1 million in 00. Testing t = 19, ou can show that the population is 11,907,01, which is under 1 million, in 019. So the population first eceeds 1 million in 00. Solving Deca Problems EXPONENTIAL DECAY FORMULA If a quantit is decaing eponentiall from initial amount b for time t and with deca rate r, then the amount remaining is = b (1 r) t. Eample 3 The sales of a popular doll decreased between the ears 000 to 005, as shown in the table. If the decrease in sales continues at this rate, what will the sales be for 010 and 00? Year 000 001 00 003 00 005 Sales 1,9,330 1,1,515 1,350,0 1,3,001 1,1,79 1,15,037 APPLICATIONS: GROWTH AND DECAY 373 Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Step 1 Determine the deca rate. Divide the sales for each ear b the sales for the previous ear. The sales have declined at a constant rate of approimatel 5% ever ear, so r =.05. Step Use the eponential deca formula to determine the equation. The initial population is 1,9,330, so b = 1,9,330. From Step 1, r = 0.05. = b(1 r) t = 1,9,330(1 0.05) t = 1,9,330(0.95) t Step 3 Substitute 10 (for 010) and 0 (for 00) for t. = 1,9,330(0.95) t = 1,9,330(0.95) t = 1,9,330(0.95) 10 = 1,9,330(0.95) 0 95,90 53,13 The sales for 010 will fall to about $95,90, and the sales in 00 will fall to about $53,13. THINK ABOUT IT If the sales were increasing at a constant rate, ou would use the eponential growth formula: = b(1 + r) t. Radioactive Deca Ever substance that is radioactive loses particles or energ from the nucleus of its atoms. As a result, radioactive elements change from one kind of atom to another over time. This process of a radioactive substance changing from one kind of atom to another is called radioactive deca. DEFINITION The half-life of a radioactive substance is the length of time it takes for one half of the substance to deca. HALF-LIFE FORMULA The amount of a radioactive substance after t time periods, where b is the initial amount and h is the half-life, is = b ( ) t h. Eample Carbon-1 has a half-life of 5730 ears. Archaeologists find bone material from what seems to be a human leg. One of the archaeologists thinks the leg is 10,000 ears old. What percentage of normal carbon-1 levels should she epect to find in the leg? 37 UNIT 10 EXPONENTIALS AND LOGARITHMS Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
The half-life is 5730, so h = 5730. The time is 10,000 ears, so t = 10,000. We don t know the original amount, so just use 1 for b. The answer for will give us a value we can treat as a percentage. After 10,000 ears, the archaeologist should epect about 9.3% of normal carbon-1 levels. = ( ) t h = ( ) 10,000 5730 = ( ) 1.75 = 0.93 Compound Interest Using compound interest is one good wa to earn mone. It is interest that is earned on both the principle and on an interest that has alread been earned. COMPOUND INTEREST FORMULA The total amount A of an investment with initial principal P, earning compound interest at an annual interest rate r and compounded n times per ear for t ears, is given b the following formula A = P ( 1 + r n ) nt. THINK ABOUT IT Eponential equations and epressions can describe man real-world phenomena. Eample 5 Find the total amount after ears of an investment of $1500 in an account paing % interest, compounded for each of the given time periods. A annuall B quarterl The initial amount invested is $1500, so P = 1500. The interest rate is %, so r = 0.0. You are finding the balance after ears, so t =. When the interest is compounded annuall, n = 1. A = P ( 1 + r n ) nt A = 1500 ( 1 + 0.0 1 ) 1 A = 1500 (1.0) A = 1500 1.01 A = 1.0 The total amount after ears is $1.0. A quarterl event happens four times a ear, so n =. A = P ( 1 + r n ) nt A = 1500 ( 1 + 0.0 ) A = 1500 (1.01) A = 1500 1.0570 A 1.9 The total amount after ears is $1.9. APPLICATIONS: GROWTH AND DECAY 375 Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Problem Set Solve each growth problem. 1. In 00, the population of a countr was 15,500,31. The population is epected to grow at a rate of 0.% each ear. Predict the population of the countr in 01.. A colon of ants has 50,000 members. Sione thinks the colon will grow at a rate of 3.1% each hour. How man ants should he epect there to be in the colon after 10 hours? 3. Juan starts a bacteria culture and records the number of bacteria in the petri dish for the first couple of hours. How man bacteria can he epect to find after 17 hours? Hours 0 1 Bacteria (in thousands) 10 30 0. A population of insects grows eponentiall, as indicated in the table. If the increase in population continues at the same rate, at the end of what hour will the insect population first eceed 100,000? At End of Hour Insect Population 1 51,500 53,05 3 5,3 5,75 5 57,93 Solve each deca problem. 9. Filipe bought a classic car in the ear 000. He believes that the value of the car depreciates eponentiall at a rate of % each ear. If the car was originall valued at $70,000, what should Filipe epect the value of the car to be in 01? 5. Shae starts a bacteria culture b swabbing the door handle in her lab. She records the number of bacteria for the first several hours. How man bacteria should she epect to find after 10 hours? Hours 0 1 Bacteria (in thousands) 9. 19. 39.. In 00, the number of people infected with a virus was 1,97. Malakai estimates that the number of infected people will grow at a rate of 1.5% each ear. What does he epect the number of infected people will be in 010? 7. In 00, the population of a count was 1,57. The population is epected to grow at a rate of 0.% each ear. Predict the population of the count in 00. *. Challenge The table shows the number of U.S. citizens that developed a certain tpe of disease from 1990 to 00. If the rate of developing the disease grows eponentiall and continues to occur at the same rate, in what ear will the number of cases first eceed 500,000? Year Number of Cases 1990 10,000 1991 1,00 199 15,5 1993 1,59 199 151,50 10. The sales of large homes decreased between the ears 005 and 009, as indicated in the table. If the decrease in sales continues at this rate, what will the large-home sales be in 015? Year Large-Home Sales 005 1,90,000 00 1,3,00 007 1,79,7 00 1,9,93 009 1,30,75 37 UNIT 10 EXPONENTIALS AND LOGARITHMS Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
11. Zhang thinks that the value of a baseball card can be modeled b eponential deca and that the value will decrease at a rate of 0.% each ear. If the card was originall valued at $50 in 000, what should Zhang epect the value of the card to be in 011? 1. Archaeologists find a bone fragment of an animal that the epect became etinct over 0,000 ears ago. What is the maimum percentage of normal carbon-1 levels that the archaeologists should find in the bone if the are right? 13. In 000, the population of a countr was 5,300,97. The population is epected to decrease at a rate of % each ear. Predict the population of the countr in 00. 1. Jessln hpothesizes that the value of antique furniture can be modeled b eponential deca and that the value of a particular piece will decrease at a rate of 0.15% each ear. If the furniture was originall valued at $15,000 in 1990, what should Jessln epect the value of the furniture to be in 010? Solve each compound interest problem. 17. Find the total amount after 10 ears of a compound interest investment of $0,000 at 5.5% interest, compounded annuall. 1. Find the total amount after 5 ears of a compound interest investment of $70,000 at % interest, compounded twice a ear. 19. Find the total amount after 15 ears of a compound interest investment of $0,000 at.% interest, compounded monthl. 0. Find the total amount after 5 ears of a compound interest investment of $50,000 at 3.5% interest, compounded quarterl. 1. Find the total amount after 5 ears of a compound interest investment of $10,000 at 3% interest, compounded monthl. 15. Uranium-3 has a half-life of about.7 billion ears. A geologist believes that a rock she found on the floor of the ocean is 0.1 billion ears old. What percentage of normal uranium-3 levels should she epect to find in the rock? * 1. Challenge The sales of SUVs decreased between the ears 00 and 009, as indicated in the table. If the decrease in sales continues at this rate, in what ear will the sale of SUVs be 0,000? Year SUV Sales 00 1,500,000 005 1,5,000 00 1,353,750 007 1,,0 00 1,1,759. Find the total amount after 30 ears of a compound interest investment of $100,000 at 5% interest, compounded annuall. * 3. Challenge The total amount of $53.5 was earned on an investment of $5000, compounded quarterl for 5 ears. What interest rate did the account pa? *. Challenge Determine the principal for an account paing 3% interest, compounded annuall for 5 ears and with the total amount of $377.. APPLICATIONS: GROWTH AND DECAY 377 Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Logarithms Just as taking a square root is the opposite of squaring a number, the opposite of raising a base to a variable power is called taking a logarithm. A logarithm is the eponent to which a base would have to be raised to result in a given value. Using logarithms, ou would write = 10 as = log 10. DEFINITION The logarithm of a with base b, log b a, where b > 0, b 1, and a > 0, is defined as follows: log b a = if and onl if b = a. NOTATION The epression log b a is read as log base b of a. Converting Between Logarithmic and Eponential Forms The definition of a logarithm can be used to convert between eponential and logarithmic forms. Eponential Form Logarithmic Form = 1 log 1 = Eample 1 The table below shows equivalent equations written in both eponential and logarithmic forms. Eponential Form 5 = 5 3 = _ 1 1 = 9 10 0 = 1 Logarithmic Form log 5 5 = log _ 1 = 3 log 9 = 1 log 10 1 = 0 37 UNIT 10 EXPONENTIALS AND LOGARITHMS Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Logarithmic Properties Man properties of logarithms are similar to properties of eponents. PROPERTIES OF LOGARITHMS Let b > 0, b 1, m > 0, n > 0, and p be an real number. Propert Product Propert Quotient Propert Power Properties Statement log b (mn) = log b m + log b n log b ( m n ) = log b m log b n log b m p = p log b m and log b ( n m ) = log m b n Eample A Write log 3 + log 3 as a single logarithm. log 3 + log 3 = log 3 ( ) Product Propert of Logarithms = log 3 3 Simplif. B Simplif log 10 log 5 + 3 log. log 10 log 5 + 3 log = log ( _ 10 5 ) + 3 log Quotient Propert of Logarithms = log + 3 log Simplif. = log + log 3 Power Propert of Logarithms = log + log Simplif. = log ( ) Product Propert of Logarithms = log 1 Simplif. = C Write log 7 (3 9 ) in epanded form. log 7 (3 9 ) = log 7 3 + log 7 9 Product Propert of Logarithms = log 7 3 + 9 log 7 Power Propert of Logarithms TIP Remember that a logarithm is an eponent. To evaluate log 1, ask, To what power do I raise in order to get a result of 1? LOGARITHMS 379 Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
D Given that log 5.319 and log 3 1.550, use logarithm _ 3 properties to estimate log 10. _ 3 log 10 = log 3 log 10 Quotient Propert of Logarithms = log 3 log ( 5) Factor 10. = log 3 (log + log 5) Product Propert of Logarithms 1.550 (1 +.319) Substitute. 1.550 3.319 Simplif. 1.739 Simplif. THINK ABOUT IT log b b = 1 since b 1 = b. Using a Calculator to Find Common and Natural Logarithms Logarithms with special bases have names. DEFINITIONS A common logarithm is a logarithm with base 10. Common logarithms, such as log 10, are usuall written without the base, as log. A natural logarithm is a logarithm with base e. Natural logarithms, such as log e, are often written using the notation ln. Calculators are programmed to calculate common and natural logarithms. Eample 3 Use a calculator to evaluate each logarithm to four decimal places. A log B ln 53 On a graphing calculator, On a graphing calculator, press LOG, tpe ), and press press LN, tpe 53), and press ENTER. ENTER. log 0.9031 ln 53 3.9703 log ().9030997 ln (53) 3.9709191 REMEMBER Like π, the number e is an irrational number. THINK ABOUT IT Pietro Mengoli and Nicholas Mercator called the natural logarithm logarithmus naturalis. TIP e.71... The LOG ke on a calculator is usuall a common logarithm with base 10. The LN ke on a calculator is a natural logarithm with base e. 30 UNIT 10 EXPONENTIALS AND LOGARITHMS Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Changing Bases of Logarithms The change of base propert can be applied to evaluate a logarithm when the base is neither 10 nor e. = log b a b = a log n b = log n a log n b = log n a _ = log a n log n b _ log b a = log a n log n b Rewrite in eponential form. Take the logarithm with base n of both sides. Power Propert of Logarithms Division Propert of Equalit Substitute for. This result gives ou the logarithmic change of base propert. LOGARITHMIC CHANGE OF BASE PROPERTY For positive real numbers a, b, and n, where n 1, b 1, and a > 0, log b a = _ log a n log n b. Eample Evaluate log 5 0 to four decimal places. Use the change of base propert. Then enter the epression in a calculator. Changing the base to 10 or e gives the same result: Base 10: Base e: log 0 ln 0 log 5 0 = log log 5 5 0 = _ ln 5 1.1 1.1 Problem Set Convert each epression to logarithmic form. 1. 3 = 1. 3 3 = _ 7 1 3. 3 =. 9 = 7 Convert each epression to eponential form. 5. log = 3. log 1 5 5 = 7. log 1 = 1 1. log 3 = 7 3 LOGARITHMS 31 Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Write as a single logarithm. 9. log 3 + log 5 10. log + log 11. log 1 log 1. log 5 1 log 5 7 13. log 3 15 log 3 3 + log 3 1. log 7 1 log 7 + log 7 3 Write in epanded form. 15. log 9 ( 5 ) 1. log (7 3 ) 17. log 5 ( ) Estimate each epression, given that log 3 = 0.309, log 3 5 = 1.50, and log 3 = 1.309. 1. log 3 19. 5 log 3 3 0. log 3 30 Evaluate each logarithm to four decimal places. 1. log 1. ln 10. log 19 5. ln 17 3. log. ln 7. log 7 17. log 3 9. log 9 30 Solve. * 30. Challenge Simplif ln + ln. * 31. Challenge Evaluate log 3 ( _ 7 9 ). 3 UNIT 10 EXPONENTIALS AND LOGARITHMS Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Using Logs to Solve Eponential Equations You can solve an eponential equation such as = simpl b rewriting it as = 3, so = 3. However, to solve an eponential equation such as = 10, ou have to use a logarithm. Just as ou use addition, multiplication, or square roots to transform and solve equations, ou can use logarithms to transform and solve equations. Using Logarithms to Solve Eponential Equations PROPERTY OF EQUALITY FOR LOGARITHMIC EQUATIONS If a,, and are positive numbers and a 1, then log a = log a if and onl if =. Eample 1 A = 10 Solve. Round our answer to four decimal places. = 10 log = log 10 Take the common (base 10) logarithm of both sides. log = log 10 Power Propert of Logarithms log = 1 log 10 10 = 1 because 10 1 = 10. _ = 1 Divide both sides b log. log 3.319 Use a calculator to approimate the epression. Check for reasonableness: Because 3 = and = 1, it is reasonable that 3.319 10. You ma use a calculator to verif the result. TIP It is customar to round the value of a logarithmic epression to four decimal places. USING LOGS TO SOLVE EXPONENTIAL EQUATIONS 33 Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
B 5 n = 101 5 n = 101 log 5 n = log 101 Take the common logarithm of both sides. n log 5 = log 101 Power Propert of Logarithms _ log 101 n = Divide both sides b log 5. log 5 n 1.33 Use a calculator to approimate the epression. When an eponential equation contains the number e, it can be easier to take the natural logarithm of both sides to solve the equation. Eample Solve each equation b using logarithms. Round our answer to four decimal places. A e 3a = 9 e 3a = 9 ln e 3a = ln 9 Take the natural logarithm of both sides. 3a ln e = ln 9 Power Propert of Logarithms 3a = ln 9 ln e = 1 because e 1 = e. a = ln 9 3 Divide both sides b 3. a 0.73 Use a calculator to approimate the epression. Check To verif the result, substitute the approimate value in the original equation. Use a calculator to evaluate the left side of the equation. e 3a = 9 e 3 0.73? 9.999 9 The solution a 0.73 is correct. B 7 = e + 1 Method Using LN Method Using LOG 7 = e + 1 7 = e + 1 ln 7 = ln e + 1 log 7 = log e + 1 ln 7 = ( + 1) ln e log 7 = ( + 1) log e ln 7 = ( + 1) 1 log 7 = log e + 1 log e ln 7 = + 1 log 7 log e = log e ln 7 = 1 (log 7 log e) = log e ( ( ln 7 ) 1 ) = 1 _ log e = log 7 log e _ = 1 (ln 7) 1 1.057 1.057 TIP Alwas solve the equation thoroughl before approimating the value of the variable. The equation log 101 n = _ gives the eact log 5 value of n; n 1.33 gives an approimation. TIP The equations in Eample 1 could also have been solved using the natural logarithm. For an equation with a variable in the eponent, ou can choose to use either the natural or the common logarithm. 3 UNIT 10 EXPONENTIALS AND LOGARITHMS Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Check To verif the result, substitute the approimate value in the original equation. Use a calculator to evaluate each side of the equation. 7 = e + 1 7 1.057? e 1.057 + 1 7. 7.0 The calculator gives the same value, if rounded to three decimal places, for each epression. Interpreting a Calculator Error Message Eample 3 Solve 10 + =. 10 + = 10 = Subtract from both sides. log 10 = log ( ) Take the common logarithm of both sides. log 10 = log ( ) Power Propert of Logarithms = log ( ) log 10 = 1 = Error Message The calculator gives an error message indicating that computing log ( ) is not possible. The reason is that it is not possible to take the logarithm of a negative number. In this case, there is no real solution to the equation. THINK ABOUT IT The statement 10 = indicates that the original equation has no real solution. It is not possible for an power of 10 to be negative. Problem Set Solve. Find each eact answer and use a calculator to approimate answers that involve logarithms to four decimal places. 1. 3 = 10. c = 100 3. 5 = 1000. 10 = 1 100 5. 7 h =. 10 + 13 = 10 7. 7 p = 50. 3 = 0 9. = 110 10. 9 = 7 11. 3 5z + 1 = 0 1. 19 + = 13. e = 1 1. e = 5 15. e 3 = 1. 3e = 1 17. e 5 + 1 = 100 1. 9e r + 3 = 5 19. = e + 0. t = e t 1 1. 0 = e + 3. = e 3 + 5 3. 9 3b = e 1 b. 3 + 1 = e 9 + 5 5. 3 = e. 3 n = e n + 7. 5 3 = e 5. 7 = e + 1 * 9. Challenge e 13 e 5 = 0 * 30. Challenge = e 1 USING LOGS TO SOLVE EXPONENTIAL EQUATIONS 35 Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Solving Logarithmic Equations You can use logarithmic and eponential properties to solve logarithmic equations. You can use the properties of equalit for eponential and logarithmic equations to solve some tpes of logarithmic equations. REMEMBER The propert of equalit for eponential equations states the following: If a is a positive number other than 1, then a = a if and onl if =. Solving Logarithmic Equations b Rewriting in Eponential Form If ou can write an equation in the form log b a = c, then ou can use the definition of logarithms to write the equivalent eponential equation. This strateg can be helpful for some problems. Eample 1 Solve and check. A log = B log 7 ( + 1) = log = = Definition of a logarithm = 3 Rewrite as 3. = 3 Propert of Equalit for Eponential Equations Check log 3 3 = The solution = 3 is correct. log 7 ( + 1) = 7 = + 1 Definition of a logarithm 9 1 = Subtraction Propert of Equalit = Subtract. Check log 7 ( + 1) log 7 9 7 9 9 = 9 The solution = is correct. 3 UNIT 10 EXPONENTIALS AND LOGARITHMS Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
C log ( + ) + log ( 5) = 3 log ( + ) + log ( 5) = 3 log ( + )( 5) = 3 Product Propert of Logarithms ( + )( 5) = 3 Definition of a logarithm 3 10 = Simplif. 3 1 = 0 Subtraction Propert of Equalit ( )( + 3) = 0 Factor the trinomial. = or = 3 Use the zero product propert to solve for. Check Substitute for : log ( + ) + log ( 5) 3 log + log 1 3 log 1 3 log = 3 The onl solution is =. Substitute 3 for : log ( 3 + ) + log ( 3 5) 3 log ( 1) + log ( ) 3 log ( 1) and log ( ) are undefined, so 3 is an etraneous solution. TIP Alwas check for etraneous solutions when ou solve logarithmic equations. Solving Logarithmic Equations b Using the Logarithmic Propert of Equalit If ou can transform an equation into the form log b a = log b c, then ou can use the logarithmic propert of equalit to write it as a = c. Eample Solve and check. A log 5 ( + ) log 5 = log 5 log 5 ( + ) log 5 = log 5 log 5 ( _ + ) = log 5 + _ = + = = 3 = Check log 5 ( + ) log 5 log 5 log 5 log 5 log 5 log 5 ( ) log 5 log 5 = log 5 SOLVING LOGARITHMIC EQUATIONS 37 Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
B log 10 r = log 10 log 10 r = log 10 log 10 r = log 10 log 10 r = log 10 r = r = ± Check Substitute for r: log 10 log 10 log 10 ( ) log 10 log 10 = log 10 The onl solution is = =. Substitute for r: log 10 ( ) log 10 log 10 ( ) is undefined, so is an etraneous solution. Problem Set Write each equation in eponential form and then solve. Check for etraneous solutions. 1. log 3 1 =. log 3 = 3. log = a. log 15 1 = n 5. log 5 5 =. log _ 1 7 9 = 7. log ( + ) = 5. log (t + ) = 3 9. log ( + ) = 10. log 3 (g 5) = 5 Solve. Check for etraneous solutions. 11. log 1 (r 5) + log 1 (r + 3) = 1. log ( ) = 13. log ( 5 v 1v ) = 1. log (k + 5) = 3 15. log 1 ( + ) log 1 ( 3 ) = 1. log (c 1) + log (c) = 7 17. log + log ( + ) = 1. log (9z + 5) log (z + ) = 1 19. log + log = 3 0. log 3 (7 ) log 3 ( ) = 3 1. log ( + 3) = log (7). log (h 11) = log (h + 3) 3. log 7 q + log 7 q = log 7 3. log 5 (5 1) log 5 ( + 1) = log 5 5. log 9 b = log 1 9. log ( 5) = 3 log 79 7. 3 log = log 10 10. log 9 (n 5) + log 9 (n + ) = 5 3 log 3 9 * 9. Challenge log 5 15 = w * 30. Challenge log 3 5 log 3 ( 15) = log 3 5 3 UNIT 10 EXPONENTIALS AND LOGARITHMS Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Graphing Logarithmic Functions A logarithmic function is an equation of the form f () = log b ( h) + k, where h > 0, b > 0, and b 1. Since f ( g()) = f (b ) = log b b = and g( f ()) = g(log b ) = b log b =, the logarithmic function f () = log b and the eponential function g() = b are inverses of each other. REMEMBER Two functions f and g are inverses of each other if and onl if f g = and g f =. Graphing Logarithmic Functions The function f () = log b is the parent function for logarithmic functions. Logarithmic Functions of the Form: f () = log b f () = log b, when b > 1. f () = log b, when 0 < b < 1. End behavior: As increases in value, f () slowl increases in value. As decreases in value, f () decreases in value. 10 = 0 (0, 1) g() = b (1, 0) f() = log b = 0 10 End behavior: As increases in value, f () decreases in value. As decreases in value, f () increases in value. g() = b = 0 (0, 1) (1, 0) f () = log b = 0 The following properties are common to both tpes of logarithmic functions: General shape: f () = log b is the inverse of g() = b. Notice that the graph of f () = log b is the reflection of g() = b across the line =. Asmptote: As decreases in value, the function nears the -ais but never meets or crosses it, so the line = 0 (the -ais) is an asmptote of the graph. Domain: all real numbers greater than 0 Range: the set of all real numbers -intercept: 1 GRAPHING LOGARITHMIC FUNCTIONS 39 Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Eample 1 Graph each of the following. A f () = log In the function f () = log, b =. Step 1 Make a table of values. Include a few positive and negative values, as well as zero. The -intercept is 1, and the point (, 1) lies on the graph. 1 THINK ABOUT IT f () = log is the inverse of g() =, so ou know that f ( g() ) = log = and g ( f () ) = log =. Step Plot the points from the table. Then connect the points in a smooth curve. As a guide, use the general shape of the parent function when b > 1. Notice that the graph gets closer to the -ais as the values of get smaller. So the line = 0 is an asmptote of the function f () = log. B f () = log f() 3 1 0 1 3 In the function f () = log, b =. Step 1 Make a table of values. The -intercept is 1, and the point (, 1 ) lies on the graph. f () = log = 0 (, 3) (, ) (1, 0) (, 1) 10 1 1 1 (, 3 1 ) (, 1 ) 1 (, ) THINK ABOUT IT Since f () = log is the inverse of g() =, if (, 3 ) is a point on the graph of f () = log, then ( 3, ) lies on the graph of g() =. In general, if (, ) is a point on f () = log, then (, ) is a point on g() =. 1 f() 3 1 0 1 3 Step Plot the points from the table. Then connect the points in a smooth curve. As a guide, use the general shape of the parent function when 0 < b < 1. 390 UNIT 10 EXPONENTIALS AND LOGARITHMS Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Notice that the graph gets closer to the -ais as the values of get smaller. So the line = 0 is an asmptote of the function f () = log. The graph of f () = log is a reflection across the -ais of the graph of g() = log. f () 3 1 1 1 3 5 7 9 1 3 5 = 0 f () = log 1 You can use the change of base formula to write an logarithmic function using common or natural logarithms. This makes it easier to check our graphs on a graphing calculator. For eample, to graph f () = log, graph _ = log log. Graphing a Function in the Famil f() = log b ( h) + k Logarithmic Functions Graph Famil: f () = log b ( h) + k Functions of the form f () = log b ( h) + k are in the famil of logarithmic functions whose parent function is g() = log b. = 0 g() = log (1, 0) = = 1 f () = log ( + ) = 0 f () = log + 1 (1, 1) 10 10 f () = log ( 1) (1, ) f () = log Parent function If h < 0, the graph of the parent function is translated to the left. If h > 0, it is translated to the right. If k < 0, the graph of the parent function is translated down. If k > 0, it is translated up. The graph of f () = log b ( h) + k has asmptote = h. The domain of the function is the set of real numbers greater than h, and the range is the set of all real numbers. GRAPHING LOGARITHMIC FUNCTIONS 391 Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Eample Graph f () = log ( + 1) 3. Then find its domain, its range, and the equation of its asmptote. Method 1 Make a table of values. log ( + 1) 3 f() 7 log ( 7 + 1 ) 3 = log 3 = 3 3 log ( + 1 ) 3 = log 3 = 1 3 TIP After = 1, the function increases much more slowl than =. So its end behavior is different from a parabola or a line. 0 log (0 + 1) 3 = log 1 3 = 0 3 3 1 log (1 + 1) 3 = log 3 = 1 3 7 log (7 + 1) 3 = log 3 = 3 3 0 To sketch the graph, use the ordered pairs and the knowledge that the function is a logarithmic function. = 1 f () = log ( + 1) 3 10 Method Transform the graph of the parent function. In the function f () = log ( + 1) 3, b =, h = 1, and k = 3. g() = log is shown in red on the graph. The graph of f () = log ( + 1) 3 gets translated down b 3 units and left b 1 unit. The domain of the function f () = log ( + 1) 3 is the set of real numbers greater than 1. The range is the set of all real numbers. The asmptote is the line = 1. f () g () = log = 1 1 1 3 f () = log ( + 1) 3 5 7 9 10 11 1 13 39 UNIT 10 EXPONENTIALS AND LOGARITHMS Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Using the Graph of a Function to Find the Equation for the Function Eample 3 What is the equation of the function shown in the graph? The asmptote is at =, so ou know the graph has been translated two units to the right. This means that h =. f () = log b ( ) + k Now, substitute the ordered pair (3, ). = log b (3 ) + k = log b 1 + k But the log of 1 is alwas zero, so ou know that k has to be : f () = log b ( ) + Now, substitute the other ordered pair to find the value for b. 10 9 7 5 3 1 f () 5 = log b (5 ) + 1 = log b 3 b 1 = 3 b = 3 So, the equation is f () = log 3 ( ) +. = (3, ) 1 (5, 5) 10 1 1 THINK ABOUT IT Changing the parameter b results in different end behavior. When b > 1, f () increases/decreases more slowl as b increases. When b < 1, f () increases/decreases more quickl as b increases. Problem Set Graph each logarithmic function. 1. f () = log 3. f () = log 3. f () = log _ 1 10. f () = log 5 5. f () = log 3. f () = log 5 + Graph each function. Then find its domain, its range, and the equation of its asmptote. 7. f () = log ( 1) +. f () = log 3 ( + ) 1 9. f () = 3 log ( + ) + 10. f () = log ( 3) 5 5 Describe the effect each parameter has on the graph, given that the parent function is f () = log. 13. f () = log ( + 9) 1. f () = log ( ) + 10 15. f () = log ( + 5) 3 1. f () = log ( 3) 7 11. f () = log ( + 3) + 1 1. f () = 3 log ( 5) 5 17. f () = 3 log ( + 11) + 1. f () = 3 log () GRAPHING LOGARITHMIC FUNCTIONS 393 Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Find the equation of the function shown in the graph. The parent function is given. 19. f () = log 3. f () = log 10 = 0 (, ) (1, 3) = 3 10 (, ) (3, 3) 0. f () = log 5. f () = log ( ) = (3, 1) ( 1, 0) = (, 0) 10 1 (, 1) 1. f () = log 3 * 5. Challenge f () = log = 7 (, 0) 10 1 1 1 1 (1, ) = 3 ( 1, ) (5, 0) 10 1. f () = log 3 ( + 1) *. Challenge f () = log b = 1 (, 0) 10 1 1 (0, ) = 0 (1, 3) (7, ) 10 1 39 UNIT 10 EXPONENTIALS AND LOGARITHMS Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Applications: Logarithms Logarithmic functions have applications in man scientific fields. Chemistr Application: ph The ph of a substance indicates its acidit. Substances that are more acidic have a lower ph than substances that are less acidic. The logarithmic function f () = log 10 models the ph of different substances. Specificall, the function ph = log 10 [H + ] measures the ph of a substance, where [H + ] is the hdrogen-ion concentration of the substance, measured in moles per liter. Eample 1 The ph of a brand of apple juice is 3.5. What is the [H + ] for the apple juice? Use the logarithmic function ph = log 10 [H + ]. The ph level of the juice is 3.5. ph = log 10 [H + ] 3.5 = log 10 [H + ] Substitute 3.5 for ph. 3.5 = log 10 [H + ] Multipl both sides b 1. [H + ] = 10 3.5 Definition of a logarithm There are eactl 10 3.5 moles of hdrogen ions in a liter of apple juice whose ph level is 3.5. B using a calculator and rounding the result to four decimal places, ou can determine that [H + ] is approimatel 0.0003 mole per liter. THINK ABOUT IT Pure water has a ph level of 7.0, which is considered neutral on the ph scale. You can use ph paper to test whether a substance is acidic (below 7.0 on the ph scale), alkaline (above 7.0), or neutral. TIP The value in the calculator window ma read 3.177E, which means 3.177 10. Biolog Application: Eponential Growth Under the right conditions, the bacteria grow eponentiall according to this formula: N t = N 0 e ct, where N 0 is the initial number of bacteria (at time t = 0), t is the number of das, and N t is the number of bacteria after t das. The constant c is the growth factor for a particular tpe of bacteria in a given environment. Eample At the beginning of an eperiment, a culture has 00 bacteria. Three das later, the culture has 000 bacteria. A Calculate the growth factor. B Predict the number of bacteria das after the beginning of the eperiment. THINK ABOUT IT This formula is the same as the one used in Applications: Growth and Deca, but written differentl: N t = N 0 (e c ) t = b(1 r) t APPLICATIONS: LOGARITHMS 395 Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
A First, find the value of c b substituting the given data: N 0 = 00, t = 3, N t = 000. N t = N 0 e ct 000 = 00 e 3c _ 000 00 = e3c Divide both sides b 00. 5 = e 3c ln 5 = ln e 3c Take the natural logarithm of both sides. ln 5 = 3c Power Propert of Logarithms ln 5 3 = c Divide both sides b 3. The growth factor of the function is ln 3 5 B Substitute the value for c to predict the number after das: N t = N 0 e ct N t = 00 e ln 5 3 N t 9,0 Use a calculator and round our answer. The prediction is that there will be about 9,0 bacteria das after the beginning of the eperiment. TIP Substitute the eact value ( ln 3 5 = c ) in the formula. Round onl at the end of the calculation. Phsics Application: Decibels The decibel scale measures the relative intensit R of a sound to the human ear. The threshold of hearing I 0 is the intensit of the faintest sound that can be heard b the human ear. Other sound intensities I are often measured b comparing them to the threshold of hearing. The logarithmic function R = 10 log I 10 I can be used to find the relative intensit of a sound in decibels. 0 Eample 3 The sound intensit of a vacuum cleaner is about 10 times greater than the threshold of hearing. What is the relative intensit of a vacuum cleaner, measured in decibels? Use the logarithmic function R = 10 log I 10 I. Since the sound 0 of the vacuum cleaner is 10 times greater than the threshold of hearing, I = 10 I 0. Substitute 10 I 0 in the equation for I: R = 10 log I 10 I 0 R = 10 log 10 I 0 10 I Substitute 10 I 0 for I. 0 R = 10 log 10 10 Simplif b using eponential form. R = 10 R = 0 The relative intensit of a vacuum cleaner is about 0 decibels. 39 UNIT 10 EXPONENTIALS AND LOGARITHMS Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.
Seismolog Application: Richter Scale The Richter scale is not linear, which means that an earthquake of magnitude does not release twice as much energ as one of magnitude. Eample An earthquake in the region of the Dominican Republic measured 3. on the Richter scale, while an earthquake near New Zealand measured.. Compare the amount of energ released b these two earthquakes. Use M = 3 log _ E to calculate the amount of energ 10 11. 10 released b each earthquake. Dominican Republic earthquake: New Zealand earthquake: 3. = 3 log _ E 10 10. = 11. 3 log _ 10 E 10 11.. = log 10 _ E 10 9. = log _ E 11. 10 10 11. 10. _ = E 10 10 11. 9. _ = E 10 11. 10. 10 11. = E 10 9. 10 11. = E 10 1. = E 10 1. = E 3.910 10 1 E.5119 10 1 E To compare the stronger quake to the weaker one, write the fraction: Think about it An earthquake with magnitude between 3.5 and 5. on the Richter scale is felt, but rarel causes major damage. One of magnitude 7 or more is major and can cause serious damage..5119 10 1 _.5119 10 = 5 3.910 10 1 3.910.31 10 The. quake released roughl 3,100 times the energ of the 3. quake. Nuclear Application: Calculating Half-Life The function f (t) = b ( ) t h gives the amount remaining, in milligrams, of a radioactive sample of b milligrams that has a half-life h after a period of time t. Eample 5 A sample of 50 mg of technetium-99m is given to a patient during a medical test. If the radioactive substance decas to. mg after 5 hours, find the half-life of this substance. t f (t) = b h. = 50 ( _ 5 h ) Substitute the given values. 0.05 = ( _ 5 h ) Divide both sides b 50. log 0.05 = log ( _ 5 h ) Take the common logarithm of both sides. _ 5 log 0.05 = h log Power Propert of Logarithms log h = 5 h Multipl both sides b log 0.05 log 0.05. Remember The half-life of a radioactive substance is the time it takes for half of the substance to deca. TIP The units for h and t must be the same. Applications: Logarithms 397 Copright 009, K1 Inc. All rights reserved. This material ma not be reproduced in whole or in part, including illustrations, without the epress prior written consent of K1 Inc.