Chapter 7: Logarithmic Functions Section 7.1 Chapter 7: Logarithmic Functions Section 7.1: Eploring Characteristics of Logarithmic Functions Terminology: Logarithmic Functions: A function of the form: y = a log b Where a 0, b > 0, b 1, and a and b are real numbers. Note: a is the coefficient, b is the base, and is the argument. Note: log 10 = log as logarithms are base 10 by default. Logarithmic Functions and Their Characteristics Graph the following eponential function and interpret its characteristics. (a) f() = log 10 Table: 1 0 1 3 5 7 8 9 10 f() Graph: y - 8 10 - - - - 8 Characteristics: 1. Number of -intercept:. Coordinates of y-intercept: 3. End Behaviour:. Domain: 5. Range:. Number of Turning Points: 158
Chapter 7: Logarithmic Functions Section 7.1 (b) f() = log 10 Table: 1 0 1 3 5 7 8 9 10 f() Graph: y - 8 10 - - - - 8 Characteristics: 7. Number of -intercept: 8. Coordinates of y-intercept: 9. End Behaviour: 10. Domain: 11. Range: 1. Number of Turning Points: 159
Chapter 7: Logarithmic Functions Section 7.1 (c) j() = log 10 Table: 1 0 1 3 5 7 8 9 10 f() Graph: y - 8 10 - - - - 8 Characteristics: 13. Number of -intercept: 1. Coordinates of y-intercept: 15. End Behaviour: 1. Domain: 17. Range: 18. Number of Turning Points: NOTE: A logarithm with no base shown is assumed to be base ten. Hence, log 10 = log. The same is true of what is known as a natural log, log e = ln. 10
Chapter 7: Logarithmic Functions Section 7.1 Comparing Eponential and Logarithmic Functions On the graph below, graph the following: y = 10 y = log 10 5 3 1 0 1 3 5 f() 1 0 1 3 5 7 8 9 10 f() y 10 5-10 - 5 5 10-5 - 10 11
Chapter 7: Logarithmic Functions Section 7.1 Matching Graphs and Equations Which function matches each graph below. Provide your reasoning. i) y = 5() ii) y = (0.1) iii) y = log iv) y = ln 8 y y 5 3 1 - - 3 - - 1 1 3 - - - 3 - - 1 1 3-1 - 8 y y - - 3 - - 1 1 3 - - - 3 - - 1 1 3-1
Chapter 7: Logarithmic Functions Section 7. Section 7.: Evaluating Logarithmic Epressions Connecting Logarithms and Eponents Use the function y = log to complete the table of values below. Then epress the function in an eponential form Amplitude of Vibrations Measured by Seismograph in eponential form base 10 Amplitude of Vibrations Measured by Seismograph, (units) 10 0 10 1 10 10 3 10 10 5 10 1 10 100 1000 10 000 100000 1000000 Richter Scale Magnitude, y y = log Conversions Between Eponential and Logarithmic Forms An eponential function of the form y = b is the equivalent to the logarithmic function = log b y and vice versa. Convert each of the following: y = b = log b Eponential Form 81 = 10 y 5 = e y 3 = 10 y Logarithmic Form y = log 1 y = log y = log 3 7 13
Chapter 7: Logarithmic Functions Section 7. Estimating and Evaluating Logarithmic Epressions Given a logarithmic epression, we can estimate it s value on a calculator using the property: y = log b = log() log(b) Determine the value of each of the following: (a) log 1 (b) log (c) log 3 7 (d) log 3 ( 1 7 ) (e) log1 (f) log (g) log 5 ( 1 5 ) 1
Chapter 7: Logarithmic Functions Section 7. (h) log 1 + log (i) log 1 log ( 1 8 ) (j) log log 8 (k) 81 = 10 y (l) 5 = e y (m) 3 = 10 y 15
Chapter 7: Logarithmic Functions Section 7. PH Scale Questions The ph scale in chemistry is used to measure the acidity of a solution. The ph scale is logarithmic, with base 10. A logarithmic scale is useful for comparing numbers that vary greatly in size. The ph, p(), is defined by the equation: p() = log Where the concentration of hydrogen ions,, in a solution is measured in moles per litre (mol/l). (a) The hydrogen ion concentration,, of a solution is 0.0001 mol/l. Calculate the ph of the solution. (b) The hydrogen ion concentration,, of a solution is 0.0000001 mol/l. Calculate the ph of the solution. (c) The ph of lemon juice is. Determine the concentration of hydrogen ion for lemon juice. 1
Chapter 7: Logarithmic Functions Section 7. (d) The ph of baking soda is 9. Determine the concentration of hydrogen ion for baking soda. (e) In terms of hydrogen ion concentration, how much more acidic is solution A with a ph of 1., than solution B with a ph of.5? Round your answer to the nearest tenth. 17
Chapter 7: Logarithmic Functions Section 7.3 Section 7.3: Laws of Logarithms Laws of Logarithms 1. Addition of Logarithms When two logarithms with the same base are added, their arguments are multiplied. log b (X) + log b (Y) = log b (X Y). Subtraction of Logarithms When two logarithms with the same base are subtracted, their arguments are divided. log b (X) log b (Y) = log b ( X Y ) 3. Power Law of Logarithms When an argument has an eponent, that eponent can be written as a coefficient of a logarithm and vice versa. log b (X c ) = c log b X Using Laws of Logarithms to Simplify and Evaluate an Epression E. Simplify and then evaluate each logarithmic epression. (a) log 5 + log. (b) log 5 100 log 5 (c) log 3 7 5 (d) log 8 log 3 18
Chapter 7: Logarithmic Functions Section 7.3 E. Write each epression as a single logarithm, and then evaluate. (a) log 3 18 + log 3 ( 3 ) (b) log 5 0 3 log 5 (c) log 3 + log 3 ( 3 ) (d) log 1 ( 1 ) log 9 (e) log 5 3 + log 5 9 1 3 log 5 (f) 1 log 3 (log 3 8 log 3 1) 19
Chapter 7: Logarithmic Functions Section 7. Section 7.: Solving Equations using Logarithms How to Solve an Eponential Equation Using Logarithms E1. Solve 1. Add or subtract any constants in the equation if any eist. Divide by any coefficient on the same side as the eponent and base 3. Ensure that there is only one base on each side of the equation. Convert to logarithmic form 5. Use the estimation method to evaluate the logarithm: log b A = log b. Multiply through by any denominator that eists from the original eponent. (a) 3() = 15 Solution: 3() 3 = 15 3 Divide both sides by the coefficient of 3 () = 5 Now that there is one base on each side convert to log form =log 5 By converting to log form we now have an equation that is rearranged to solve for. We will now apply the estimation rule to evaluate. = log(5) log() =.319 log A Put this into your calculator to determine the estimated value Always give your answer to decimal places when dealing with logarithms 170
Chapter 7: Logarithmic Functions Section 7. (b) 180 = 5(1.5) Solution: 180 = 5(1.5) 5 5 Divide both sides by the coefficient of 5 3 = (1.5) Now that there is one base on each side convert to log form = log 1.5 3 By converting to log form we now have an equation that is rearranged to solve for. We will now apply the estimation rule to evaluate. Note the / is still attached to the. = log(3) log(1.5) Put this into your calculator to determine the estimated value = 8.8380 Always give your answer to decimal places when dealing with logarithms ( ) = (8.8380) Now we multiply both sides by the denominator of, this will provide us with the value of. = 17.70 Final Answer Word Problems: Sometimes you will have to take the same approach to word problems. In such situations you may have to create the equation such as we did in chapter or the equation will be given to you. When dealing with a word problem, we approach it in eactly the same way with eactly the same steps as those highlighted above. 171
Chapter 7: Logarithmic Functions Section 7. E1. The half life of substance is 15 hours. Given that there is originally 150 mg of the substance, how long will it take for there to be 5 mg of the substance remaining. Keep t in mind that the equation for the half life of a substance is given by A = A o ( 1 ) h. Step 1: Set up the problem We know the original amount is 150 this represents A o, the half life is 15 hours, this represents h, and the amount remaining is 5 mg, this will represent A. So our equation becomes: A = A o ( 1 t ) h 5 = 150 ( 1 ) t 15 Step : Work it out using the same steps we applied in the previous eamples 5 = 150 ( 1 ) t 15 t 5 = 150( 1 ) 15 150 150 t 1 = (1) 15 15 = log1 = log( 1 ) 15 log( 1 ) 15 Divide by 150 Ensure there is only one base on each side. Note I kept the answers as fractions to reduce the amount of estimation needed. ( 1 ) Converted to log. The /15 stays with the until the end. Use the rule for estimating logs. =.5850 State the answer to decimal places when estimating logs = 15(.5850) Multiply through by 15 to determine the value of. = 38.7750 Step 3: Write a statement. It would take 39 hours for the substance to decay to 5 mg. 17
Chapter 7: Logarithmic Functions Section 7. E. An investment of $00 was placed into a savings account that provided %/a interest compounded annually. Determine how long it will take for the investment to reach a value of $1000. Step 1: Set up the problem This is a compounded interest problem, so we must set up the equation using A = P(1 + i) n, and we know the principal amount (P) is $00, we know that it is compounded annually at % interest, so i = 0.0, and we know the amount we are looking for, A, is $1000. So our equation is: A = P(1 + i) n 1000 = 00(1 + 0.0) n 1000 = 00(1.0) n Step : Work it out using the same steps we applied in the previous eamples 1000 = 00(1.0) n 1000 = 00(1.0)n Divide by 00 00 00 5=(1.0) n Ensure there is only one base on each side. n=log 1.0 (5) Converted to log. n= log(5) log(1.0) n= 7.09 Use the rule for estimating logs. State the answer to decimal places when estimating logs Step 3: Write a statement. It will take 8 years for the investment to reach a value of $1000 173
Chapter 7: Logarithmic Functions Section 7. Solving Eponentials Using Common Logarithms Solve: (a) 3 +1 = 0 (b) 10 = 5 + 17
Chapter 7: Logarithmic Functions Section 7. (c) 5 7 +1 = 0 (d) 1 = 3 +1 175
Chapter 7: Logarithmic Functions Section 7. (e) Allen currently has $000 in credit card debt. The interest rate on his credit card is 18.5% yearly. Determine the number of years it would take for hid debt to double if he made no payments against his balance. (f) $1000 is invested at %/a compounded monthly. How long, in months will it take for the investment to reach a value of $3000 17
Chapter 7: Logarithmic Functions Section 7.5 Section 7.5: Modeling Data Using Logarithmic Functions Using Logarithmic Regression to Solve a Problem Graphically E1. The flash on most digital cameras requires a charged capacitor in order to operate. The percent charge, Q, remaining on a capacitor was recorded at different times, t, after the flash had gone off. The t5 flash duration represents the time until a capacitor has only 50% of its initial charge. The t5 flash duration also represents the length of time that the flash is effective, to ensure that the object being photographed is properly lit. Percentage Charge, Q (%) Time, t (s) 100.00 0 90. 0.01 73.90 0.03 0.51 0.05 9.5 0.07 0.5 0.09 (a) Construct a scatter plot for the given data using graphing technology (b) Determine a logarithmic model for the data (c) Use your logarithmic model to determine the t5 flash duration to the nearest hundredth of a second. (d) Use your logarithmic model to determine the duration to the nearest hundredth of a second until the capacitor has just 10% of the initial charge remaining. 177
Chapter 7: Logarithmic Functions Section 7.5 E. Caffeine is found in coffee, tea, and soft drinks. Many people find that caffeine makes it difficult for them to sleep. The following data was collected in a study to determine how quickly the human body metabolizes caffeine. Each person started with 00 mg of caffeine in her or his bloodstream, and the caffeine level was measured at various times. (a) Determine the equation of the logarithmic regression function for the data representing time as a function of caffeine level. Caffeine Level in Bloodstream, m (mg) Time after Ingesting, t(h) Caffeine Level in Bloodstream, m (mg) Time after Ingesting, t(h) 18 1.0 33 1.0 17 1.5 80 7.5 113 5.0 15 3.0 15 3.0 100.0 90.5 71 8.5 15.0 15.0 138 3.5 153.5 77 8.0 130.0 83 7.0 90.5 50 1.0 11 5.0 150.5 3 1.0 55 1.0 3 18.0 11 5.0 5 17.5 8 7.0 5 13.0 13 3.5 7 18.5 180 1.0 18 0.0 110 5.0 9 15.0 75 8.0 3 1.0 7 9.0 5 17.5 9 1.5 1 19.0 (b) Determine the time it takes for an average person to metabolize 50% of the caffeine in their bloodstream. (c) Estimate how much caffeine would be in Paula s bloodstream at 9:00pm if she ingested 00 mg at 10:00 am. 178
Chapter 7: Logarithmic Functions Section 7.5 179