SECTION 4.1 Exponential Functions 5 Define an exponential function. Graph exponential functions. Use transformations on exponential functions. Define simple interest. Develop a compound interest formula. 6 Understand the number e. 1 2 3 4 EXPONENTIAL FUNCTION A function f of the form f x a x , a 0 and a 1, is called an exponential function with base a. Its domain is (–∞, ∞). EXAMPLE 1 Evaluating Exponential Functions a. Let f x 3 . Find f 4 . x2 b. Let g x 2 10 x. Find g 2 . x 1 3 c. Let h x . Find h . 9 2 d. Let F(x) = 4x. Find F(3.2). EXAMPLE 1 Evaluating Exponential Functions Solution a. f 4 3 4 2 3 9 2 1 1 b. g 2 2 10 2 2 2 0.02 10 100 2 3 1 c. h 2 9 3 2 9 3 1 2 3 2 9 27 d. F(3.2) = 43.2 ≈ 84.44850629 RULES OF EXPONENTS Let a, b, x, and y be real numbers with a > 0 and b > 0. Then a a a x y x y a , x a x y a , y a ab x y a xy , a 1, 0 x x a b , x x a x 1 1 x . a a EXAMPLE 2 Graphing an Exponential Function with Base a > 1 – Exponential Growth Graph the exponential function f x 3 . x Solution Make a table of values. Plot the points and draw a smooth curve. EXAMPLE 2 Graphing an Exponential Function with Base a>1 Solution continued This graph is typical for exponential functions when a > 1. EXAMPLE 3 Graphing an Exponential Function with Base 0 < a < 1 – Exponential Decay x 1 Sketch the graph of y . 2 Solution Make a table of values. Plot the points and draw a smooth curve. EXAMPLE 3 Graphing an Exponential Function with Base 0 < a < 1 Solution continued As x increases in the positive direction, y decreases towards 0. PROPERTIES OF EXPONENTIAL FUNCTIONS Let f (x) = ax, a > 0, a ≠ 1. 1. The domain of f (x) = ax is (–∞, ∞). 2. The range of f (x) = ax is (0, ∞); the entire graph lies above the x-axis. 3. For a > 1, Exponential Growth (i) f is an increasing function, so the graph rises to the right. (ii) as x → ∞, y → ∞. (iii) as x → –∞, y → 0. 4. For 0 < a < 1, - Exponential Decay (i) f is a decreasing function, so the graph falls to the right. (ii) as x → – ∞, y → ∞. (iii) 5. 6. 7. as x → ∞, y → 0. The graph of f (x) = ax has no x-intercepts, so it never crosses the x-axis. No value of x will cause f (x) = ax to equal 0. The graph of is a smooth and continuous curve, and it passes through the points The x-axis is a horizontal asymptote for every exponential function of the form f (x) = ax. TRANSFORMATIONS ON EXPONENTIAL FUNCTION f (x) = ax Transformation Equation Horizontal y = ax+b Shift Effect on Equation Shift the graph of y = ax, |b| units (i) left if b > 0. (ii) right if b < 0. y = ax + b Shift the graph of y = ax, |b| units (i) up if b > 0. (ii) down if b < 0. Vertical Shift TRANSFORMATIONS ON EXPONENTIAL FUNCTION f (x) = ax Transformation Equation Stretching or Compressing (Vertically) y = cax Effect on Equation Multiply the y coordinates by c. The graph of y = ax is vertically (i) stretched if c > 1. (ii) compressed if 0 < c < 1. TRANSFORMATIONS ON EXPONENTIAL FUNCTION f (x) = ax Transformation Equation Reflection Effect on Equation y = –ax Reflect the graph of y = ax in the x-axis. y = a–x Reflect the graph of y = ax in the y-axis. EXAMPLE 6 Sketching Graphs Use transformations to sketch the graph of each function. a. f x 3x 4 b. f x 3x1 c. f x 3x d. f x 3x 2 State the domain and range of each function and the horizontal asymptote of its graph. EXAMPLE 6 Sketching Graphs Solution a. f x 3x 4 Domain: (–∞, ∞) Range: (–4, ∞) Horizontal Asymptote: y = –4 EXAMPLE 6 Sketching Graphs Solution continued b. f x 3x1 Domain: (–∞, ∞) Range: (0, ∞) Horizontal Asymptote: y = 0 EXAMPLE 6 Sketching Graphs Solution continued c. f x 3 x Domain: (–∞, ∞) Range: (–∞, 0) Horizontal Asymptote: y = 0 EXAMPLE 6 Sketching Graphs Solution continued d. f x 3 2 x Domain: (–∞, ∞) Range: (–∞, 2) Horizontal Asymptote: y = 2 General Exponential Growth/Decay Model Rate of decay (r < 0),Growth (r > 0) Amount after t time periods Original amount Number of time periods COMPOUND INTEREST – Growth Compound interest is the interest paid on both the principal and the accrued (previously earned) interest. It is an application of exponential growth. Interest that is compounded annually is paid once a year. For interest compounded annually, the amount A in the account after t years is given by Rate of decay (r < 0),Growth (r > 0) Amount after t time periods Number of time periods Original amount EXAMPLE 2 Calculating Compound Interest Juanita deposits $8000 in a bank at the interest rate of 6% compounded annually for five years. a. How much money will she have in her account after five years? b. How much interest will she receive? EXAMPLE 2 Calculating Compound Interest Solution a. Here P = $8000, r = 0.06, and t = 5. b. Interest = A P = $10,705.80 $8000 = $2705.80. COMPOUND INTEREST FORMULA r A P 1 n nt A = amount after t years P = principal r = annual interest rate (expressed as a decimal) n = number of times interest is compounded each year t = number of years EXAMPLE 3 Using Different Compounding Periods to Compare Future Values If $100 is deposited in a bank that pays 5% annual interest, find the future value A after one year if the interest is compounded (i) (ii) (iii) (iv) (v) annually. semiannually. quarterly. monthly. daily. EXAMPLE 3 Using Different Compounding Periods to Compare Future Values Solution In the following computations, P = 100, r = 0.05 and t = 1. Only n, the number of times interest is compounded each year, changes. Since t = 1, nt = n(1) = n. nt (i) Annual Compounding: r A P 1 n A 100 1 0.05 $105.00 EXAMPLE 3 Using Different Compounding Periods to Compare Future Values (ii) Semiannual Compounding: 2 r A P 1 n 2 0.05 A 100 1 $105.06 2 (iii) Quarterly Compounding: 4 r A P 1 4 4 0.05 A 100 1 $105.09 4 EXAMPLE 3 Using Different Compounding Periods to Compare Future Values (iv) Monthly Compounding: 12 r A P 1 12 12 0.05 A 100 1 $105.12 12 (v) Daily Compounding: 365 r A P 1 365 0.05 A 100 1 365 365 $105.13 EXAMPLE 8 Bacterial Growth A technician to the French microbiologist Louis Pasteur noticed that a certain culture of bacteria in milk doubles every hour. If the bacteria count B(t) is modeled by the equation B t 2000 2t , with t in hours, find a. the initial number of bacteria, b. the number of bacteria after 10 hours; and c. the time when the number of bacteria will be 32,000. EXAMPLE 8 Bacterial Growth Solution a. Initial size B0 B 0 2000 20 2000 1 2000 b. B 10 2000 2 2, 048,000 10 c. Find t when B(t) = 32,000 t 4 t 32,000 2000 2 2 2 t 4t 16 2 After 4 hours, the number of bacteria will be 32,000. THE VALUE OF e As h gets larger and larger, 1 1 h h gets closer and closer to a fixed number. This irrational number is denoted by e and is sometimes called the Euler number. The value of e to 15 places is e = 2.718281828459045. CONTINUOUS COMPOUND FORMULA rt A Pe A = amount after t years P = principal r = annual rate (expressed as a decimal) t = number of years EXAMPLE 4 Calculating Continuous Compound Interest Find the amount when a principal of $8300 is invested at a 7.5% annual rate of interest compounded continuously for eight years and three months. Solution P = $8300 and r = 0.075. Convert eight years and three months to 8.25 years. A Pe rt 0.0758.25 A $8300e $15, 409.83 EXAMPLE 5 Calculating the Amount of Repaying a Loan How much money did the government owe DeHaven’s descendants for 213 years on a $450,000 loan at the interest rate of 6%? Solution a. With simple interest, A P Prt P 1 rt A $450,000 1 0.06 213 $6.201 million. EXAMPLE 5 Calculating the Amount of Repaying a Loan Solution continued b. With interest compounded yearly, t 213 A P 1 r $450,000 1 0.06 A $1.105 10 $110.500 million. c. With interest compounded quarterly, 4t 4 213 r 0.06 A P 1 $450, 000 1 4 4 11 A $1.45305 10 $145.305 billion. 11 EXAMPLE 5 Calculating the Amount of Repaying a Loan Solution continued d. With interest compounded continuously, 0.06 213 rt A Pe $450,000e A $1.5977 1011 $159.77 billion. Notice the dramatic difference between quarterly and continuous compounding and the dramatic difference between simple interest and compound interest. THE NATURAL EXPONENTIAL FUNCTION The exponential function f x e x with base e is so prevalent in the sciences that it is often referred to as the exponential function or the natural exponential function. EXAMPLE 6 Sketching a Graph Use transformations to sketch the graph of Solution Start with the graph of y = ex. EXAMPLE 6 Sketching a Graph Use transformations to sketch the graph of Solution coninued Shift the graph of y = ex one unit right. EXAMPLE 6 Sketching a Graph Use transformations to sketch the graph of Solution continued Shift the graph of y = ex – 1 two units up. MODEL FOR EXPONENTIAL GROWTH OR DECAY A t A0 e kt A(t) = amount at time t A0 = A(0), the initial amount k = relative rate of growth (k > 0) or decay (k < 0) t = time EXAMPLE 7 Modeling Exponential Growth and Decay In the year 2000, the human population of the world was approximately 6 billion and the annual rate of growth was about 2.1%. Using the model on the previous slide, estimate the population of the world in the following years. a. 2030 b. 1990 EXAMPLE 7 Modeling Exponential Growth and Decay Solution a. The year 2000 corresponds to t = 0. So A0 = 6 (billion), k = 0.021, and 2030 corresponds to t = 30. A 30 6 e 0.021 30 11.265663 The model predicts that if the rate of growth is 2.1% per year, over 11.26 billion people will be in the world in 2030. EXAMPLE 7 Modeling Exponential Growth and Decay Solution b. The year 1990 corresponds to t = 10. A 10 6e 0.021 10 4.8635055 The model predicts that the world had over 4.86 billion people in 1990. (The actual population in 1990 was 5.28 billion.) SECTION 4.3 Logarithmic Functions 1 2 3 4 5 6 7 Define logarithmic functions. Inverse Functions Evaluate logarithms. Rules of Logarithms Find the domains of logarithmic functions. Graph logarithmic functions. Use logarithms to evaluate exponential equations. DEFINITION OF THE LOGARITHMIC FUNCTION For x > 0, a > 0, and a ≠ 1, y log a x if and only if xa . y The function f (x) = loga x, is called the logarithmic function with base a. The logarithmic function is the inverse function of the exponential function. Inverse Functions Certain pairs of one-to-one functions “undo” one another. For example, if x 5 f ( x ) 8 x 5 and g ( x ) , 8 then f (10) 8(10) 5 85 (85) 5 g (85) 10 8 Inverse Functions Starting with 10, we “applied” function and then “applied” function g to the result, which returned the number 10. Inverse Functions As further examples, check that f (3) 29 and g ( 29) 3, f ( 5) 35 and g ( 35) 5, 3 g (2) 8 3 and g 2, 8 Inverse Functions In particular, for this pair of functions, f ( g (2)) 2 and g ( f ( 2)) 2. In fact, for any value of x, f ( g ( x )) x and g ( f ( x )) x, or ( f g )( x ) x and (g f )( x ) x. Because of this property, g is called the inverse of . Inverse Function Let be a one-to-one function. Then g is the inverse function of if (f g )( x ) x for every x in the domain of g, and (g f )( x ) x for every x in the domain of . EXAMPLE 1 Converting from Exponential to Logarithmic Form Write each exponential equation in logarithmic form. 4 2 1 1 3 c. a 7 a. 4 64 b. 2 16 Solution a. 4 64 log 4 64 3 3 4 1 1 1 b. log1 2 4 16 2 16 c. a 2 7 log a 7 2 EXAMPLE 2 Converting from Logarithmic Form to Exponential Form Write each logarithmic equation in exponential form. a. log 3 243 5 b. log 2 5 x c. log a N x Solution a. log3 243 5 243 3 5 b. log 2 5 x 5 2 x c. log a N x N a x EXAMPLE 3 Evaluating Logarithms Find the value of each of the following logarithms. c. log1 3 9 a. log 5 25 b. log 2 16 1 d. log 7 7 e. log 6 1 f. log 4 2 Solution a. log 5 25 y 25 5 or 5 5 y y2 b. log 2 16 y 16 2 or 2 2 y y4 y y 2 4 EXAMPLE 3 Evaluating Logarithms Solution continued y 1 c. log1 3 9 y 9 or 32 3 y 3 d. log 7 7 y 7 7 y or 71 7 y e. log 6 1 y 1 6 or 6 6 y 0 y y 2 y 1 y0 1 1 y 1 2y f. log 4 y 4 or 2 2 2 2 1 y 2 EXAMPLE 4 Using the Definition of Logarithm Solve each equation. a. log 5 x 3 c. log z 1000 3 Solution a. log 5 x 3 x 53 1 1 x 3 5 125 1 b. log 3 y 27 d. log 2 x 2 6 x 10 1 EXAMPLE 4 Using the Definition of Logarithm Solution continued 1 b. log 3 y 27 1 3y 27 3 y 3 3 3 y c. log z 1000 3 1000 z 3 103 z 3 10 z EXAMPLE 4 Using the Definition of Logarithm Solution continued d. log 2 x 6 x 10 1 2 x 6 x 10 2 2 2 1 x 6x 8 0 2 x 2 x 4 0 x 2 0 or x 4 0 x 2 or x 4 Rules of Logarithms Rules of Logarithms with Base a If M, N, and a are positive real numbers with a ≠ 1, and x is any real number, then 1. loga(a) = 1 3. loga (ax) =x 2. loga(1) = 0 These relationships are used to solve exponential or logarithmic equations 4. a loga ( N ) N 5. loga(MN) = loga(M) + loga(N) 6. loga(M/N) = loga(M) – loga(N) 7. loga(Mx) = x · loga(M) 8. loga(1/N) = – loga(N) COMMON LOGARITHMS The logarithm with base 10 is called the common logarithm and is denoted by omitting the base: log x = log10 x. Thus, y = log x if and only if x = 10 y. Applying the basic properties of logarithms 1. log 10 = 1 2. log 1 = 0 3. log 10x = x log x 4. 10 x NATURAL LOGARITHMS The logarithm with base e is called the natural logarithm and is denoted by ln x. That is, ln x = loge x. Thus, y = ln x if and only if x = e y. Applying the basic properties of logarithms 1. ln e = 1 2. ln 1 = 0 3. log ex = x ln x 4. e x DOMAIN OF LOGARITHMIC FUNCTION Domain of y = loga x is (0, ∞) Range of y = loga x is (–∞, ∞) Logarithms of 0 and negative numbers are not defined. EXAMPLE 5 Finding the Domain Find the domain of f x log3 2 x Solution Domain of a logarithmic function must be positive, that is, 2 x 0 2x The domain of f is (–∞, 2). EXAMPLE 6 Sketching a Graph Sketch the graph of y = log3 x. Solution by plotting points (Method 1) Make a table of values. EXAMPLE 6 Sketching a Graph Solution continued Plot the ordered pairs and connect with a smooth curve to obtain the graph of y = log3 x. EXAMPLE 6 Sketching a Graph Solution by using the inverse function (Method 2) Graph y = f (x) = 3x. Reflect the graph of y = 3x in the line y = x to obtain the graph of y = f –1(x) = log3 x. 66 GRAPHS OF LOGARITHMIC FUNCTIONS PROPERTIES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS Exponential Function f (x) = ax Logarithmic Function f (x) = loga x 1. Domain (–∞, ∞) Range (0, ∞) Domain (0, ∞) Range (–∞, ∞) 2. y-intercept is 1 No x-intercept x-intercept is 1 No y-intercept 3. x-axis (y = 0) is the y-axis (x = 0) is the horizontal asymptote vertical asymptote PROPERTIES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS Exponential Function f (x) = ax 4. The graph is a continuous smooth curve that passes through the points 1 1, , (0, 1), and a (1, a). Logarithmic Function f (x) = loga x The graph is a continuous smooth curve that passes through the points 1 (1, 0), and , 1 , a (a, 1). PROPERTIES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS Exponential Function f (x) = ax 5. Is one-to-one , that is, au = av if and only if u = v. Logarithmic Function f (x) = loga x Is one-to-one, that is, logau = logav if and only if u = v. 6. Increasing if a > 1 Increasing if a > 1 Decreasing if 0 < a < 1 Decreasing if 0 < a < 1 EXAMPLE 7 Using Transformations Start with the graph of f (x) = log3 x and use transformations to sketch the graph of each function. a. f x log 3 x 2 b. f x log 3 x 1 c. f x log 3 x d. f x log 3 x State the domain and range and the vertical asymptote for the graph of each function. EXAMPLE 7 Using Transformations Solution a. f x log3 x 2 Shift up 2 Domain (0, ∞) Range (–∞, ∞) Vertical asymptote x = 0 EXAMPLE 7 Using Transformations Solution continued b. f x log3 x 1 Shift right 1 Domain (1, ∞) Range (–∞, ∞) Vertical asymptote x = 1 EXAMPLE 7 Using Transformations Solution continued c. f x log3 x Reflect graph of y = log3 x in the x-axis Domain (0, ∞) Range (–∞, ∞) Vertical asymptote x = 0 EXAMPLE 7 Using Transformations Solution continued d. f x log3 x Reflect graph of y = log3 x in the y-axis Domain (∞, 0) Range (–∞, ∞) Vertical asymptote x = 0 EXAMPLE 8 Using Transformations to Sketch a Graph Sketch the graph of y 2 log x 2 . Solution Start with the graph of f (x) = log x. Step 1: Replacing x with x – 2 shifts the graph two units right. EXAMPLE 8 Using Transformations to Sketch a Graph Solution continued Step 2: Multiplying by 1 reflects the graph in the x-axis. Step 3: Adding 2 shifts the graph two units up. Rules of Logarithms Rules of Logarithms with Base a If M, N, and a are positive real numbers with a ≠ 1, and x is any real number, then 1. loga(a) = 1 2. loga(1) = 0 3. loga(ax) = x 4. a loga ( N ) N 5. loga(MN) = loga(M) + loga(N) 6. loga(M/N) = loga(M) – loga(N) 7. loga(Mx) = x · loga(M) 8. loga(1/N) = – loga(N) EXAMPLE 1 Using Rules of Logarithms to Evaluate Expressions Given that log 5 z = 3 and log 5 y = 2, evaluate each expression. 7 b. log 5 125y a. log5 yz c. log 5 d. log 5 z1/30 y 5 z y Solution a. log 5 yz log 5 y log 5 z 23 5 EXAMPLE 1 Using Rules of Logarithms to Evaluate Expressions Solution continued b. log 5 125 y 7 log 5 125 log 5 y 7 log 5 53 7 log 5 y 3 7 2 17 c. log 5 z log 5 y 1/2 z y 1 log 5 z log 5 y 2 1 1 3 2 2 2 EXAMPLE 1 Using Rules of Logarithms to Evaluate Expressions Solution continued d. log 5 z1/30 y 5 log 5 z1/30 log 5 y 5 1 log 5 z 5 log 5 y 30 1 3 5 2 30 0.1 10 10.1 EXAMPLE 2 Writing Expressions In Expanded Form Write each expression in expanded form. x x 1 2 a. log 2 2 x 1 3 3 2 5 b. ln x y z 4 Solution x x 1 2 a. log 2 2 x 1 3 4 log 2 x x 1 log 2 2 x 1 2 3 log 2 x log 2 x 1 log 2 2 x 1 2 3 4 4 2log 2 x 3log 2 x 1 4log 2 2 x 1 EXAMPLE 2 Writing Expressions In Expanded Form Solution continued b. ln x y z ln x y z 3 2 5 3 2 5 1/2 1 3 2 5 ln x y z 2 1 ln x 3 ln y 2 ln z 5 2 1 3ln x 2ln y 5ln z 2 3 5 ln x ln y ln z 2 2 EXAMPLE 3 Writing Expressions in Condensed Form Write each expression in condensed form. a. log3 x log 4 y 1 b. 2ln x ln x 2 1 2 c. 2log 2 5 log 2 9 log 2 75 1 2 d. ln x ln x 1 ln x 1 3 EXAMPLE 3 Writing Expressions in Condensed Form Solution 3x a. log3 x log 4 y log 4y 1/2 1 2 2 2 b. 2 ln x ln x 1 ln x ln x 1 2 ln x 2 x 2 1 EXAMPLE 3 Writing Expressions in Condensed Form Solution continued c. 2log 2 5 log 2 9 log 2 75 log 2 5 log 2 9 log 2 75 2 log 2 25 9 log 2 75 25 9 log 2 75 log 2 3 EXAMPLE 3 Writing Expressions in Condensed Form Solution continued 1 d. ln x ln x 1 ln x 2 1 3 1 2 ln x x 1 ln x 1 3 1 x x 1 ln 2 3 x 1 ln 3 x x 1 x 1 2 CHANGE-OF-BASE FORMULA Let a, b, and x be positive real numbers with a ≠ 1 and b ≠ 1. Then logb x can be converted to a different base as follows: log a x log b x log a b (base a ) log x log b ln x ln b (base 10) (base e) EXAMPLE 4 Using a Change of Base to Compute Logarithms Compute log513 by changing to a. common logarithms and b. natural logarithms. Solution log13 a. log 5 13 log 5 1.59369 ln 13 b. log 5 13 ln 5 1.59369 EXAMPLE 9 Evaluating the Natural Logarithm Evaluate each expression. 1 4 b. ln 2.5 a. ln e e c. ln 3 Solution a. ln e 4 4 1 2.5 b. ln 2.5 ln e 2.5 e c. ln 3 1.0986123 Use a calculator. EXAMPLE 10 Doubling Your Money a. How long will it take to double your money if it earns 6.5% compounded continuously? b. At what rate of return, compounded continuously, would your money double in 5 years? Solution a. If P is the original amount invested, A = 2P. It will take 11 years to double your money. EXAMPLE 10 Doubling Your Money Solution continued b. Your investment will double in 5 years at the rate of 13.86%. Solving Exponential Or Logarithmic Equations To solve an exponential or logarithmic equation, change the given equation into one of the following forms, where a and b are real numbers, a > 0 and a ≠ 1, and follow the guidelines. 1.ax = b Solve by taking logarithms on both sides. 2. Loga x = b Solve by changing to exponential form ab = x. SOLVING AN EXPONENTIAL EQUATION Solve 7x = 12. Give the solution to the nearest thousandth. Solution While any appropriate base b can be used, the best practical base is base 10 or base e. We choose base e (natural) logarithms here. SOLVING AN EXPONENTIAL EQUATION Solve 7x = 12. Give the solution to the nearest thousandth. Solution x 7 12 In 7 In 12 x Property of logarithms xIn 7 In 12 Power of logarithms In 12 x Divide by In 7. In 7 x 1.277 Use a calculator. The solution set is {1.277}. SOLVING AN EXPONENTIAL EQUATION Solve 32x – 1 = .4x+2 . Give the solution to the nearest thousandth. Solution 32 x 1 .4 x 2 In 3 2 x 1 In .4 x 2 (2 x 1) In 3 ( x 2) In .4 2 x In 3 In 3 x In .4 2 In .4 Take natural logarithms on both sides. Property power Distributive property SOLVING AN EXPONENTIAL EQUATION Solve 32x – 1 =.4x+2 . Give the solution to the nearest thousandth. Solution 2 x In 3 x In .4 2 In .4 In 3 Write the terms with x on one side x(2 In 3 In .4) 2 In .4 In 3 Factor out x. 2 In .4 3 x 2 In 3 .4 In .42 In 3 x In 32 In .4 Divide by 2 In 3 – In .4. Power property SOLVING AN EXPONENTIAL EQUATION Solve 32x – 1 =.4x+2 . Give the solution to the nearest thousandth. Solution This is exact. In .16 In 3 x In 9 In .4 In .48 x 9 In .4 x .236 Apply the exponents. Product property; Quotient property This is approximate. The solution set is { –.236}. SOLVING BASE e EXPONENTIAL EQUATIONS Solve the equation. Give solutions to the nearest thousandth. a. e x2 200 Solution e In e x2 x2 200 In 200 x In 200 2 Take natural logarithms on both sides. In e x2 = x2 SOLVING BASE e EXPONENTIAL EQUATIONS Solve the equation. Give solutions to the nearest thousandth. a. e x2 200 Remember both roots. Solution x In 200 Square root property x 2.302 Use a calculator. The solution set is { 2.302}. SOLVING BASE e EXPONENTIAL EQUATIONS Solve the equation. Give solutions to the nearest thousandth. b. e 2 x 1 e 4 x 3e Solution e 2 x 1 e e 4 x 2 x 1 e 3e 3e am an amn 2 x 3 am Divide by e; n a m n . a 2 x In 3 Take natural logarithms on both sides. In e 2 x In e In 3 Power property SOLVING BASE e EXPONENTIAL EQUATIONS Solve the equation. Give solutions to the nearest thousandth. 2 x 1 4 x e e 3e b. Solution 2 x In 3 1 x In 3 2 In e = 1 Multiply by – ½ x .549 The solution set is {–.549}. SOLVING A LOGARITHMIC EQUATION Solve log(x + 6) – log(x + 2) = log x. Solution log( x 6) log( x 2) log x x6 log log x x2 x6 x x2 x 6 x ( x 2) Quotient property Property of logarithms SOLVING A LOGARITHMIC EQUATION Solve log(x + 6) – log(x + 2) = log x. Solution x 6 x 2x 2 x x 60 2 Standard form ( x 3)( x 2) 0 x 3 or Distributive property Factor. x2 Zero-factor property The proposed negative solution (x = – 3) is not in the domain of the log x in the original equation, so the only valid solution is the positive number 2, giving the solution set {2}. SOLVING A LOGARITHMIC EQUATION Solve log(3x + 2) + log(x – 1 ) = 1. Give the exact value(s) of the solution(s). Solution log(3 x 2) log( x 1) 1 log(3 x 2) log( x 1) log10 log[(3 x 2)( x 1)] log10 (3 x 2)( x 1) 10 Substitute. Product property Property of logarithms SOLVING A LOGARITMIC EQUATION Solve log(3x + 2) + log(x – 1 ) = 1. Give the exact value(s) of the solution(s). Solution 3 x x 2 10 2 3 x 2 x 12 0 1 1 144 x 6 Multiply. Subtract 10. Quadratic formula SOLVING A LOGARITMIC EQUATION Solve log(3x + 2) + log(x – 1 ) = 1. Give the exact value(s) of the solution(s). Solution 1 145 6 The number is negative, so x – 1 is negative. Therefore, log(x – 1) is not defined and this proposed solution must be discarded. 1 145 6 Since > 1, both 3x + 2 and x – 1 are positive and the solution set is 1 145 . 6 NEWTON’S LAW OF COOLING Newton’s Law of Cooling states that T Ts T0 Ts e , kt where T is the temperature of the object at time t, Ts is the surrounding temperature, and T0 is the value of T at t = 0. EXAMPLE 11 McDonald’s Hot Coffee The local McDonald’s franchise has discovered that when coffee is poured from a coffeemaker whose contents are 180ºF into a noninsulated pot, after 1 minute, the coffee cools to 165ºF if the room temperature is 72ºF. How long should the employees wait before pouring the coffee from this noninsulated pot into cups to deliver it to customers at 125ºF? EXAMPLE 11 McDonald’s Hot Coffee Solution Use Newton’s Law of Cooling with T0 = 180 and Ts = 72 to obtain kt T 72 180 72 e T 72 108e kt We have T = 165 and t = 1. 165 72 108e 93 k e 108 k 93 ln k 108 k 0.1495317 EXAMPLE 11 McDonald’s Hot Coffee Solution continued Substitute this value for k. 0.1495317t T 72 108e Solve for t when T = 125. 0.1495317 t 125 72 108e 1 53 t ln 125 72 0.1495317 t 0.1495317 108 e 108 t 4.76 53 ln 0.1495317t The employee should 108 wait about 5 minutes. GROWTH AND DECAY MODEL A A0 e rt A is the quantity after time t. A0 is the initial (original) quantity (when t = 0). r is the growth or decay rate per period. t is the time elapsed from t = 0. EXAMPLE 12 Chemical Toxins in a Lake A chemical spill deposits 60,000 cubic meters of soluble toxic waste into a large lake. If 20% of the waste is removed every year, how many years will it take to reduce the toxin to 1000 cubic meters? Solution In the equation A = A0ert, we need to find A0, r, and the time when A = 1000. EXAMPLE 12 Chemical Toxins in a Lake Solution continued 1. Find A0. Initially (t = 0), we are given A0 = 60,000. So A 60,000e rt 2. Find r. When t = 1 year, the amount of toxin will be 80% of its initial value, or EXAMPLE 12 Chemical Toxins in a Lake Solution continued 2. continued So EXAMPLE 12 Chemical Toxins in a Lake Solution continued 3. Find t when A = 1000. It will take approximately 18 years to reduce toxin to 1000 m3.