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x. f ( x ) = 3 x. g ( x ) = 3 x+1. g( x ) = 3 x +1. f ( x ) = 3 x. -2. 3 -2 = 1/9. 3 -2+1 = 3 -1 = 1/3. -1. 3 -1 = 1/3. 3 -1+1 = 3 0 = 1. 0. 3 0 = 1. 3 0+1 = 3 1 = 3. 1. 3 1 = 3. 3 1+1 = 3 2 = 9. (-1, 1). (0, 1). 2. 3 2 = 9. 3 2+1 = 3 3 = 27. 1. 2. 3. 4.
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x f (x) = 3x g(x) = 3x+1 g(x) = 3x+1 f (x) = 3x -2 3-2 = 1/9 3-2+1 = 3-1 = 1/3 -1 3-1 = 1/3 3-1+1 = 30 = 1 0 30 = 1 30+1 = 31 = 3 1 31 = 3 31+1 = 32 = 9 (-1, 1) (0, 1) 2 32 = 9 32+1 = 33 = 27 1 2 3 4 5 6 -5 -4 -3 -2 -1 3.1 Exponential Functions Exponential function – any function whose equation contains a variable in the exponent. [measures rapid increase or decrease (Example: epidemic growth)] f(x) = bx f – exponential function b - constant base (b > 0, b 1)x = any real number -- domain is (-∞ , ∞) f(x) = 2x g(x) = 10x h(x) = 3x+1 Shift up c units f(x) = bx+ c Shift down c units f(x) = bx– c Shift left c units f(x) = bx+c Shift right c units f(x) = bx-c Graphing Exponential Functions:
See P.193 All properties above are the same Except #4 which is ‘decreasing’
3.2 Investment and Interest Examples Compound Interest Karen Estes just received an inheritance of $10000 And plans to place all money in a savings account That pays 5% compounded quarterly to help her son Go to college in 3 years. How much money will be In the account in 3 years? Use the formula: A = P(1 + r/n)nt Simple Interest Juanita deposited $8000 in a bank for 5 years at a simple interest rate of 6% 1. How much interest will she get? 2. How much money will she have at the end of 5 years Use the formula: I = Prt I = (8000)(.06)(5) = $2400 (interest earned in 5 years) In 5 years she will have a total of A = P + I = 8000 + 2400 = $10,400 • A = amount in account after t years • P = principal or amount invested • t = time in years • r = annual rate of interest • n = number of times compounded • per year • Annual => n = 1 • Semiannual => n = 2 • Quarterly => n = 4 • Monthly => n = 12 • Daily => n = 365 As n increases (n ∞), A does NOT increase indefinitely. It gets closer and closer to a fixed number “e”, and gives us the continuous compound interest formula: A = Pert
Compound Interest related to Simple Interest If a principal of P dollars is borrowed for a period of t years at a perannum interest rate r, the interest I charged is: I = Prt Annually -> Once per year Semiannually -> Twice per year Quarterly -> Four times per year Monthly -> 12 times per year Daily -> 365 times per year Example: A credit union pays interest of 8% per annum compounded quarterly on a Certain savings plan. If $1000 is deposited in such a plan and the interest is Left to accumulate, how much is in the account after 1 year? I = (1000)(.08)(1/4) = $20 => New principal now is: $1020 I = (1020)(.08)(1/4) = $20.40 => New principal now is: $1040.40 I = (1040.40)(.08)(1/4) = $20.81 => New principal now is: $1061.21 I = (1061.21)(.08)(1/4) = $21.22 => New principal now is: $1082.43 Compound Interest Formula: The amount, A, after t years due to a principal P Invested at an annual interest rate r compounded n times per year is: A = P (1 + r/n)nt
f (x) = 3x f (x) = ex 4 (1, 3) f (x) = 2x 3 (1, e) 2 (1, 2) (0, 1) 1 An irrational number, symbolized by the letter e, appears as the base in many applied exponential functions. This irrational number is approximately equal to 2.72. More accurately, The number e is called the natural base. The function f (x) = ex is called the natural exponential function. Analogous to the continuous compound interest formula, A = Pert The model for continuous growth and decay is: A(t) = A0ekt A(t) = the amount at time t A0 = A(0), the initial amount k = rate of growth (k>0) or decay (k < 0) In year 2000, population of the world was about 6 billion; annual rate of growth 2.1%. Estimate the population in 2030 and 1990. 2000=year 0, 2030=year 30, year 1990=year -10 A(30) = 6e(.021)(30) = 11.265663 (billion people) A(-10) = 6e(.021)(-10) = 4.8635055 (billion people) ** Note these are just estimates. The actual population in 1990 was 5.28 billion The Natural Base e -1
Exponential Growth & Decay(Law of Uninhibited Growth/Decay) A0 is the original amount of ‘substance’ (drug, cells, bacteria, people, etc…)
Solving Exponential Equations • 25 = 32, so • 23x-1 = 25 • 3x – 1 = 5 • 3x = 6 • X = 2 • e2x-1 = e-4x • e3x • e2x-1 = e-7x • 2x – 1 = -7x • -1 = -9x • X = 1/9
Evaluate the Logarithmic Expression Question Needed for Evaluation Logarithmic Expression Evaluated log2 16 2 to what power is 16? log2 16 = 4 because 24 = 16. log3 9 3 to what power is 9? log3 9 = 2 because 32 = 9. log25 5 25 to what power is 5? log25 5 = 1/2 because 251/2 = 5. 3.3 Logarithmic Functions A logarithm is an exponent such that for b > 0, b 1 and x > 0 y = logb x if and only if by = x • Logarithmic equations Corresponding exponential forms • 2 = log5 x 1) 52 = x • 3 = logb 64 2) b3 = 64 • log3 7 = y 3) 3y = 7 • y = loge 9 4) ey = 9
x -2 -1 0 1 2 3 x 1/4 1/2 1 2 4 8 f (x) = 2x 1/4 1/2 1 2 4 8 g(x) = log2 x -2 -1 0 1 2 3 y = x f (x) = 2x 6 5 4 3 f (x) = log2x 2 -2 -1 2 3 4 5 6 -1 -2 Logarithmic Properties Logbb = 1 1 is the exponent to which b must be raised to obtain b. (b1 = b). Logb 1 = 0 0 is the exponent to which b must be raised to obtain 1. (b0 = 1). logb bx = x The logarithm with base b of b raised to a power equals that power. b logb x = x b raised to the logarithm with base b of a number equals that number. Graphs of f (x) = 2x and g(x) = log2x [Logarithm is the inverse of the exponential function] Reverse coordinates. • Properties of f(x) = logb x • Domain = (0, ∞) • Range = (-∞, +∞) • X intercept = 1 ; No y-intercept • Vertical asymptote on y-axis • Decreasing on 0<b<1; increasing if b>1 • Contains points: (1, 0), (b, 1), (1/b, -1) • Graph is smooth and continuous
Common Logs and Natural Logs A logarithm with a base of 10 is a ‘common log’ log10 1000 = ______ because 103 = 1000 If a log is written with no base it is assumed to be 10. log 1000 = log10 1000 = 3 3 A logarithm with a base of e is a ‘natural log’ loge 1 = ______ because e0 = 1 If a log is written as ‘ln’ instead of ‘log’ it is a natural log ln 1 = loge 1 = 0 0
Doubling Your Investments How long will it take to double your Money if it earns 6.5% compounded continuously? Recall the continuous compound interest formula: A = Pert If P is the principal and we want P to double, the amount A will be 2P. 2P = Pe(.065)t 2 = e(.065)t Ln 2 = ln e(.065)t Ln 2 = .065t T = ln 2 = .6931 ≈ 10.66 ≈ 11 years .065 .065 At what rate of return compounded continuously would your money double in 5 years? A = Pert 2P = Per(5) 2 = er(5) ln 2 = ln e5r .6931 = 5r r = .6931/5 R = .1386 So, annual interest rate needed is:13.86%
Newton’s Law of Cooling T = Ts + (T0 – Ts)e-kt T = Temperature of object a time t Ts = surrounding temperature T0 = T at t = 0 A McDonald’s franchise discovered that when coffee is poured from a CoffeeMaker whose contents are 180° F into a noninsulated pot, after 1 minute, the Coffee cools to 165° F if the room temperature is 72° F. How long should employees wait before pouring coffee from this noninsulated pot into cups to deliver it to customers at 125°F? T0 = 180 and Ts = 72 Now, T = 72 + 108e-.1495t T = 72 + (180 – 72)e-kt 125 = 72 + 108e-.1495t T = 72 + 108e-kt 53 = 108e-.1495t .4907 = e-.1495t Since T = 165 when t = 1 ln .4907 = ln e-.1495t 165 = 72 + 108e-k(1) -.7119 = -.1495t 93 = 108e-k t = 4.76 93 = 108e-k Employees should wait about 5 min Ln .8611 = ln e-k to deliver the coffee at the desired -.1495 = -k temperature. K = .1495
Basic Properties Logbb = 1 1 is the exponent to which b must be raised to obtain b. (b1 = b). Logb 1 = 0 0 is the exponent to which b must be raised to obtain 1. (b0 = 1). Inverse Properties logb bx = x The logarithm with base b of b raised to a power equals that power. b logb x = x b raised to the logarithm with base b of a number equals that number. 3.4 Properties & Rules of Logarithms For M>0 and N > 0 Product Rule logb(MN) = logb M + logb N Quotient Rule logb M = logb M - logb N N Power Rule p logb M = p logb M
p logb M = p logb M logb M = logb M - logb N N Logarithmic Property Practice Quotient Rule Product Rule logb(MN) = logb M + logb N • log3 (27 • 81) = • 2) log (100x) = • 3) Ln (7x) = 1) log8 23 x 2) Ln e5 11 = = Power Rule • log5 74 • Log (4x)5 • Ln x2 = • Ln x = = =
p logb M = p logb M logb M = logb M - logb N N x x 2) log6 3) log5 3 36y4 25y3 1) Logb (x2 y ) 4) log2 5x2 3 Expanding Logarithms logb(MN) = logb M + logb N
p logb M = p logb M logb M = logb M - logb N N Condensing Logarithms logb(MN) = logb M + logb N Note: Logarithm coefficients Must be 1 to condense. (Use power rule 1st) • log4 2 + log4 32 4) 2 ln x + ln (x + 1) • Log 25 + log 4 5) 2 log (x – 3) – log x • Log (7x + 6) – log x 6) ¼ logb x – 2 logb 5 – 10 logb y
The Change-of-Base Property Example:Evaluate log3 7 • Most calculators only use: • Common Log [LOG] (base 10) • Natural Log [LN] (base e) • It is necessary to use the change • Of base property to convert to • A base the calculator can use.
Matching Data to an Exponential Curve Find the exponential function of the form f(x) = aebx that passes through the points (0,2) and (3,8) F(0) = 2 f(3) = 8 2 = aeb(0) 8 = aeb(3) f(x) = 2e((ln4)/3)x 2 = ae0 8 = 2e3b 2 = a(1) 4 = e3b a = 2 ln 4 = ln e3b ln 4 = 3b b = (ln4)/3
3.5 Exponential & Logarithmic Equations Step 1: Isolate the exponential expression. Step 2: Take the natural logarithm on both sides of the equation. Step 3: Simplify using one of the following properties: ln bx = x ln b or ln ex = x. Step 4: Solve for the variable using proper algebraic rules. Examples (Solve for x): log4(x + 3) = 2. log 2 (3x-1) = 18 54x – 7 – 3 = 10 3x+2-7 = 27