Notes - Exponential and Logarithmic Functions Exponential Functions Def: A function whose equation contains a variable in the exponent is called an exponential function. f(x) = bx or y = bx b is called a base and is a positive constant other than 1. Ex1 The exponential function f(x) = 13.49(0.967)x - 1 describes the number of O-rings expected to fail, f(x), when the temperature is xoF. If the temperature is 31oF, how many O-rings are expected to fail? f(31) = 13.49(0.967)31 – 1 f(31) ≈ 4 You Try: How many O-rings are expected to fail at 60oF? f(60) = 13.49(0.967)60 – 1 f(60) ≈ 1 Ex2 The function below describes the number of people in the audience, f(x), who have heard a rumor x minutes after 8:00. Evaluate f(10) and describe what this means in terms of the problem? !"" f(x) = !!!""(!.!")! Since there is so much going on in the denominator of this function, I must !"" f(10) = !!!""(!.!")!" put extra parentheses around it. It would look like this in the calculator: 400/(1 + 399(0.67)^10) f(10) ≈ 48 → approximately 48 people heard the rumor by 8:10. You Try - Evaluate f(20) and describe what this means. !"" f(20) = !!!""(!.!")!" f(20) ≈ 353 → approximately 353 people heard the rumor by 8:20. Def: The number e is called the natural base and is approximately = 2.72... Just like π is called pi and is approximately = 3.14... Ex3 push 2nd ln to get e in the calculator. The function f(x) = 6e0.013x describes world population, f(x), in billions, x years after 2000. Use the function to find world population in 2050. 0.013*50 f(50) = 6e f(50) = 11.5 billion Interest formulas are exponential functions: ***compounding interest: A = P(1 + r/n)nt ***continuous compounding: A = Pert A = balance (what you will make) P = principal (what you are investing) r = interest rate (in decimal form) t = time (years) n = number of compounding I plug in 50 instead of 2050 because the problem indicates that x is years after 2000. The year 2050 is 50 years after 2000, so I must plug in 50 for x. Compounded: monthly: n = 12 yearly: n = 1 quarterly: n = 4 semiannually: n = 2 Ex4 You want to invest $8000 for 6 years and you have a choice between two accounts. The first pays 7% per year, compounded monthly. The second pays 6.85% per year, compounded continuously. Which is the better investment? I have to use both formulas because I must choose the best account. A = P(1 + r/n)nt A = what I’m finding P = 8000 r = 0.07 → 7% converts to 0.07 n = 12 → compounded monthly, n = 12 t=6 0.07 A = 8000(1 + A = 12,160.84 A = Pert A = what I’m finding P = 8000 r = 0.0685 → 6.85% converts to 0.0685 t=6 A = 8000e(0.0685*6) A = 12,066.60 (12*6) /12) The better investment is the account that is compounded monthly. You Try A sum of $10,000 is invested at an annual rate of 8%. Find the balance in the account after 5 years subject to quarterly compounding. P = 10000 r = 0.08 A = 10000(1 + 0.08/4)(4*5) A = 14,859.47 t=5 n=4 Logarithmic Functions Def: The inverse function of the exponential function is called the logarithmic function. y = logbx is equivalent to by = x Ex5 The percentage of adult height attained by a boy who is x years old can be modeled by: f(x) = 29 + 48.8 log (x + 1) where x represents the boy's age and f(x) represents the percentage of his adult height. Approximately what percent of his adult height is a boy at age 8? f(8) = 29 + 48.8 log (8 + 1) f(8) ≈ 76% Approximately what percent of his adult height is a boy at age 10? f(10) = 29 + 48.8 log (10 + 1) f(10) ≈ 80% Def: The inverse of the natural base e is called the natural log, ln. Ex6 The annual amount that we spend to attend sporting events can be modeled by f(x) = 2.05 + 1.3 ln x where x represents the number of years after 1984 and f(x) represents the total amount in billions of dollars. About how much money was spent in 2000? f(16) = 2.05 + 1.3 ln 16 I plug in 16 because the problem indicates that x is the years after 1984. The year 2000 is 16 years f(16) = 5.65 billion after 1984. Estimate how much will be spent in 2015. f(31) = 2.05 + 1.3 ln 31 = 6.51 billion

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