Modeling with Exponential and Logarithmic Functions

Exponential Growth and Decay Models The mathematical model for exponential growth or decay is given by f (t) = A0ekt or A = A0ekt. If k > 0, the function models the amount or size of a growing entity. A0 is the original amount or size of the growing entity at time t = 0. A is the amount at time t, and k is a constant representing the growth rate. If k < 0, the function models the amount or size of a decaying entity. A0 is the original amount or size of the decaying entity at time t = 0. A is the amount at time t, and k is a constant representing the decay rate. x y increasing y = A0ekt k > 0 k < 0 A0 decreasing A0

Example The graph below shows the growth of the Mexico City metropolitan area from 1970 through 2000. In 1970, the population of Mexico City was 9.4 million. By 1990, it had grown to 20.2 million. 20 15 10 5 25 30 1970 1980 1990 2000 Population (millions) Year Find the exponential growth function that models the data. By what year will the population reach 40 million?

Example cont. Solution a. We use the exponential growth model A = A0ekt in which t is the number of years since 1970. This means that 1970 corresponds to t = 0. At that time there were 9.4 million inhabitants, so we substitute 9.4 for A0 in the growth model. A = 9.4 ekt We are given that there were 20.2 million inhabitants in 1990. Because 1990 is 20 years after 1970, when t = 20 the value of A is 20.2. Substituting these numbers into the growth model will enable us to find k, the growth rate. We know that k > 0 because the problem involves growth. A = 9.4 ekt Use the growth model with A0 = 9.4. 20.2 = 9.4 ek•20 When t = 20, A = 20.2. Substitute these values.

Example cont. Solution 20.2/ 9.4 = ek•20 Isolate the exponential factor by dividing both sides by 9.4. ln(20.2/ 9.4) = lnek•20 Take the natural logarithm on both sides. 20.2/ 9.4 = 20k Simplify the right side by using ln ex = x. 0.038 = k Divide both sides by 20 and solve for k. We substitute 0.038 for k in the growth model to obtain the exponential growth function for Mexico City. It is A = 9.4 e0.038t where t is measured in years since 1970.

Example cont. Solution b. To find the year in which the population will grow to 40 million, we substitute 40 in for A in the model from part (a) and solve for t. A = 9.4 e0.038t This is the model from part (a). 40 = 9.4 e0.038t Substitute 40 for A. 40/9.4 = e0.038t Divide both sides by 9.4. ln(40/9.4) = lne0.038t Take the natural logarithm on both sides. ln(40/9.4) =0.038t Simplify the right side by using ln ex = x. ln(40/9.4)/0.038 =t Solve for t by dividing both sides by 0.038 Because 38 is the number of years after 1970, the model indicates that the population of Mexico City will reach 40 million by 2008 (1970 + 38).

More Examples Use the fact that after 5715 years a given amount of carbon-14 will have decayed to half the original amount to find the exponential decay model for carbon-14. In 1947, earthenware jars containing what are known as the Dead Sea Scrolls were found by an Arab Bedouin herdsman. Analysis indicated that the scroll wrappings contained 76% of their original carbon-14. Estimate the age of the Dead Sea Scrolls. In 1952, syringoporid coral fossils were found in the Bakken formation in North Dakota. Analysis indicated that these coral fossils contained .02% of their original carbon-14. Estimate the age of the coral fossils. Solution We begin with the exponential decay model A = A0ekt. We know that k < 0 because the problem involves the decay of carbon-14. After 5715 years (t = 5715), the amount of carbon-14 present, A, is half of the original amount A0. Thus we can substitute A0/2 for A in the exponential decay model. This will enable us to find k, the decay rate.

Example cont. Solution A0/2= A0ek5715 After 5715 years, A = A0/2 1/2= ekt5715 Divide both sides of the equation by A0. ln(1/2) = ln ek5715 Take the natural logarithm on both sides. ln(1/2) = 5715k ln ex = x. k = ln(1/2)/5715=-0.000121 Solve for k. Substituting for k in the decay model, the model for carbon-14 is A = A0e –0.000121t.

Dead Sea Scroll Example Solution A = A0e-0.000121t This is the decay model for carbon-14. 0.76A0 = A0e-0.000121t A = .76A0 since 76% of the initial amount remains. 0.76 = e-0.000121t Divide both sides of the equation by A0. ln 0.76 = ln e-0.000121t Take the natural logarithm on both sides. ln 0.76 = -0.000121t ln ex = x. t=ln(0.76)/(-0.000121) Solve for t. The Dead Sea Scrolls are approximately 2268 years old plus the number of years between 1947 and the current year.

The Bakken Formation Fossil Example Solution A = A0e-0.000121t This is the decay model for carbon-14. 0.0002A0 = A0e-0.000121t A = .0002A0 since .02% of the initial amount remains. 0.0002 = e-0.000121t Divide both sides of the equation by A0. ln 0.0002 = ln e-0.000121t Take the natural logarithm on both sides. ln 0.0002 = -0.000121t ln ex = x. t=ln(0.0002)/(-0.000121) Solve for t. The syringoporid coral fossils found in the Bakken formation in North Dakota are approximately 70390 years old plus the number of years between 1952 and the current year.