Modeling with Exponential and Logarithmic Functions. y. y. increasing. y = A 0 e kt k > 0. y = A 0 e kt k < 0. A 0. x. x. Exponential Growth and Decay Models. The mathematical model for exponential growth or decay is given by f ( t ) = A 0 e kt or A = A 0 e kt .
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y = A0ekt
k > 0
y = A0ekt
k < 0
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.
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•20When t = 20, A = 20.2. Substitute these values.
20.2/ 9.4 = ek•20Isolate the exponential factor by dividing both sides by 9.4.
ln(20.2/ 9.4) = lnek•20Take the natural logarithm on both sides.
20.2/ 9.4 = 20kSimplify the right side by using ln ex = x.
0.038 = kDivide 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.
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).
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.
A0/2= A0ek5715After 5715 years, A = A0/2
1/2= ekt5715 Divide both sides of the equation by A0.
ln(1/2) = ln ek5715Take the natural logarithm on both sides.
ln(1/2) = 5715kln ex = x.
k = ln(1/2)/5715=-0.000121Solve for k.
Substituting for k in the decay model, the model for carbon-14 is
A = A0e–0.000121t.
A = A0e-0.000121tThis is the decay model for carbon-14.
0.76A0 = A0e-0.000121tA = .76A0 since 76% of the initial amount remains.
0.76 = e-0.000121tDivide both sides of the equation by A0.
ln 0.76 = ln e-0.000121tTake the natural logarithm on both sides.
ln 0.76 = -0.000121tln ex = x.
t=ln(0.76)/(-0.000121)Solver for t.
The Dead Sea Scrolls are approximately 2268 years old plus the number of years between 1947 and the current year.
V = 125e.07t
200 = 125e.07t
1.6 = e.07t
ln1.6 = ln e.07t
ln 1.6 = .07t
ln 1.6 / .07 = t
6.71 = tExample