Constructing Exponential Models: Solutions

Here are some solutions to the Exercises to accompany the section Constructing Exponential Models.

Modeling Exponential Functions

For each of the following tables, determine if it represents an exponential function, and, if so, determine the corresponding formula for that function.

x	y = f(x)
-1 0 1	-2 -6 -18

Solution: The initial value is -6 and the change factor between each pair of values is, b = (-6/-2)^{1/(0 - -1)} = 3 = (-18/-6)^{1/(1 - 0)}, so f(x) = -6·3^t.

x	y = g(x)
0 1 3	36 12 4

Solution: The change factor between the first pair of values is, b = (12/36)^{1/(1 - 0)} = 1/3, but the change factor between the second pair is b = (4/12)^{1/(3 - 1)} » 0.57735, so g is not given by an exponential function.

x	y = h(x)
0 1 3	17.01 22.68 40.32

Solution: The initial value is 17.01 and the change factor between each pair of values is, b = (22.68/17.01)^{1/(1 - 0)} = 1.33... = (40.32/22.68)^{1/(3 - 1)}. Writing this as a fraction, b = 4/3, so h(x) = 17.01(4/3)^t.

x	y = k(x)
1 4 6	-92.16 -38.88 -21.87

Solution: While we cannot original see the initial value, the change factor between each pair of values is, b = (-38.88/-92.16)^{1/(4 - 1)} = .75 = (-21.87/-38.88)^{1/(6 -
4)}, that is, b = 3/4, so k(x) = C(3/4)^t. Plugging in x = 1 gives -92.16 = k(1) = C(3/4)¹ = 3/4C, so C = -92.16 (4/3) = -122.88, and k(x) = -122.88(3/4)^t.

Back to Exercises.

Have your partner fill-in-the-blanks of f(x) = __·__^x to create a formula for an exponential function, and keep it hidden from you. Then have your partner make a table of three pairs of values for his/her function, and show it to you. Make sure to verify that the values do, in fact, correspond to an exponential function and try to guess your partner's function.
Repeat this Exercise as often as necessary until you are confident in your ability to discover the formula for an exponential function.

Solution

Your partner should correct your work.

Back to Exercises.

Growth, Decay, and Exponential Functions

Have your partner fill-in-the-blanks of f(x) = __·__^x + __ to create a formula with an exponential "core", and keep it hidden from you. Then have your partner make define a new function with this formula in XFunctions:

Try to guess your partner's function by looking only at its graph, and using the "x =" box to read off values from the graph. If you cannot see the value of the horizontal asymptote, your partner should tell it to you.
Repeat this Exercise as often as necessary until you are confident in your ability to discover the formula for an exponential function.

Solution

Look at the function definition in XFunctions, or ask your partner.

Back to Exercises.

The following situations can be modeled by a function with an exponential "core". Determine the functional relationship between the indicated variables.

You have a sore throat, so your doctor takes a culture of the bacteria in you throat with a swab, and lets it grow in a petri dish. If there were originally 500 germs in the dish and they grow continuously by 150% per day, give the relationship, f, between N = the number of germs in the dish (in hundreds), in terms of t = the time since the culture was taken (in days).

Solution

Using the continuous compound interest formula, with I = 5 and r = 1.50, gives N = f(t) = 5e^1.5^t.
Radioactive iodine, I¹³¹, used to test for thyroid problems, has a half-life of 8 days. If you start with 20 g. of radioactive iodine, give the relationship, h, between A = the amount of I¹³¹ (in g.), in terms of t = the time since you first measured the sample (in days).

Solution

Using the half-life formula, with I = 20 and h = 8, gives A = h(t) = 20(1/2)^t/8.
Xenon gas, Xe¹³³, is used in medical imaging to study blood flow in the heart and brain. When breathed, it is quickly absorbed into the bloodstream, and is gradually eliminated from the body (through exhalation) so that there is half has much in the body after about 5 min. (i.e., Xe¹³³ has a biological half-life of 5 min.). If you originally breathe in and absorb 3 mL. of Xe¹³³, give the relationship, k, between A = the amount of Xe¹³³ in your body (in mL.), in terms of t = the time since you breathed it in (in min.).

Solution

Using the half-life formula, with I = 3 and h = 5, gives A = h(t) = 3(1/2)^t/5.

You are hiking and want to cool your bottle of water by setting it in a cold mountain stream. Assume that your bottle of water starts at 70ºF and the stream is is 34ºF, and after 1 minute your water cools to 62ºF. Give the relationship, s, between T = the temperature of your water (in ºF), in terms of t = the time since you placed your bottle in the stream (in min.).

Solution

If we leave the bottle in the stream long enough, they will have the same temperature. This means that s has a horizontal asymptote of T = 34. Newton's Law of Cooling says that s(t) = Cb^t + 34. We know that:

t	T = s(t)
0 1	70 62

so that s(t) - 34 is an exponential function with:

t	s(t) - 34 = Cb^t
0 1	36 28

This has initial value, C = 36, and change factor, b = 28/36 = 7/9, so that s(t) = 36(7/9)^t + 34.

Have your partner imagine an applied situation similar to one given in the text. Have him/her describe this situation to you, and see if you can write down a correct equation.

Solution

Your partner should correct your work.

Back to Exercises.

Go to Logarithmic Functions.

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