Andrew M. answered • 06/19/15

Mathematics - Algebra a Specialty / F.I.T. Grad - B.S. w/Honors

_{o}e

^{kt}

_{o}is the starting amount = $231

^{12k}

^{12k}= 455/231

^{12k}) = ln(455/231)

Billy B.

asked • 06/18/15A toy tractor sold for $231 in 1975 and was sold again in 1987 for $455. Assume that the growth in the value V of the collectors item was exponential.

Find the value K of the exponential growth rate. Assume V=$231

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Andrew M. answered • 06/19/15

Tutor

New to Wyzant
Mathematics - Algebra a Specialty / F.I.T. Grad - B.S. w/Honors

We need to use the exponential growth formula:

P(t) = P_{o}e^{kt}

P(t) is the amount over time = $455

P_{o} is the starting amount = $231

k is our rate of growth (to be found)

t = time = 12 years (1987-1975 = 12 years)

455 = 231e^{12k}

e^{12k} = 455/231

Take the natural log of both sides

ln(e^{12k}) = ln(455/231)

12k ln(e) = ln(455/231)

Note that ln(e) = 1 so we have

12k = ln(455/231)

k = ln(455/231)/12

k = .0565 or 5.65%

Our exponential growth rate is approximately 5.65%

Stephanie M. answered • 06/19/15

Tutor

5.0
(699)
Private Tutor - English, Mathematics, and Study Skills

In general, V(t) = V_{0}K^{t}, where t is time that has passed, V(t) is value after t time has passed, V_{0} is initial value, and K is rate of growth.

You're told that V_{0} = 231. After t = 12 years have passed, V(12) = 455. Plug those values in and solve for K:

455 = 231K^{12}

1.9697 = K^{12}

1.058 = K

Andrew M.

Stephanie,

We are told to assume the growth rate is exponential.

Thus, we have to use the exponential growth rate formula P(t) = P_{o}e^{kt}

For this problem our original equation to solve becomes 455 = 231e^{12k}

From this we solve for k

Report

06/19/15

Andrew M.

Stephanie,

This problem states we assume the growth is exponential; thus we have to use the exponential growth rate formula

P(t) = P_{o}e^{kt}

For this problem we have 455 = 231e^{12k} and we must solve for k

Report

06/19/15

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Stephanie M.

I've been looking through my old answers and, although this one is several months old, I feel I ought to explain a bit about exponential growth just in case another student happens upon it.

First, it's important to note that there's no specific "exponential growth rate formula." Exponential growth describes any situation in which something is increasing exponentially. That is,

exponential growth occurs whenever you see the variable in the exponent. So, a basic exponential growth formula looks like this:P(t) =a(r)^{t}. Replacingrwith the constantegives youoneexponential growth formula, usually used for computing continuously-compounded interest. It's not really applicable in this particular case.Second, in any exponential growth formula, you can immediately pick out the rate of growth

rand the initial amounta. So,P(t) = 50(1.2)^{t}models a situation where we started with 50 of something and it grew at a rate of 1.2, or 120%. That is, it increased 20% each year. You can tell when a quantity isincreasingbecause the rate will be over 1. Anything under 1 indicates an exponentialdecrease.Since the

rate always appears in a specific placein the formula (in the base, not in the exponent), you should not solve for a variable in the exponent and call it the rate, as Andrew does. And, if the rate of growth for something you know isincreasingcomes out to beless than one(as in Andrew's answer, where k=0.0565), you ought to check your work---this is a good indication that you've done something wrong.Therefore, please use the second answer (below) as a guide for problems like this, not the first.

04/23/16