Exponential Growth and Decay

The formula for exponential growth and decay is given by

$$y=Ce^{kt},$$

where $C$ is the initial amount (the amount when $t=0$) since $t=0\Rightarrow y=Ce^0=C(1)=C$, and

the constant $k$ is the growth constant (if $k>0$) or the decay constant (if $k<0$).

An example of exponential growth is given by continuous compounding of interest. In this case, the constant $C$ is the principal (the initial amount), and the constant $k$ is the annual interest rate.

Notice that in exponential growth or decay, the proportional amount of change in two time intervals of the same length is the same:

If $[a,a+h]$ and $[b,b+h]$ are two time intervals of length $h$, then

$\frac{y(a+h)}{y(a)}=\frac{Ce^{a+h}}{Ce^a}=e^h$ and

$\frac{y(b+h)}{y(b)}=\frac{Ce^{b+h}}{Ce^b}=e^h$, so

$$\frac{y(a+h)}{y(a)}=\frac{y(b+h)}{y(b)}.$$

Ex 1 A bacterial culture is growing exponentially, and its mass increases from 2 mg to 5 mg in 30 hours. Find the mass of the culture after 25 hours.

Sol Since $C=2$, we get that $y=Ce^{kt}=2e^{kt}$.

Since $y=5$ when $t=30$, $2e^{30k}=5$; so $e^{30k}=5/2$. Taking natural logarithms on both sides yields $30k=\ln 5/2$, so $k=\frac{1}{30}\ln 5/2$.

Substituting back in the formula for $y$ gives $y=2e^{(\frac{1}{30}\ln 5/2)t}=2(e^{\ln 5/2})^{t/30}=2(5/2)^{t/30}$;

so when $t=25$, $y=2(5/2)^{25/30}=2(5/2)^{5/6}$ mg.

Ex 2 A radioactive isotope has a half-life of 450 years. Find how long it will take a sample of the isotope to decrease from 6 mg to 2 mg.

Sol Since $C=6$, we get that $y=Ce^{kt}=6e^{kt}$.

Since the half-life is 450 years, $y=3$ when $t=450$; so

$6e^{450k}=3$ and $e^{450k}=3/6=1/2$. Taking natural logarithms on both sides gives $450k=\ln 1/2$, so $k=\frac{1}{450}\ln 1/2$.

Substituting back in the formula for $y$ gives $y=6e^{(\frac{1}{450}\ln 1/2)t}=6(e^{\ln 1/2})^{t/450}=6(1/2)^{t/450}$;

so setting $y=2$ gives

$6(1/2)^{t/450}=2$ or $(1/2)^{t/450}=2/6=1/3$.

Taking natural logarithms on both sides gives

$\frac{t}{450}\ln 1/2=\ln 1/3$, so $t\ln 1/2=450\ln 1/3$ and therefore

$$t=\frac{450\ln 1/3}{\ln 1/2}=\frac{450(-\ln 3)}{-\ln 2}=\frac{450\ln 3}{\ln 2}\mbox{ years.}$$

In the following problems, leave your answers in exact form:

Pr 1 A bacterial culture which is growing exponentially increases from 5 g to 7 g in 11 hours. How long does it take the culture to double its mass?

Pr 2 A sample of a radioactive isotope decreases from 5 mg to 4 mg in 24 years. Find the half-life of the isotope.

Pr 3 The population of a town increases from 15,000 to 25,000 in 10 years. Assuming that the population is growing exponentially, how long does it take the population to increase by 40% ?

Pr 4 A bank account earns interest at an annual rate of 5% compounded continuously. How long will it take the amount of money in the account to triple?

Pr 5 Suppose Rex has learned a list of 500 vocabulary words, and that the number of words he remembers after $t$ weeks decreases exponentially. If he remembers 450 words after 3 weeks, how long does it take for the number of words he remembers to decrease by 40% ?

Pr 6 Assume that the number of people infected by a newly discovered virus is growing exponentially. If the number of people infected increases from 300 to 600 in 5 weeks, how much additional time will it take before 4800 people are infected?

Pr 7 Suppose that the number of grasshoppers in a town is growing exponentially. If there were 8,000 grasshoppers initially and 18,000 grasshoppers after 30 days, find the number of grasshoppers in the town after 15 days.

Pr 8 Suppose that the number of bugs in a rainforest is growing exponentially. If there were 2880 bugs after 3 weeks and 5120 bugs after 5 weeks, how many bugs were there initially?

Pr 9 The populations of two states are growing exponentially. If state A currently has a population of 9 million and state B currently has a population of 11 million, and if the populations of the two states increase annually by 3% and 2%, respectively, when will their populations be equal?

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