Math 121.
Applications of Exponetials and Logarithms
Fall 2016
Instructions. Work in groups of 3 to solve the following problems. Turn them in at the end
of class for credit.
Names.
1. (Interest Income) Use the properties of logarithms and exponentials, along with the compound interest formula
r nt
A=P 1+
n
to answer the following questions.
(a) Suppose $73000 is invested at an annual interest rate of 3.5% compounded monthly. How
much will it be worth after 4 years? Express answer to nearest penny.
(b) How long will it take until the investment is worth $340000? Express your answer in years
rounded to one decimal place.
Solution: (a) We use P = 73000 (the amount invested), n = 12 (the number of compounding periods per year), r = 3.5/100 = 0.035 (the interest rate in decimal form) and
t = 4 (the number of years the money is invested). Then
0.035
A = 73000 1 +
12
(12)(4)
≈ 83952.88
Thus after 4 years the investment balance will be $83952.88
(b) We use P = 73000 (the amount invested), n = 12 (the number of compounding periods
per year), r = 3.5/100 = 0.035 (the interest rate in decimal form), t (the number of years
the money is invested is an unknown we wish to determine), and A = 340000 the amount
of the investment after t years. Thus we solve
12t
12t
340000
0.035
0.035
or
= 1+
340000 = 73000 1 +
12
73000
12
by taking logs of both sides of the equation to obtain
0.035
log(340000/73000)
≈ 44.02082
log(340000/73000) = 12t log 1 +
⇒ t=
12
12 log 1 + 0.035
12
Thus it will take approximately 44.0 years for an investment of $73000 to reach a balance
of $340000 when invested at 3.5% compounded monthly.
2. (Exponential Growth) The population of bacteria in a vat of potato salad at Bob’s All Day
Buffet is modeled by P (t) = P0 ekt . At noon there were 750 bacteria present and at 1:30 pm
there were 1500 bacteria present.
(a) Find the specific model for P (t) (i.e., use the information given to find P0 and k, and plug
those values into P (t) = P0 ekt ), and then use it to answer the following questions.
(b) How may bacteria were present in the potato salad at 11:30 am when was placed in the
buffet? Express answer to nearest whole number.
(c) How may bacteria were present in the potato salad at 3:30 pm when the potato salad was
removed from the buffet?
(d) If the potato salad had been allowed to remain in the buffet indefinitely, and the model
for the bacteria remained valid, how many hours after noon, when there were 750 bacteria
present would it have taken for the bacteria population to reach 39750. Express answer in
hours, rounded to nearest decimal place.
(For all of these questions, assume no one ate took any potato salad because of its funny smell
and hence the population of bacteria remained in tact and followed the given growth model).
Solution: (a) Let t be measured in hours and let t0 = 0 at 12 noon. Now P0 = 750 is
the population when t = 0. Thus P (t) = 750ekt and so we need to find k. Also, we know
P (1.5) = 1500 because we are told there were 1500 bacteria present at 1:30 pm, and t = 1.5
at 1:30 pm. Therefore, 1500 = 750e1.5k . Then ln(1500/750) = ln(e1.5k ) and so
1.5k = ln(1500/750)
⇒
k=
ln(1500/750)
≈ 0.46209812
1.5
Therefore, P (t) = 750e0.46209812t .
(b) At 11:30am, t = −1/2 and so there were approximately 750e(0.46209812)(−.5) ≈ 595
bacteria present in the vat of potato salad.
(c) At 3:30 pm, t = 3.5 and so there were approximately 750e(0.46209812)(3.5) ≈ 3780 bacteria
in the vat of potato salad.
(d) For this, solve P (t) = 39750 for t. That is 750e0.46209812t = 39750 and so
39750
39750
0.46209812t
0.46209812t
e
=
⇒ ln e
= ln
⇒
750
750
39750
39750
1
ln
0.46209812t = ln
⇒
t=
≈ 8.59188069
750
0.46209812
750
and so it would take approximately 8.6 hours, after noon, for the bacteria population to
reach 39750.
3. The population of a small city is currently 52000 and is growing at 3 percent per year. Thus
the population is given by
P (t) = 52000(1.03)t
where t is time measured in years from the present.
(a) What will the population of the city be in one year?
(b) According to this model, what will the population of the city be in 11 years from now?
Express answer to the nearest whole number.
(c) If this rate of growth continues, how long will it take for the population to quadruple.
Express answer in years rounded to two decimal places.
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Solution: (a) P (1) = 52000(1.03) = 53560 which represents an increase of 3 percent over
the current populaton.
(b) Assuming the current growth rate continues, the population will be
P (t) = 52000(1.03)11 = 71980.
(c) In this case we solve 4(52000) = 52000(1.03)t and so 4 = (1.03)t this implies
ln(4) = t ln(1.03)
⇒
t=
ln(4)
≈ 46.90
ln(1.03)
The population will quadruple in approximately 46.90 years.
4. (Carbon Dating) Carbon-14 has a half-life of 5730 years, and satisfies the exponential-decay
t/5730
1
.
equation N (t) = N0
2
(a) If an ancient scroll is discovered to have 24.5% of its original Carbon-14, how old is the
scroll? Round answer to nearest year.
(b) What percentage of a bone’s original Carbon-14 would you expect to find remaining in a
bone that is 3250 years old? Round to the nearest tenth of one percent.
Solution: (a) Solve 0.245N0 = N0 (.5)t/5730 and so 0.245 = (.5)t/5730 and so
ln(0.245) =
5730 ln(0.245)
t ln(.5)
or t =
≈ 11627 years
5730
ln(.5)
Thus the scroll is approximately 11627 years old.
(b) N (t) = N0 (.5)3250/5730 ≈ 0.675N0 . This implies that we would expect to find approximatly 67.5% of the original amount of the Carbon-14 remaining in the bone.
5. An unknown radioactive element decays into non-radioactive substances. In 740 days the
radioactivity of a sample decreases by 30 percent, that is 70 percent of the original substance
remains.
(a) What is the half-life of the element?
(b) How long will it take for a sample of 100 mg to decay to 90 mg?
t/H
1
Solution: The radioactive decay equation is A(t) = A0
where H is the half-life in
2
the units corresponding to t.
a) We are given that after 740 days, only 70 percent of the original sample would remain.
Thus we solve
740/H
1
0.70A0 = A0
2
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This implies
740
ln(70)
740 ln(.5)
=
and so H =
H
ln(.5)
ln(0.70)
and so the half-life is approximately 1438.085075 days.
b) The amount of time for a 100 mg sample to decay to a 90 mg sample would be given by
t/1438.085075
1
90 = 100
2
or
t=
1438.085075 ln(0.90)
≈ 218.6 days
ln(.5)
6. According to Newton’s Law of Cooling an object placed in a refigerator with a constant
temperature of 37◦ F has its temperature (in degrees Fahrenheit) given by
T (t) = 37 + Ce−kt
where C and k are constants. Suppose a can of soda (the soda is rather warm because the can
was in a warm car) had a temperature of 95◦ F when it was placed in the refrigerator and 20
minutes later, the soda has cooled to 81◦ F.
(a) Find the C and k for the temperature equation above. Use t in minutes with t = 0 being
the time when the can of soda was placed in the refrigerator.
(b) What will be the temperature of the can of soda after 30 minutes? Express answer to the
nearest degree.
(c) How long will it take for the can of soda to reach 45◦ F. Express answer to the nearest
minute
Solution: We know T (0) = 95 and so 37 + Ce0 = 95 and then C = 95 − 37 = 58. Then
T (20) = 81 implies
81 = 37 + 58e−20k
⇒
e−20k =
44
58
k=−
1
ln(44/95)
20
and so k ≈ 0.01381267.
(b) In 30 minutes, the temperature will be approximately
T (30) = 37 + 58e−(30)(0.01381267) ≈ 75◦ F
(c) Solve 45 = 37 + 58e−0.01381267t and so e−0.01381267t = 8/58 and then
t=−
1
ln(8/58) ≈ 143.
0.01381267
The soda temperature will reach 45◦ F approximately 143 minutes after being placed in the
refrigerator.
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