1. No category

chapter 2
exponential growth and decay – solutions
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Exercise 2A
1.
2.
Simplify each of the following.
a)
( xy )3 ⋅ x 4 y 5 = x 7 y 8
b)
( 4mn )
c)
6 p 8 m 4 ⋅ 2 p 7 m 6 4 10 8
= p m
9 p5 m2
3
d)
34 ⋅ 27 2
3
=
4
5
6 ⋅3
16
4 2
÷ 14n 3 =
8 2 5
mn
7
Find x in each of the following.
a) 4 2 x = 8 x−1
x = −3
b) 16 4−3x =
x=
1
8 x+3
25
9
c) 34 x ⋅ 27 x+3 = 81
x=−
5
7
( )
d) 32 x − 4 3x + 3 = 0
L = {0;1}
( )
( )
e) 2 25 x + 10 = 12 5 x
L = {0;1}
3.
The value of a car bought for $25,000 decreases by a fixed amount each year. If the value after
4 years is $17,400 and this trend continues, what will be the car’s value after 7 years?
$11,700
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chapter 2
exponential growth and decay – solutions
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Ch. 2.3 Demographic development in the USA between 1800 and 1860
year
population (in Mio.)
variation (in Mio.)
variation (percentage)
1800
5.3
-
-
1810
7.2
+1.9
35.8%
1820
9.6
+2.4
33.3%
1830
12.9
+3.3
34.4%
1840
17.1
+4.2
32.6%
1850
23.2
+6.1
35.7%
1860
31.4
+8.2
35.3%
Task A:
Determine the variation (in Mio.). Is this growth linear? Explain!
Obviously no linear growth, as variation not constant.
Task B1:
Determine the percentage variation. What can you recognise?
see table
Task B2:
In a book you can read that the arithmetic mean 34.5% is a reasonable approximation
for all decades. What do you think? Explain again!
With the suggested variation of 34.5% we get:
1800
5.3
Task C1:
1810
7.1
1820
9.6
1830
12.9
1840
17.3
1850
23.3
1860
31.4
What is the factor for the population to get from one row to the other?
The factor is: 1.345
Task C2:
What would the population have been in 2010 if it had continued to grow at a rate of
34.5%? How can you calculate this? What do you think about your result?
It is about finding a formula. Reducing the difficulties, we take:
B1 = B0 + B0 ⋅ 0.345 = B0 ⋅ (1+ 0.345 ) = B0 ⋅1.345
B2 = B1 + B1 ⋅ 0.345 = B1 ⋅ (1+ 0.345 ) = ( B0 ⋅1.345 ) ⋅1.345 = B0 ⋅1.345 2
(
)
B3 = B2 + B2 ⋅ 0.345 = B2 ⋅ (1+ 0.345 ) = B0 ⋅1.345 2 ⋅1.345 = B0 ⋅1.345 3
In general: Bt = B0 ⋅1.345
t
There are 21 decades. Therefore B2010 = B1800 ⋅1.345 21 = 5.3⋅1.345 21 = 2675.74 . This is
about 40% of the world population.
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chapter 2
Task D:
exponential growth and decay – solutions
V1.0
In the table, the variation is calculated per decade. Someone would like to know the
population in the year 1835.
Bradley declares the following formula to be the right one to solve this problem:
B1835 = B1800 ⋅1.1725 7 , where B1835 is the population in the year 1835 and B1800 the one in
1800. Moreover, there are seven 5-year-steps needed to get from 1800 to 1835 – hence
“to the power of 7”. 0.1725 is half of the decade variation which is 0.345. Why is
Bradley’s reasoning wrong? What would the correct calculation be?
1800 – 1810 is a decade: variation 0.345
1800 – 1805 and 1805 – 1810: look for a variation a , so that a 2 = 0.345 .
7
Therefore it should be: B1835 = B1800 ⋅1.345 2 .
Task E:
We try to generalise:
We are looking for a formula which describes the population growth for a percentage
variation of r% . B0 should stand for the population at the beginning, Bt for the
population after 𝑡 years. What is the formula?
t
r ⎞
⎛
(compare with C2): Bt = B0 ⋅ ⎜ 1+
⎝ 100 ⎟⎠
where
B0 = population at the beginning
t = number of growing periods
Bt = final or total population after t periods
r = growing rate per period
Exercise 2B
review questions
1.
Using 𝑡 to represent time in years, write an equation to describe the increase in value of a
painting if it was bought for $5,000 and its value increased by a factor of 1.05 each year.
V ( t ) = 5,000 ⋅1.05 t
2.
Using 𝑡 to represent time in hours, write an equation to describe the increase in number of a
population of bacteria if initially there are 860 bacteria and the number increases by a factor of
1.25 each hour.
P ( t ) = 860 ⋅1.25 t
3.
Using 𝑡 to represent time in hours, write an equation to describe the decrease in number of a
population of bacteria if initially there are 8,200 bacteria and the number decreases by 12%
each hour.
P ( t ) = 8,200 ⋅ 0.88 t
4.
The value of a machine bought for $120,000 decreases by 15% each year. Find how much the
machine is worth after 6 years.
$45,257.90
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chapter 2
exponential growth and decay – solutions
V1.0
basic skills
5.
Lillian wishes to have $24,000 in a bank account after 5 years so that she can buy a new car.
The account pays interest at 2.5% p.a. How much (to the nearest dollar) should Lillian deposit in
the account now, if she is to reach her target?
$21,213
6.
The half-life of iodine-131 is 8 days. How much is left after 10 days?
0.420=42%
7.
The half-life of cobalt-60 is 5.3 years. How long do we have to wait if we want to have 90% left?
4.3 years
8.
Sophie has been told that if you invest $500 at 8% p.a. compounded annually then its value will
double in 9 years. Is this true?
yes
9.
The value of a car bought for $25,000 decreases by a fixed factor each year. If the value after 4
years is $13,675 and this trend continues, what will be the car’s value after 7 years?
$8,698
10.
A gallery owner sells one of his artworks for $62,000 to an art lover. Assume an increase in
worth of 5% per year. What was the value of the artwork 50 years ago?
$5,406
11.
The population of a certain type of bacteria grows by 7% each hour. How long do you have to
wait for the population being doubled?
10.25 hours
12.
Linda says that her car depreciated 30% each year and now its worth is $8,800. How much did
she pay if she owns the car for five years?
$52,360
13.
Carbon dating in archaeology is based on the decay of the isotope carbon-14, which has a halflife of 5715 years. By what percentage does carbon-14 decay in 100 years?
1.2%
Ch. 2.6 Logarithm (questions)
1.
Find x :
a) 8 x = 2 ⇒ x =
1
3
b) 8 = 4 x ⇒ x =
3
2
4
chapter 2
exponential growth and decay – solutions
c) 2 x =
1
⇒ x = −1
2
d) 9 x = 1 ⇒ x = 0
e) 16 − x = 210 ⇒ x = −
f)
1
h) 2 x =
2.
1
2
9 0.25 = 3x ⇒ x =
g) 815 = 3x ⇒ x =
5
2
4
5
1
1
⇒x=−
2
2
Find x :
a) 32 x+1 = 81 ⇒ x =
3
2
b) 5 2 x+1 = 3125 ⇒ x = 2
c) 0.5 2 x+2 = 2 ⇒ x = −
3
2
d) 34 x = 9 x+5 ⇒ x = 5
Exercise 2C
basic skills
1.
Write each of the following in the form a x = b .
a) log 3 ( 81) = 4 ⇔ 34 = 81
b) log 7 ( x ) = 4 ⇔ 7 4 = x
c) log x ( 5 ) = t ⇔ x t = 5
d) log p ( q ) = r ⇔ p r = q
2.
Write each of the following in the form x = log a ( b ) .
a) 36 = 729 ⇔ log 3 ( 729 ) = 6
b) a 8 = 20 ⇔ log a ( 20 ) = 8
c) h 9 = k ⇔ log h ( k ) = 9
d) m n = p ⇔ log m ( p ) = n
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chapter 2
3.
exponential growth and decay – solutions
Evaluate the following.
a) log 4 (16 ) = 2
1
⎛ 1⎞
b) log 27 ⎜ ⎟ = −
⎝ 3⎠
3
(
)
2
(8 2 ) = 7
c) log 2 2 2 =
d) log
4.
3
2
Find the value of x in each of the following.
a) log x ( 49 ) = 2 ⇒ x = 7
b) log 3 ( 81) = x ⇒ x = 4
c) log 2 ( x ) = 2.5 ⇒ x = 4 2
d) log x ( 27 ) = −6 ⇒ x =
5.
1
3
3
If a is a positive number, express the following in their simplest form.
( )
a) log a a 5 = 5
b) log
(a ) = 8
4
a
( a ) = 12
c) log a
d) log
a
(a) = 2
Exercise 2D
basic skills
1.
Write each of the following in terms of log ( p ) , log ( q ) and log ( r ) .
(
)
a) log pq 2 r 3 = log ( p ) + 2 log ( q ) + 3log ( r )
⎛
b) log ⎜
⎝
p2q ⎞
1
1
⎟ = log ( p ) + log ( q ) − log ( r )
r ⎠
2
2
1
⎛ p⎞
= log ( p ) − log ( r )
c) log ⎜
⎟
⎝ r⎠
2
⎛ q 2 rp ⎞
= log 2 ( p ) + 2 log 2 ( q ) + log 2 ( r ) − 1
d) log 2 ⎜
⎝ 2 ⎟⎠
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V1.0
chapter 2
2.
exponential growth and decay – solutions
V1.0
Express as a single logarithm, simplifying where possible. All the logarithms have base 10.
a) 2 log ( 5 ) + log ( 4 ) = 2
1
b) log ( 24 ) − log ( 9 ) + log (125 ) = 3
2
( )
c) 3log ( 2 ) + 3log ( 5 ) − log 10 6 = −3
d)
3.
1
1
log (16 ) + log ( 8 ) = log ( 8 )
2
3
Suppose that log b ( c ) = r and log c ( b ) = s . Write these equations in their equivalent exponential
forms, and use the rules for indices to show that rs = 1 . Write this result as an identity
connecting log b ( c ) and log c ( b ) .
c = b r ; b = c s ; log b ( c ) ⋅ log c ( b ) = 1
4.
Solve the following equations, giving those answers which are inexact correct to 3 significant
figures.
a) 7 x = 21 ⇒ x ≈ 1.56
b) 5 2 x−1 = 10 ⇒ x ≈ 1.22
c) 2 x+1 = 3x ⇒ x ≈ 1.71
⎛ 1⎞
d) ⎜ ⎟
⎝ 4⎠
5.
2 x−1
= 7 ⇒ x ≈ −0.202
Solve the equation:
a) log 2 ( x + 2 ) − log 2 ( x ) = 3 ⇒ x =
2
7
b) 2 log ( x + 5 ) = 1+ log ( 2x + 10 ) ⇒ x = 15
6.
A radioactive isotope decays so that after t days an amount 0.82 t units remains. How many
days does it take for the amount to fall to less than 0.15 units?
9.56
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chapter 2
7.
exponential growth and decay – solutions
(
V1.0
)
The number of bacteria in a culture, N , is given by N t = 1500 2 0.18t , where t is the number of
days.
a) Find the initial number of bacteria in the culture.
1,500
b) Find the number of bacteria (to the nearest 100) after: i) 5 days ii) 10 days.
i) 2,800 ii) 5,200
c) How many days does it take for the number of bacteria to reach 9000?
14.36 days
further explorations
8.
st
Bank accounts: On the 1 of January, Brendan has $48,000 on a bank account which pays
interest at 1.2% p.a. However, he also has another account with $52,000 which pays interest at
0.875%. How long does he have to wait for both accounts being equal?
25 years
9.
Laboratory: In a laboratory there are two types of bacteria. The first one consists of an initial
population of 482 and the number increases by 4% every hour. The other one consists of an
initial population of 4’620 and decreases by 6% every hour. How long does lab assistant have to
wait for both populations being equal?
22.4 hours
10.
⎛ P2 ⎞
Decibels: The intensity of sound in decibels is equal to 10 log10 ⎜ 2 ⎟ , where P is the
⎝ PR ⎠
amplitude of the sound pressure wave and PR is a reference pressure of 0.0002 microbars,
which is chosen as the pressure of an undetectable sound.
a) Write this formula in a simpler form.
20 log10 ( 5000P )
b) An airport imposes a restriction of 110 decibels on aircraft taking off. To what pressure does
this correspond?
63.2 microbars
c) A man and a woman are talking on the radio. The pressure on the listener’s ear of the man’s
voice is double that of the woman’s. What is the relationship between the sound intensity of
their voices?
The man’s sound intensity is 6.02 decibels more than the woman’s.
d) A new design of silencer reduces the noise of an engine by 10 decibels. What effect does
this have on the amplitude of the sound pressure waves?
The amplitude is reduced by a factor of 0.316.
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chapter 2
11.
exponential growth and decay – solutions
V1.0
If log r ( p ) = q and log q ( r ) = p . Prove that log q ( p ) = pq .
log r ( p ) = q ⇔ r q = p (1)
log q ( r ) = p ⇔ q p = r (2)
substitute r in (1):
( )
⇒ qp
q
= p ⇔ q pq = p ⇔ log q ( p ) = pq
Ch. 2.6.5 Logarithm (Richter scale)
Task A:
Find the Richter scale value for an earthquake that releases the following amounts of
energy:
1.
1,000 kJ
1.1
2.
2,000 kJ
1.3
3.
10,000 kJ
1.77
4.
1,000,000 kJ
3.1
Task B:
Does doubling the energy released double the Richter scale value?
No
Task C:
Find the energy released by an earthquake of:
1.
magnitude 4 on the Richter scale
22, 387,211kJ
2.
magnitude 5 on the Richter scale
707,945, 784 kJ
3.
magnitude 6 on the Richter scale
2.24 ⋅1010 kJ
Task D:
What is the effect (on the amount of energy released) of increasing the Richter scale
value by 1?
Energy is multiplied by 32
Task E:
Why is an earthquake measuring 8 on the Richter scale so much more devastating than
one that measures 5?
The amount of energy in a magnitude 8 quake is 2.24 ∙ 10!" kJ, which is 30,000 times
more than for a magnitude 5 quake.
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chapter 2
exponential growth and decay – solutions
V1.0
Ch. 2.7 Exponential and logarithmic functions
Task A:
Fill in the table and sketch the graph of each function.
x
-4
-3
-2
-1
0
1
2
3
4
𝑓! 𝑥 = 2 !
1
16
1
8
1
4
1
2
1
2
4
8
16
𝑓! 𝑥 = 3 !
1
81
1
27
1
9
1
3
1
3
9
27
81
16
8
4
2
1
1
2
1
4
1
8
1
16
81
27
9
3
1
1
3
1
9
1
27
1
81
1
2
!
𝑓! 𝑥 =
1
3
!
𝑓! 𝑥 =
Task B:
Which properties of exponential functions can you recognise?
The graph is increasing if a > 1
The graph is decreasing if a < 1
In both cases:
- the y − intercept is ( 0 /1)
- the domain is !
- the asymptote is y = 0 ( x − axis)
- the range is ! +
Task C:
The definition for exponential functions states that a must be a positive number. Why is
it important that 𝑎 is positive? What would happen otherwise?
a < 0 causes problems as a flipping of the functions values (positive / negative)
10
chapter 2
Task A:
exponential growth and decay – solutions
V1.0
Fill in the table and sketch the graph of each function.
x
-4
-3
-2
-1
0
1
2
3
4
𝑓! 𝑥 = log ! (𝑥)
-
-
-
-
-
0
1
1.58
2
𝑓! 𝑥 = log ! (𝑥)
-
-
-
-
-
0
0.63
1
1.26
𝑓! 𝑥 = log(𝑥)
-
-
-
-
-
0
0.30
0.48
0.60
𝑓! 𝑥 = ln(𝑥)
-
-
-
-
-
0
0.693
1.099
1.386
Task B:
Which properties of logarithmic functions can you recognise?
The graph is increasing if a > 1
The graph is decreasing if a < 1
In both cases:
- the x − intercept (=zero) is (1 / 0 )
- the domain is ! +
- the asymptote is x = 0 ( y − axis)
- the range is !
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chapter 2
exponential growth and decay – solutions
V1.0
Exercise 2E
basic skills
1.
P is a point on the graph of f ( x ) = a x . Determine a .
a) P (1.5 / 27 )
a=9
(
)
a=
1
≈ 0.564
π
b) P −6 / π 3
2.
In the first five sections we worked out the equation Bt = B0 ⋅ Rt . Explain why this equation
describes an exponential function and draw the graphs for Exercise 2B / No.1-4. Do not forget
to label the axes appropriately.
open
3.
Radioactive decay: The strength of a radioactive source is said to “decay exponentially”.
Explain briefly what is meant by exponential decay, and illustrate your answer by means of a
sketch-graph.
After t years the strength S of a particular radioactive source, in appropriate units, is given by
S = 10,000 ⋅ 3−0.0014t .
State the value of S when t = 0 , and find the value of t when the source has decayed to onehalf of its initial strength, giving your answer correct to 3 significant figures.
S ( 0 ) = 10,000 ; t = 451
4.
Finch population: It is feared that the population of a species of finch on a remote island is
endangered. At the beginning of the year 2000 it was estimated that there were 800 breeding
pairs. By the beginning of 2005 this number had dropped to 640. If the size of the population is
decreasing exponentially, in which year would you expect the number of breeding pairs to drop
below 200?
2031
further explorations
5.
Music: An orchestra tunes to a frequency of 440, which sound the A which is 9 semitones
above middle C. Each octave higher doubles the frequency, and each of the 12 semitones in
the octave increases the frequency in the same ratio.
a) What is the ratio?
12
2 ≈ 1.059
b) Find the frequency of middle C.
262
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chapter 2
exponential growth and decay – solutions
V1.0
c) How many semitones above the tuning A is a note with a frequency of 600? Where is this on
the scale?
between 5 and 6; between D and D sharp (= dis)
6.
Fishing: The population, 𝑃, of a certain fish 𝑡 months after being introduced to a reservoir is
modelled by
⎧ 400 ⋅10 0.08t
⎪
P (t ) = ⎨
⎪⎩ 15,000 + 924 log10 (10 ( t − 19 ))
for 0 ≤ t ≤ 20
for t ≥ 20
as 20 months after the introduction, fishing is allowed.
a) Draw the graph of P ( t ) .
b) Find the population
i.
at the beginning
400
ii.
after 5 months
1,005
iii.
after 15 months
6,340
iv.
after 25 months
16,643
v.
after 40 months
17,146
c) How long does it take the population to pass 10,000?
17.48 months
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chapter 2
7.
exponential growth and decay – solutions
V1.0
Tea-time: A cup of tea is poured out at a temperature of 99°C. Two minutes later its
temperature is 96°C. The temperature in the room is 15°C. If the difference between the
temperature of the tea and the room temperature decreases exponentially with time, write an
expression for the temperature of the tea in the cup t minutes after it is poured.
If you like to drink your tea at a temperature between 90°C and 85°C, how long after it is poured
would you drink it?
(
)
Tt = T ( t ) = 15 + 84 ⋅ 0.982 t º C ; between 6.2 and 10.0 minutes
8.
Intersecting graphs: Given the functions f1 ( x ) = 2 x and f2 ( x ) = −2x + 3 .
a) Determine the type of function of f1 and f2 .
f1 is an exponential function
f2 is a linear function
b) Sketch the two functions.
open
c) Determine the point of intersection of the two functions. How could you do this without
calculator?
solve the equation 2 x = −2x + 3
using the calculator:
x ≈ 0.692153
without calculator (looking at the sketched graph):
be x = 0 :
RHS: 2 0 = 1 ; LHS: −2 ⋅ ( 0 ) + 3 = 3
be x = 1 :
RHS: 21 = 2 ; LHS: −2 ⋅ (1) + 3 = 1
conclusion: value must be 0 < x < 1
be x = 0.5 :
RHS: 2 0.5 ≈ 1.41421 ; LHS: −2 ⋅ ( 0.5 ) + 3 = 2
conclusion: value must be 0.5 < x < 1
be x = 0.75 :
RHS: 2 0.75 ≈ 1.68179 ; LHS: −2 ⋅ ( 0.75 ) + 3 = 1.5
conclusion: value must be 0.5 < x < 0.75
go on…
14