Chapter 7: Exponential and Logarithmic Fynctions I 5

Chapter 7: Exponential and Logarithmic Fynctions I 5
Chapter 7: Exponential and
Logarithmic F y n c t i o n s
L o s s e s . /.I
hrplormrj
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the C h a r a c t e r i s t i c s
of txpoii-'rr.ti.jl I ijr.ct'CJrts, p a g e
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439
4o, linear
fes
40, quadratic
slo. cubic
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Jo, quadratic
2. b) No xontercepts; y-^intercept: y = 1
End behaviour; Qll to Ql
Domain; {x | x e R}; Range; (y | y > 0, y e R)
e) No x-intercepts; yontercept; y = 1
End behaviour; Qll to Ql
Domain; { x | x e R}; Range; {y | y > 0, y o R}
C...'
cl) Number of xontercepts; 0
yontercept: y = 4
Domain: {x | x e R}
flange; {y | y > 0, y e R}
End Behaviour; Qll to Ql
10
y
y > 0, y e R}
jhaviour Ql
' - oc/
4'
.
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L e s s o n 7.2: R e l a t i n g t h e C h a r a c t e r i s t i c s
J4.4
of a n E x p o n e n t i a l F u n c t i o n to Its
-N
E q u a t i o n , page 448
l!l
1.a)
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H4IS:
Term
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b) Number of xdntercepts; 0
yontercept; y = 6
Domain; {x | x o R}
Range: fv'1 K-'O, y.- R}
End Behoviour Ojil to 04!
28^14
Ui
2-;
Foundations of Mathematics 12 Solutions Manual
1
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I noticed that the difference between consecutive
y-values doubles from one pair of values to the
next. I know that there must be a base value of 2
being raised to an exponent and then multiplied
by some constant term in the eguation. The
equation looks like y = a(2'). Thus, the answer is
yes, because for each unit increase in x, the
value of y doubles.
1
Terrri
c) Number of x-interccpts: 0
yintercept; y = 27
Domain; {x | x o R}; Range: { y ! y > 0 , yc. R}
End Behaviour; Qll to Ql
r
j
Firi,l
Difference
2304
57 n
144
36
9
Second
Difference
3072 - 768
1/28
768 - 192
432
192 - 4 8
108
48 12
27
12 - 3
I noticed that the difference between consecutive
y-values is not constant. The function is not
7-1
4 . a) The yontercept is 5 and the function is
increasing.
y = 5(2)',
y = 5{2f»
1 noticed each y-^value c-. ^* the previous y-value
as X increases by 1, The function must be an
exponential function with a base value of -1
4
being raised to an exponent and then multiplied
by some constant term in the equation. Thus, the
answer is yes because for each unit increase in
X , the value of y is divided by 4.
2 a) Number of xontercepts; 0
yontercept; y = 4 ; Domain- {x | x o R}
Range; {y | y > 0, y o R}
End Behaviour; Qll to Ql
b) Number of xontercepts; 0
yontercept: y = 2; Domain: {x | x o R}
Range; {y | y > 0, y o R};
End Behaviour: Qll to Ql
c) Number of xontercepts; 0; yonterc<
Domain; { x j x e R); Range: (y | y > 0,
End Behaviour; Qll to Ql
d) Number of xontercepts; 0; yontercept; y = 3
Domain: { x | x e R}; Range: {y | y > 0, y G R )
End Behaviour; Qll to Ql
3, e.g. Increasing exponential functions increase
as X increases, whereas decreasing exponential
functions decrease as x increases.
fix rt'.iMiig
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1
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-
4
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The function is increasing because the base is
greater than 1,
b) The yontercept is 2 and the function is
decreasing.
y = 2(0.6)"
y = 2(0.5f^
y=2(1)
The function is decreasing because the base is
less than 1.
c) The yontercept is 10 and the function is
increasing.
y=
10(1.5)'
y=
10(1.5)^°^
y=
10(1,5)
Jo
The function is increasing because the base is
greater than 1.
d) The yontercept is 1 and the function is
decreasing
y=(0.4);
y= 1
The function is decreasing because the base is
less than 1.
'%
1 j
L
y=5(1)
2'"*
5. a) i) Yes, an exponential equation can be used
to model the function because the rate of change
in y-values doubles for each unit increase in x
ii) yontercept: y = 1, the function is increasing
b) i) No, y increases by 2 as x increases by 1
li) y-intercept; y = 3, the function is increasing.
c| i) Yes. an exponential equation can be used to
model the function because the rate of change in
y-values get divided by 4 for each unit increase in
X.
ii) yontercept; y = 64, the function is decreasing
d) i) No. e.g., y decreases, then increases then
decreases again.
ii) yontercept: y = 1, the function is first
decreasing than increasing as it reaches the
yontercept.
7-2
Chaptet 7 Exponential a n d L o i a r i t h m i c F u n c t i o n s
:-i:i-=H,tH- •„ x-intercepts: 0
i.^i. u'-pi V
'.; Domain: {x I X 6 R}
i. pv ' i
•=,]/€ R}
r =.o t;.-h..-'M)Ui Oil 1'. f H
f , } f y , ; , , j , : -4 - H P - . -ph. '.mrr.-.iv
I-:;
I t..'l C< h,ivi--tH • 41
•4p
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fo.nor
V intercept; y = 25
fv'!y>0,yeR}
')!
1 1
'.fl
- . 4
.P
i;.
• If)
rpiro
f.. p:.;.r.| < ;
, ,ro.K..:)'-~ 0 V intercept; y = 12
f x . n . - << y i r -
r
r c ••Of-
fj. j y > 0 , y o
R}
F nd Behaviour: Ql I to Ql
b) Number of xontercepts; 0; yontercept: y = 4
Domain: (x | x G R ) ; Range; f^ | y > 0, y o R }
End Behaviour; Qll to Ql
c) Number of xontercepts: 0; y-intercept: y = 2
Df.i..:4c fx I y-- P}. Px.cgo' tv j •/ > 0, y e R}
d l Number of xontercepts; 0; yontercept; y = 3.5
Domain; {x | x e R}; Range: {y | y > 0. y e R}
f e d B o h w i o o r Oil to Of
l i
F o u n d a t i o n s of Mathematics 12 S o i u t i o n s Manual
.
The y-^intercept is positive and since the
base is 8, it must mean the function is increasing
because the base is larger than 1.
b| The yontercept is positive and since the base
is 0.6, it must mean the function is decreasing
because the base is less than 1.
c) The yontercept is positive and since the base
is e. it must mean the function is increasing
because e is greater than 1.
8. a) Number of xontercepts; 0; yontercept; y = 4
Domain: {x | x e R}; Range: {y | y > 0, y o R}
End Behaviour: Qll to Q!
b) Number of xontercepts; 0; yontercept; y = 8
Domain; {x | x e R}; Range; {y | y > 0, y o R}
End Behaviour; Qll to Ql
c) Number of xontercepts; 0; yontercept: y = 3
Domain; {x | x e R}; Range: {y | y > 0, y e R}
End Behaviour: Qll to Ql
d | Number of xontercepts; 0; yontercept; y = 10
Domain; {x | x o R}; Range; {y | y > 0, y G R}
End Behaviour; Qll to Ql
e) Number of xontercepts; 0; yontercept; y = 30
Domain: {x | x G R } ; Range; {y | y > 0, y G R }
End Behaviour: Qll to Ql
f) Number of xontercepts; 0; yontercept; y = 1
Domain; {x | x G R } ; Range; {y | y > 0, y e R }
End Behaviour: Qll to Ql
g) Number of xontercepts: 0; yontercept: y = 3
Domain: (x | x o R ) ; Range: {y | y > 0, y e R }
End Behaviour; Qll to Ql
h) Number of xontercepts; 0; yontercept: y = 45
Domain; {x | x e R } ; Range; {y i y > 0, y G R }
End Behaviour: Qll to Ql
7-3
i . a) It is a d e c r e a s i n g e x p o n e n t i a l f u n c t i o n
b e c a u s e it i s a f u n c t i o n w i t h a n e x p o n e n t of x a n d
the base is less than 1,
b ) It is n o t a d e c r e a s i n g e x p o n e n t i a l f u n c t i o n
b e c a u s e a l t h o u g h it h a s a n e x p o n e n t of x , t h e
base is g r e a t e r than 1,
c) It is a d e c r e a s i n g e x p o n e n t i a l
function
because
it is a f u n c t i o n with an e x p o n e n t of x and the base
is less than 1.
1 1 . a) It is an increasing function with a
yontercept of (0.6). It matches with iii).
b) It is a decreasing function with a yontercept of
(0, 4), It matches with i).
c) It is a decreasing function with a yontercept of
(0. 2). It matches with ii).
d) It is an increasing function with a yontercept of
(0, 3). It matches with iv).
i ) It is not a d e c r e a s i n g e x p o n e n t i a l f u n c t i o n
1? ,»| y ii.hji. ...pl 5,
because a l t h o u g h it has an
base is greater than 1.
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exponent
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of x, the
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b| It is a decreasing function because its base is
m
b} yontercept; y = 7; Base: 4
Domain- {x I x < R). Range; (y | y > 0, y c R}
Infjoasinq or Decreasing- increasing
l e s s than 1.
-A
c) yontercept: y = 6; Base; | -
Ik
Domain; {x | x G R ) ; Range: {y | y > 0, y e R}
Increasing or Decreasing: Decreasing
mm
c) It is an increasing function because its base is
more than 1.
d) It is a decreasing function because its base is
less than 1.
7-4
Chaptof 7. t x p o n e n t i a l a n d L o g a r i t h m i c F u n c t i o n s
15. Yes. because all eiponentiai functions have
the form y=a(bf\ Polynomial functions are
characterized by having a numerical exponent
while exponential functions have variable
exponents.
R}
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or
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rcept; y=2
^ 3/ > 0. y G
R}
il! > 1) or
jreasing
liiii)
'Moreasing
re
le
yontercept and whether the function increases or
decreases are unique to the function.
IIP a;«;. • -'^o'}., -4 4 '
? = R} a; 2 b: 0.5
1J>; IC ' O -< f'.CS...'j UjlirhO'.O
ii) R a n g e 4 y I y > 0, /
Y; 1: 1 ft; 3
This is an increasing function.
iii) Range; {y | y > 0, y e R} a; 3 fe; 0 5
This is a decreasing function.
iv) Range; {y | y > 0, y e R} a; 2 ft; 4
This is an increasing function.
b) i) This function matches with the graph B.
because it is a decreasing function with a
yontercept of 2.
ii) This function matches with grop' I ; ^ .ocause it
is an increasing function with a y r m rr r.pt of 1.
iii) This function matches with the graph A.
because it is a decreasing function with a
yontercept of 3.
iv) This function matches with the graph C.
because it is an increasing function with a
yontercept of 2.
14. a) e.g. An example of an increasing
exponential function with a yontercept of 5 would
be y = 5(2)'' while an example of a decreasing
exponential function with a yontercept of 5 would
b e y =5(0.25)1
b) e.g. Same; number of xontercepts,
ydntercepts, and behaviour, domain, and range.
Different: rate of change (increasing vs.
decreasing function.
Foundations of Mathematics 12 Solutions Manual
18. a) Student A;
yontercept; y = 80
domain; {x | x > 0, x c R}
range; {y | 0 < y < 8, y o R)
Student B;
yontercept; y = 100
domain; {x | x > 0, x e R}
range; { y | 0 < y < 100, y e R }
b) e.g., Concentration of caffeine in blood
naturally decreases over time as the kidneys filter
it from the blood into the urine.
c) Student B consumed more caffeine. Student B
processed the caffeine more guickly. This can be
seen from how much they initially started with
and how much they consumed in four hours.
Student A had 80 mg of caffeine at the start and
consumed 60 mg after four hours. Student B
started with 100 mg of caffeine and within the
same time penod, had consumed 80 mg.
d) Both Student A and B had about 20 mg of
caffeine in their body after four hours.
e) The y-intercepts are the values representing
the initial amount of caffeine in the students'
bodies. The values are different because the
students drank different amounts of the same
dnnk, or they drank different dnnks that did not
have the same concentration of caffeine.
7-5
19 a)
i
Houi
i
"
I e s s o n 7 3: Modollincj D a t a U s i n g
E x p o n e n t i a l F u n c t i o n s , p a g e 461
I
'In;, d.ito :.(d docs m,l involve exponential
<lo(..iy bHc.uinc
differences are
t-un;>Liol t! v, a
tuiK.tic.ri
exponential growth
tur.
between the
V
unit value o f x .
StiicJonts who are TolcJ
(jiowlh Ol
the
linssof
b) ihr; thiLi -.^ii dfu-.s involve
licr:<uici: il
c.oritJ.Hil rnhos
volu.'-s wilh oa'.h '.onooguf-ot
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rzziiizzz izzzzizi^
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•.Iii4f;(il , wh<i .tn; lol4 H, thai i-rnji oh.ool liu-.
Oi'^..f'i4.4iuo Oouijlf.-'.
Cf V
I h'J fliili.llfHl I-., y
(/ ;,|
l.'H)
90
H;-')' finr.infn fv j x o x «o
ciorjf {./) y , 3, V k;
/()
cJ) AflOl 'j h, /riH
Sfurj,-.,-.!;,
CO
WoolO ho tcilO
50
40
II
1
i;^^
30
±t
A. I
• w
c) This data set does not involve exponential
growth or decay because the differences are not
constant, or a constant ratio.
d) This data set does not involve exponential
growth or decay the differences are not constant,
or a constant ratio.
X4
1
4- " 4 - 1
?-
2. a) Using a graphing calculator, the exponential
regression function for the data is
y = 10.097...(0.200...)''.
2©. a) Yes. e.g., Any data with a constant
doubling time can be expressed with an
exponential function.
b| Y = 4 ( 1 . 2 6 ) 1 a represents the initial number of
reguests. b represents the rate of growth of the
number of reguests, x represents the amount of
time in hours since the news broke, y represents
the total number of interview reguests.
c) domain: {x | x > 0, x e N}
range: {y I y > 4, y e N}
d)
]
rime
o m.
jryCOajin
1 U J O ami
1 2 OOpxTj.
_ 1 0 0 () m .
2 J ) 0 p rrc
3 0 0 p m.
P
rn.
Domain: {x I x o R}; Range: { y | y > 0 , yc
40
5 0
2
4
0
R}
y-intercept: y = 10.097...;
End Behaviour: Qll to Ql
This function is decreasing and shows exponential
decay.
b) Using a graphing calculator, the exponential
regression function for the data is
y = 2.780...{1.054...)'',
Domain: {x | x e R); Range: {y | y > 0, y e R)
y-intercept: y = 2 . 7 8 0 . . . ; End Behaviour: Qll to Ql
This function is increasing and shows exponential
growth.
>'.
')0
Y+YJO
U 1 L 4
4
3. a)
0
01 ^ „
1 0
1
12
'
J6JJ
2 0
2
i
1
Years sinci^
Retirement
Rent ($)
IZZZiZZZI
\AZ
2
3
Ratios
)
9 9 6 0
1
------- 1
1 0
3 4 4
1 . 0 3 8 .
1 0
752
1 . 0 3 9 .
.
Yes, an exponential model can represent the
data since the ratio of consecutive pairs of
y-values are close. ,
7-6
Chapter / Exponential and Logarithmic Functions
b) The equation will be of the form f(x) = a{bf.
Thus, We can create the formula by setting a to
the first rent value and b to the average ratio of
consecutive pairs of Rvalues, x will be the
number of years since retirement, f(x) will be the
rent after x years.
c) Using graphing technology, the regression
equation i s y = 9595,433...(1.038...f.
f(x) = 9595.433...(1038...f
f(x) = 9595.433.,.(1,038...f
f(x) = 9595.433...(1.459...)
f(x) = 14 000.638...
1 predict the Belands will pay about $14 000 in
10 years.
6. a) Using a graphing calculator, the exponential
regression eguation for the data is
y = 14.429...(1.065...)'.
b) y = 14.429...(1.065,.,)™
y = 14.429...(1,879...)
4. a)
Rainbow Tro'.«t l.rj.gtl, vi. >Xa%'
900
y = 27.119... cm
The height of the sunflower on day 10 was about
27,1 cm.
800
c) y = 14.429...(1.065...)"
1000'
700
I
c) The population is estimated to be 8.05 billion.
• . c . t this number, 50 (for 2020) was substituted
for X in the regression equation so that y could be
determined.
d) The population is expected to reach
9.50 billion 61.5 years after 1970. that is, dunng
the year 2031. I plotted y = 9.50 and the
regression function on a graphing calculator.
The x-coordinate of the point at which these two
functions intersect is the point at which the
population is expected to reach 9.50 billion.
y = 14.429...(6.638...)
y = 95.792... cm
The height of the sunflower on day 30 was about
95.8 cm. The answer does not make sense since
the height was 98 0 cm on day 28.
rf| I wnuld expect tho sunflower to reach a height
(}i
(.m oo d.ov 20
500
400
300
200
100
0
150
250
Length (mm)
50
3
b) Using a graphing calculator, the exponential
regression function for the data is
y = 26.934...(1.008...)!
c) e.g.. 690 g; I identified the point on the curve
that had an x-value of 400.
d) I would expect this fish to be about 446 mm
long. I identified the point on the curve that had a
y-value of 1000.
5. •:.)
Populatioi; bf Y*.,.r
1
h 4
4 I .4
\
2
7. a) Using graphing technology. I input the x and
y values into the spreadsheet. I then used the
exponential regression function to determine the
eguation that models this growth. The eguation is
y=2^.
b) Under this context, the domain is {x | x o N),
and the range is (y | y > 0, y c N}.
c) If we extend the first and last column of the
table, wc get the following:
Number of Possible
Number of Digits
Codes
1
0
'0
20
50
40
Years since 1970
'
SO
b) Using a graphing calculator, the exponential
regression function for the data is
y = 3.911...(1.014...)T
Foundations of Mathematics 12 Solutions Manual
J
4
8
2 .16
.32
64
128
7-7
Based on the table, 104 codes lies in between
the codes for 6 digits and the codes for 7 digits.
6 digits would not provide enough codes.
Therefore, 7 digits would be needed to have
104 codes.
8. a) Here is the graph.
1000 T
900
9. a)
Time
(years)
0
1
2
3
4
5
Frog
Population
575
460
368
294
235
188
Decrease in
Population (%)
20%
20%
20.1%
20.1%
20%
800
I" 700
V 600
I 500
^ 400
A
300
Therefore, the frog population is decreasing by
20% every year.
b) Using a graphing calculator, the exponential
regression function for the data is
y = 575.228...(0.799^..)'.
575
c) e.g., Halfthe population is
200
100
0
2
X
4 6 8 10 12
Altitude (km)
b) Using a graphing calculator, the exponential
regression function for the data is
y= 1050.311...(0.873...)'
s15
c) y = 1050.311...(0.873...)^
y = 1050.311...(0.130...)
y = 137.249... mb
The pressure at 15 km is about 137.3 mb.
d) Using graphing technology, the pressure
reaches 500.0 mb at about 5 km.
^ T J l ^ pmnMw k
tS£3
Using graphing technology, the pressure reaches
50.0 mb at about 22 km.
or about
288 frogs. One possible estimate could be that
the population is halved in about 3 years, since
the size of the population at 3 years is 294,
almost egual to halfthe initial population. Using
the regression equation, we determine that the
time to halve the population is about 3.1 years.
10. a)
Year
1960
1965
1970
1975
1980
1985
1990
1995
2000
2005
Atmospheric
CO2 (ppm)
319
322
328
333
341
348
356
364
372
381
First
Differences
3
6
5
8
7
8
8
8
9
e.g., No, the rate of change of atmosphenc CO2
is increasing.
b) x: The number of years since 1960.
y: The atmosphenc CO2; a = 319
Solve for b using the a-value and one of the
points in the table. Using (45, 381)
381 =319d^^
1.194... =6^005
b= 1.003...
Therefore, the exponential function is
y = 319(1.003...)''.
c) Using a graphing calculator, the exponential
regression function for the data is
y = 315.609...(1.004...)'.
7-8
Chapter 7: Exponential and Logarithmic Functions
d) Predicted Proportion in 2010
y = 315.609...(1.004.
y = 315.609...(1.225...)
y = 386.609... cm
The predicted proportion in 2010 is 387 ppm.
Predicted Proportion in 2020
y = 315.609...(1.004...)''°
y = 315.609...(1.275...)
y = 402.687... cm
The predicted proportion in 2020 is 403 ppm.
11. a) Using a graphing calculator, the
exponential regression function for the data is
y = 83.933...(0.951...)!
b) It will take the sandwich 14 minutes to have an
internal temperature of half its initial temperature
(42 °C).
b) Tnstan will have less than 3 mg of caffeine in
his body after 9 hours.
•
-112 I
•lkBul}s12t«a,,.12 IT
-flO
raC3f)-40.7986022432fo 747i«2W3247)'''"
c) Half the initial amount of caffeine in Tnstan's
body is 20 mg. Half the initial amount of caffeine
in Tnstan's body remains after about 2 h.
J5(^)»4Ct, 7986022432 fo, 747392903247)^
^ ^ \ . (2-'i4S49«3S3.ao)
\
c) The sandwich will need to sit for 29 minutes to
reach an internal temperature of 20 °C.
|28 93518SS312,20)
12. a) Using graphing technology to determine
the regression function, a graphical exponential
model is shown below.
pml2-12a
A function that models this situation algebraically
is y = 40.799...(0.747...)!
Foundations of Mathematics 12 Solutions Manual
13. a) Using graphing technology, the equation is
y = 4120.075...(1.499...)".
b) y = 4120.075...(1.499...)^°
y = 4120.075...(57.659...)
y = 237 559.646...
There will be 237 560 bactena.
c) Assuming that the obsen/ations are made
every hour, the function that models the growth of
these bacteria is y = 4120.075...(1.2)T The base
has changed to 1.2 because the bactena are
growing at a rate of 20% per hour. Since the
population is growing, the value must be greater
than 1.
14. a
Week
Pond Surface Covered
by Algae (%)
25.0
27.5
30.3
33.3
36.6
40.3
Growth
(%)
0
1
2
3
4
5
(10.0 + 10.2 + 9.9 + 9.9 + 10.1) ^ ^^ ^^^^^
10.0
10.2
9.9
9.9
10.1
5
The algae are growing at a rate of about 10%
every week.
7-9
b) Use the equation y = 25(1.100...f and a
graphing calculator to solve this.
I|
mi>th12t06».I4b v
-JgQ
0,2 1806123617,
fl(x)-2s(l
1002)^
16. a) y = 3000(0.9)! where x is the number of
weeks that the well has operated for and y is the
weekly amount of oil produced, in barrels.
b) y = 3000(0.9)^°
y = 3000(0.121...)
y = 364.729...
At the end of 20 weeks of operation, 365 barrels
of oil will be extracted from the well.
c) The well will be closed dunng the 61*' week.
It will take 12.2 weeks for the algae to cover 80%
of the pond.
15. a) i) P(26) = 200
P(26)-200
P ( 2 6 ) = 200(0.686...)
P ( 2 6 ) - 1 3 7 . 3 9 5 . . . ban-els/week
The expected production of this well, 26 weeks
after production begins to drop is
137 barrels/week.
ii) P(26) = 200
2
2
P(26)-200l I
P(26) = 200(0.235...)
P(26) = 47.193... ban-els/week
The expected production of this well, 100 weeks
after production begins to drop is
47 barrels/week.
b) Production will reach this level dunng the
132nd week.
17. a) e.g., Calculate the rate of change in the
decrease (b) and use 100 for a. Use software or
a calculator to perform exponential regression to
get those values.
b) Using the method from part a) to create an
algebraic model, we get our equation as so97 = 1005^
b = 0.97
y = 100(0.97)'
Now substitute 7 for x:
y = 100(0.97)^
y = 100(0.807...)
y = 80.798...%
The light intensity would be 80.80%.
c) 1 0 0 - 6 0 = 40
The light intensity has to go below 40%.
The minimum of gels that would be needed is 3 1 .
g
|
T
t
mUhUms17e
•
*2{3c)-2«{i
12/
O£)c)-30
(131:374348'>2,30)^
X
c) The company should cap the well dunng the
256th week.
I-12 h
7-10
k
ma9i12e06a,.be
18. a) Penmeter: 4(4) = 16; Inner Rows: 4(3) = 12
Inner Columns: 4(3) = 12; 16 + 12 + 12 = 40
40 toothpicks will be needed to create the next
figure in the pattern.
Chapter 7: Exponential and Logarithmic Functions
b)
Number of
Toothpicks
4
12
24
40
Figure
1
2
3
4
1st Diff.
8
12
16
2nd Diff.
d) Using graphing technology, the exponential
regression equation i s y = 100.001 ...(0.951 ...)*
It is a good model for the data because the
function lies very close to the scatter plot.
1
-'J|>
flmUISd*
^
€ B
4
4
The second differences are constant, so the data
is best modelled by a quadratic function,
c) Continuing the table:
Number of
Toothpicks
4
12
24
40
60
84
112
Figure
1
2
3
4
5
6
7
1st Diff.
8
12
16
20
24
28
2nd Diff.
4
4
4
4
4
112 toothpicks will be needed to make the 7th
figure.
19. a) Using point (9, 64.0), calculate h as shown:
1
64.0 =
100.01 -
64 ^
1 V/i
9
100"
9
11^
16 ^
25"
Use a graphing calculator to solve this equation.
: >
Ik
rnatti12eC*»..«d •»
The equation determined by hand and the one
determined using graphing technology are very
close because the difference in numbers start
showing up in the thousandth decimal place.
History Connection, page 468
A. For example, uranium, thonum, radon, and
radium.
B. The radioactive decay of an element can be
modeled by a halfdife equation, which is an
exponential decay function.
The halfdives of the four elements listed in
prompt A is given below.
• U-235: 7.1 X 10^ years
• U-238: 4.51 x 10^ years
• Rn-220: 55 s
• Rn-222: 3.823 days
• Ra-226: 1600 years
• Th-227: 18.5 days
• Th-230: 8.0 x lO'* years
• T h - 2 3 1 : 25.5 h
• Th-234: 24.1 days
To determine the halfdife equations for the
elements shown above, I substituted each halflife into the halfdife equation. A{x) ^ A -
,
\IJ
where A{x) is the amount of the element at time x
and Ao is the onginal amount. The half-life
equations for the elements are (where x
represents time in years, except as noted):
U-235:
The value of h is 13.978... or 14 days.
13.978.
b) A{t) =
( 1 ITTXIO"
A x )
=
A ) k
m
c ) a = 100
Use the same point used to calculate h in a) to
calculate b for this part. Using (9, 64.0), b is
calculated as shown:
64.0-1005'
16 = b'
25
5 = 0.951...
The exponential regression equation is
y = 100(0.951...)!
Foundations of Mathematics 12 Solutions Manual
u-238:
A x )
= A,l
2
Rn-220:
(1
V55
V2/
where x represents time in seconds
7-11
Rn-222:
M i d - C h a p t e r R e v i e w , p a g e 472
{ i Y3T823
A{x) =
A^[l
\ zJ
where x represents time in days
Ra-226:
f -( Y-ieoo
2)
Th-227:
A{x) =
\
2
where x represents time in days
Th-230:
A{x) = A,
2
2
Th-231:
l¥
where x represents time in hours
Th-234:
1
12.41
where x represents time in days
Applying Problem-Solving Strategies,
page 469
A . - G . Answers will vary, based on the decisions
that players make.
H. Possible vanations:
• Increase the size of the playing square, or use
another shape.
• If a player hits a square, she or he gets to go
again immediately.
• Once a function is announced, the opposing
player must tell the person if the function passes
above, below, or between her or his squares.
• Assign points to each hit, or start out with a
given number of points and deduct points for
misses or questions (for example, "Does the
function pass above your squares2").
1. a) This graph does not represent an exponential
function because it does not have a horizontal
asymptote. In addition, its end behaviour is unlike
that of an exponential equation since it extends
from Qll to QIV.
b) This graph does represent an exponential
function because it has all of the basic
charactenstics of an exponential function. It is
constantly decreasing at an unsteady rate like an
exponential decay function. It has a y-intercept but
no x-intercept. It has a domain of {x | x e R} and
a > 0 and 0 < i ) < 1. Also, it decreases from Qll to
Ql.
c) This graph does represent an exponential
function because it has all of the basic
charactenstics of an exponential function It is
constantly increasing at an unsteady rate like an
exponential growth function. It has a y-intercept but
no x-intercept. It has a domain of {x | x o R} with
a < 0 and 0 < d < 1. Also, it decreases from Qll to
Ql.
d) This graph does represent an exponential
function because it has all of the basic
charactenstics of an exponential function. It has
only one asymptote that is honzontal. It is
constantly decreasing at an unsteady rate like an
exponential decay function. It has a y-intercept but
no x-intercept. it has a domain of {x | x e R}, with
a < 0 and 5 > 1. Also, it decreases from Qll to Ql.
2. a) i)
8
1
fiW-34'=y^
[I
, , , IJ
1, ,
, ,1 T - f j
•8
IS
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-2'
i#
ii)
y
zJ
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1
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1
t
1I
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ll1
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i
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t \
h- \\
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N
7-12
X
4
Chapter 7: Exponential and Logarithmic Functions
b) Number of x-intercepts: 0; y-intercept: y = 90
domain: {x | x e R}; range: {y | y > 0, y e R}
The function is decreasing.
Hi)
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1
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c) number of x-intercepts: 0
y-intercept: 8
domain: {x | x € R}
range: {y\y>0,ye
R}
The function is increasing.
b) i) number of x intercepts: 0; y-intercept: y = 3
domain: {x | x e R}; range: {y | y < 0, y o R)
end behaviour: Qll to Ql
ii) number of x-intercepts: 0; y-intercept: y = 8
domain: {x | x e R}; range: {y | y > 0, y € R}
end behaviour: Qll to Ql
iii) number of x-intercepts: 0; y-intercept: y=5
domain: {x | x o R}; range: {y | y > 0, y e R}
end behaviour: Qll to Ql
iv) number of x-intercepts: 0; y-intercept: y = 2
domain: {x j x e R}; range: {y | y > 0, y e R}
end behaviour: Qll to Ql
3. a) Number of x-intercepts: 0; y-intercept: y = 5
domain: {x | x o R}; range: (y | y > 0, y o R}
The function is increasing.
44
A'
•"t
1
2
4
r—n
6
8
4. a) b) a = 2; c) a = 5; d) a = 3
b) b) It is between 0 < fe< 1 because the function
is decreasing.
c) It is between 0 < 5 < 1 because the function is
decreasing.
d) b is greater than 1 because the function is
increasing.
c) b) domain: {x | x o R}; range: {y I y > 0, y e R}
c) domain: {x | x e R}; range: {y\y>0,ye
R}
d) domain: {x | x e R}; range: (y | y > 0, y e R}
5. a) e.g., It is a decreasing function because the
base is less than 1.
4^
>
il
1' )
\ 5>
2
4
6
1
4
S
]
J
n1
1t
1
i
1
\
-
•,\
'\
s
>
\
—
r-
Foundations of Mathematics 12 Solutions Manual
\
Ji
0
7-13
b) e.g., It is an increasing function because the
base is greater than 1.
1 f<
/ )
1
13
1
1'J
I1
1r
11
1'.
1\
-
1J
1
r
V
K
0
•1
•
c) e.g., It is a decreasing function because the
base is less than 1.
9. a) e.g., First divide amounts in consecutive
rows to determine the base, then use the initial
amount and the base to write an exponential
formula,
b)
84H» •< -4 '2
2
4
b) i ) y = (7.628...)(1.742...)°^
y = (7.628...)(1.320)
y = $10.069...
The card had a value of about $10.07 6 months
after purchase.
ii) y = (7.628...)(1.742...)'^
y = (7.628...)(2.300...)
y = $17.546...
The card had a value of about $17.55 18 months
after purchase.
iii) y = (7.628...)(1.742...)^^
y = (7.628...)(4.008...)
y = $30.575...
The card had a value of about $30.58 30 months
after purchase.
iv) y = (7.628...)(1.742...)^^
y = (7.628...)(6.984...)
y = $53.279...
The card had a value of about $53.28 42 months
after purchase.
6 S
d) e.g., It is an increasing function because the
base is greater than 1.
Time (years)
Amount ($)
0
1
2
3
4
1500.00
1560.00
1622.40
1687.30
1754.79
04-1
0^
^
-6 '4
-2
2
4
6
»
-oaH
6. a) Assuming that Lauren does not count as a
student who has received a handout, the
following is a table of values that models this
situation:
Round (x)
0
1
2
3
4
Number of Students Who
Received Handouts (y)
7
49
343
2401
16807
b) Yes. y= 7{7f ; e.g., The ratio of consecutive
numbers of students who received handouts in
each round is constant.
Increase in
Amount (%)
4
4
4.000...
3.999...
The annual interest rate is 4 % .
c) The function, which we must use, is
y = 1500(1.04)'.
y = 1500(1.04)'°
y= 1500(1.480...)
y = $2220.366...
Paula will have $2220.37 in her account after
10 years.
10. a ) y = 140 (0.80)°
y = 140(1)
y = 140 cm
The ball was first dropped from a height of
140 cm.
b) y = 37 cm
rr
7. Using a graphing calculator, the function is
y = (36.871...)(0.663...)!
8. a) Using a graphing calculator, the function is
y = (7.628...)(1.742...f.
7-14
1 I r ,
—+—+--f
-
^
,
Mf,,,,j|, ,.»„.,,,._,
.y.
m
Chapter 7: Exponential and Logarithmic Functions
c ) x = 3.106... when y = 70 cm
L e s s o n 7.4: C h a r a c t e r i s t i c s of
L o g a r i t h m i c F u n c t i o n s w i t h B a s e 10 a n d
B a s e e, p a g e 482
1. x-intercept: 1
Number of y-intercepts: 0
End Behaviour: QIV to Ql
Domain: {x | x > 0, x e R}
Range: {y | y e R}
The height was less than half the initial drop
height on the 4th bounce.
11. a) The function is decreasing. It has one
y-intercept and no x-intercepts.
b) The domain is time, so it must be positive:
{f I f > 0, X G R}
The range is all the possible temperatures from
90 °C to 21 X : {C(f) | 21 < C{t) < 90, C(f) e R}
c) C(f) = 90(0.977/
C(10) = 9 0 ( 0 . 9 7 7 r
C(10) = 71.316...
The temperature of the coffee after 10 min is
71 °C.
d) C{t) = 60 °C when f = 17 minutes.
4
- I 1 2^;.
>
t ma!h12c(«» .-10c •
C
at) = 30 °C when t = 47 minutes.
100 y
2. a) No. e.g., no x-intercept and one y-intercept
whereas logarithmic functions have no
y-intercepts and one x-intercept. It is an
exponential function.
b) No. e.g., two x-intercepts and one y-intercept
whereas logarithmic functions have no
y-intercepts and one x-intercept. It is a quadratic
function.
c) Yes. e.g., one x-intercept and no y-intercept.
d) No. e.g., one x-intercept and one y-intercept
whereas logarithmic functions have x-intercepts
ofi.
e) Yes. e.g., one x-intercept and no y-intercept.
f) No. e.g., no x-intercept and one y-intercept
whereas loganthmic functions have no
y-intercepts and one x-intercept.
3. c: x-intercept: 1
Number of y-intercepts: 0
End Behaviour: QIV to Ql
Domain: {x | x > 0, x o R}
Range: {y | y e R}
a: a > 0 e.g., since the graph is increasing
e: x-intercept: 1
Number of y-intercepts: 0
End Behaviour: Ql to QIV
Domain: {x | x > 0, x e R)
Range: {y | y e R}
a: a < 0 e.g., since the graph is decreasing
4. y = log x:
^rni:
IB
e) C(f) = 21 °C when t = 63 minutes.
fllxP^O (O <>77Y
62 5428924706, 2
1
Foundations of Mathematics 12 Solutions Manual
X-intercept: 1
Number of y-intercepts: 0
End Behaviour: QIV to Ql
Domain: {x | x > 0, x e R}
Range: {y | y e R}
Increasing or decreasing: increasing
y = - l o g x:
7-15
*Uni»ved
40
20
/
5
,.fi(C"-tor~-4L-_.
x-intercept: 1
Number of y-intercepts: 0
End Behaviour: Ql to QIV
Domain: {x | x > 0, x e R}
Range: {y | y o R}
Increasing or decreasing: decreasing
m tt 20 23 50 iS 40
d) x-intercept; 1
Number of y-intercepts; 0
End Behaviour; QIV to Ql
Domain: {x ( x > 0, x G R }
Range; {y I y e R}
Increasing or decreasing; increasing
hHt'Mr
I
,Mi!h44
5. a) x-intercept: 1
Number of y-intercepts; 0
End Behaviour: QIV to Ql
Domain: {x | x > 0, x e R}
Range: {y | y e R}
Increasing or decreasing: increasing
e) x-intercept; 1
Number of y-intercepts; 0
End Behaviour; Ql to QIV
Domain; {x | x > 0, x G R }
Range: {y j y G R }
Increasing or decreasing: decreasing
21-
-1-
1
2
-20«b) x-intercept: 1
Number of y-intercepts: 0
End Behaviour: Ql to QIV
Domain: {x | x > 0, x G R }
Range: {y | y G R }
Increasing or decreasing; decreasing
0.4-
0.2-
r
-
a i 0 4
0.6.0.8
IJO
t>r f
0.4-
4
2H
0
-2-
<X5 X6<k^
1.0 23 IG
3.5
4.<
-4
c) X-intercept; 1
Number of y-intercepts: 0
End Behaviour; QIV to Ql
Domain; {x | x > 0, x e R}
Range: {y | y G R }
Increasing or decreasing; increasing
f) x-intercept; 1
Number of y-intercepts; 0
End Behaviour; QIV to Ql
Domain; {x | x > 0, x e R}
Range: {y | y G R }
Increasing or decreasing: increasing
5O0H
400
300
2O0
100H
-KK)
too
200
300
40O
-200
7-16
Chapter 7: Exponential and Logarithmic Functions
6. e.g.. one xontercept of 1. no yontercepts,
I'P f\. litt; f;n<;igy " i ihr. • A-ahvI ifti.s.,>.:i%.' s, the
It'.i!Ilh<':t s>[ :Jf;Olh<'f. II1! .o,;; I',<j
domain; (x | x > 0, x o R } .
"/
d ).g., y = 5 log x
12. a l PdLicM.--f.'
• ' intercept: none.
R ) .
function: decreasing,
p = "I ; f =
o. .
0.
•
12 log P
;'.f ,.,!2 loom Ofi)
It would take abooi i '
< v ars for only 5% of the
original amount to if •
( '
)0.422 log P
)0.422 log(0-5)
230...
The half-life of cesium-137 is about 30 years.
8. i) b, e.g., x-intercept is 1, no y o n t o i o c p i l , graph
extends from QIV to Ql. Thus, the function is
logarithmic so b and c are the only options. The
function is increasing so the correct graph is b.
ii) c, e.g . X intercept is 1, no yontercept, graph
extends from Ql to QIV. Thus, the function is
loganthmic so b and c are the only options. The
function is decreasing so the correct graph is c.
jjj) d. e.g., no xontercept, y-mtercept is 1, graph
extends from Qll to Ql. Thus, the function is
exponential so a and d are the only options. The
function is increasing so the correct graph is d.
iv) a. e.g., no x-intercept, y-intercept is 2, graph
extends from Qll to Ql Thus, the function is
exponential a and d are the only options. The
function is decreasing so the correct graph is a.
9. Yes, e.g.. An exponential function has no
x-intercepts, and a loganthmic function has one
x-intercept.
13. a) When a < 0, each function will bo
decreasing.
b) The domain of each function is restncted as
for each function x must be greater than 0.
c) The range is unrestricted as all values of y are
possible.
14 a)
10. As hydrogen ion concentration increases, pH
decreases.
The Tontercepl is 10 '''' and there is no
A/fdntercept In context, the domain of this graph
•s {71
ICr^-^ T G R } , and the range is
[M \ M 0, M>: R ) . The graph is increasing, and
when 7 - i, M = 5.5.
b)M=
log T+ 5.5
M = log(50) + 5.5
M = 7.198...
An earthguake that is 50 times more intense than
the Saskatchewan earthquake would have a
magnitude of about 7.2.
Foundations of Mathemati
'Iutions Manual
7-17
ifili'ir., t|.,,ri Uw '..r.katf tif •Well) > ;.irth(jiJ.ikc^
1f> ,1) I lH< A inlf
i;, M)(}ij . j i „ | Uit^-H I-, IK,
/ iiilor* f'f>t In < uii!<;yi, Ihc ilmiumi <if Ihr. 'ir.-lf.h is
.A '• A
;U(J<i, A
IVf Jilt* Iho
IS
i' I '
b) I
<h /
R } Ihi'. hi!uJu,u r. !fiu<..j.'.i!KJ
'.H / ItHj 4
VIM 1
'
'H> / iOf}| I I I 0 0 0 )
'
'Ul /
Xfl4 '
It Will take about 31 years for Christie's
investment to reach $10 000.
c) f = 58.7 log ^ ^ 2 0 4 . 1
f = 58.7 iog(6000)-^ 204,1
t= 17.677...
It will take about 18 years for Christie's
investment to double.
b) I hf 'Ml. •. ; i o ( . . j u a t i o n that models the data
I: /
;4o ()0(i
i 23-957... In P
P inh.-ic-pi 1
,
0
0 0 / 0 023...
r llllcn.ept liuiu;
I nd t)oh.ovi.-.!a «jlV lo Ql
t ioi,,Hll! >P I
0 / ^ G W}
H.oog«^ \i I 0 /
W}
I !.ii(.fini> iiK.ieaciog
cH
•'xlOOOO
; 28,957... In P
i - :>4o 000
^ 23,957... In (2 000 000)
/ / 4 i)A I
The population exceeded 2 000 000 in 1974.
4. a) The independent variable is seismographic
reading and the dependent variable is Richter
Scale magnitude,
b)
• .t»-fi(j:i.)k>
L e s s o n 7.5: M o d e l l i n g D a t a U s i n g
Logarithmic F u n c t i o n s , page 494
4
1. a | e.g., Graph is logarithmic because graph is
increasing, xontercept is 1, there is no
yontercept, end behaviour is QIV to Ql. domain is
{ X I X > 0, X G R}, and the range is {y | y G R}.
b) When the decibel level increases by 10,
the relative sound intensity increases by a factor
of 10.
c) When the relative sound intensity doubles, the
decibel level increases by a factor of 3.
2. The logarithmic regression equation that
models this data is y = ^6.653... + 108.491. .In X.
xontercept; 1,063...
y-intercept: none
End behaviour: QIV to Ql
Domain: {x j x > 0, x
R}
Range: (y I y r- R}
Function: increasing
3. a)
' i
,o,\.3^^o\J>./y,,3^^vfx?
4>
^
^
^
^
^
Seismographic reading (microns)
c) The regression eguation for the data is
M = -0.006... + 0.434... In r.
d | M = ^0.006... + 0 , 4 3 4 . . . In r
Let M = 5.7
5.7 = ^0.006... + 0.434... In n
5.706... = 0.434. .. In n
13.122
-Hi r-i
.122...
Let M
4 5 - - 0 006... + 0.434... In r?
4.50b
- 0 4348 In f2
10.362
-In/2
r, =
e''^-''''"
d5,794...
7-18
Chapter 7 L x o o n e n t i a l and L o g a r i t h m i c F u n c t i o n s
A magnitude 5.7 earthquake is about 15.8 times
more intense than a magnitude 4.5 earthquake.
5. a) The independent vanable is pressure and
the dependent variable is altitude.
b) The regression equation for this data is
h = 30 665.960... - 6640.436... In P.
c) Pdntercept; 101.297...
Ii-intercept; none
End behaviour: Ql to QIV
Dom?
Range; {li | fi e R}
Function; decreasing
d)
y = 30 6 6 5 . 9 6 0 . . . - 6 6 4 0 , 4 3 6 . . . In X
139 = 30 665.960... 6640.436.,. In x
- 3 0 526,960... = -6640.436... In x
X - 99.199...
The pressure setting that Michael will need to use
is 99.2 kPa.
e)
y = 30 6 6 5 . 9 6 0 . . . - 6 6 4 0 . 4 3 6 . . . In X
8848 = 30 665.960... - 6640.436... In x
-21817,960... = -6640.436... In x
3 285... = In X
X = 26.725..
The atmospheric pressure at the summit of Mt.
Everest is 26.7 kPa.
6. a) Here is the graph.
/ a- Fhe exponential regression eguation that
represents this data is 71 = 14
999.826...(1.045..,)'.
b) The logarithmic regression equation that
represents this data is f = -218.442... + 22.717.
In A.
c) Exponential:
A = 14 999.826...(1.045.,.)'
25 000 = 14 999.826..,(1.045...)'
1.666... = (1.045...)'
log 1.666,, = log (1.045,..)'
log 1.666. - t log 1.045...
togijea.
logioisl
= t
11 604.
Logarithmic:
f = .^218.442... • V 71/
In A
f = ^218.442...
In (25 000)
f = 11.604...
It will take about 12 years for the balance to
egual $25 000 according to both equations,
e.g., I prefer using the loganthmic because it
makes the calculations simpler.
8. a) The independent variable ts distance from
Earth and the dependent vanable is distance
modulus.
b) Here is the graph.
2
ICon !C
20 00(
ttmttr
18 OCX
16 0(X
14 00(
% 12 OCX
^ 10 000
<
8 000
6 000
4 OOi
2 000
10
A) J O 10 W cc 71 < HV, 190
00
Percent of carbon-14 {%)
The regression eguation that represents this data
i s f = 38 0 7 0 . 3 3 2 . . . - 8 2 6 6 . 9 6 1 . . . In P
b) f = 38 070.332. . - 8266.961... In P
t = 38 070.332... - 8266.961. . In (96.8)
f = 268.437...
The age of these fragments is about 268 years.
c) t = 38 070.332... - 8266.961... In P
f = 38 070.332 .. - 8266.961... In (50)
f = 5729.790,,.
The halfdife of carbon-14 is 5730 years.
Foundations of Mathematics 12 Solutions Manual
c) The logarithmic regression equation for this
data is rn = -5.086... + 2.174... In d.
d) m = - 5 . 0 8 6 . . . + 2.174. . In d
m = - 5 . 0 8 6 . . . +2.174... In (2.39)
r?i = - 3 . 1 9 1 . . .
The distance modulus of Wolf 359 is - 3 . 1 9 .
e)
m =-5.086...+2.174... Ind
- 1 . 6 6 = -5.086... + 2.174... In d
3.426.,. = 2.174... In d
1 575, . = In d
d = 4.834...
Gliese 876 is 4.8 pc from Earth.
7-19
A = 5000 1 +
A
5000
In
A
0.05
= 1.0125"*"
= In 1.0125'
5000
l n A - l n 5 0 0 0 = 4filn1.0125
\nA
4n
In 5000
Ini 0125
In 7\
In 5000
'4Tnl"oT25
iv)
b) fh-: icwjanlliniic ro;jiessif)ii .;q,i;i|ion U;: tivs
•'f;d,i i '
h.'58. .
friO In n
t'l / 3 mm
' li;..H.>:j. in n
/ ';..'i8{/>
^10h>;3. Iiif;'7r,i
I - Ciy./30...
Facebook first surpassed 275 million registered
users dunng the 66th month after February 2004.
which is August 2009.
10. Enter the data in my calculator, perform a
loganthmic regression, graph the data points in a
scatter plot and the logarithmic function on the
same axes, and identify the point with the known
x-value and the unknown y-value.
11-a)
1)
A = P{1 + if
A=5000
5000
In
A
5000
1+
0.05
365
• = 1.000136.-.*''"
= In 1.000136.
In7\ - In5000 = 365nIni.000 136.
365n =
n=
In ^ - I n 5000
In 1.000136...
In 4 - I n 5000
365ln1,000136...
b) e.g.. The denominator approaches In 1 = 0 as
the number of compounding penods per year
^ = 5000(1 + 0,051
5000
in
A
5000
= 1.05"
Math in A c t i o n , p a g e 500
= In 1.05"
• It looks as if you need to ask no more than 10
questions to guess a number from 1 to 1000.
In 4 - I n 5000 = /? In 1.05
n=
In 4 - I n 5000
In 1.05
ii)
A
• Let n represent the numbers remaining and let q
represent the number of guestions asked.
Since the secret number is from 1 to 1000, it
means that there are 1000 possible numbers.
Each time you do a binary search, you are
dividing the set of possible numbers in half.
5000 1 ^
"I
5000
In
A
5000
I n / \ - I n 5000
2n
1.025'-"
In 1.025'
2:? In 1.025
In 4 - I n 5000
In 1.025
In ^ - I n 5000
2ln1.025
7-20
Chapter / '-hfoonential and Logarithmic Functions
I wrote a table of values for the first few guesses.
Then I substituted o = 1 and solved for q:
1 = 1 000 000 00011
1000
1000[ 2-
1
1 000
ooc
1
00 000 000
You would 0..OU :o ^c!. -0 questions to guess a
number from 1 to 1 000 000 000.
I noticed that the exponent on the
equal to q, the number of questions asked. So, I
set up an eguation expressing n in terms of q to
mode! the situation.
0 = 1000
• Answers will vary. e.g.. One possibility is to
allow the guesser to divide the list of remaining
numbers into three groups, instead of two.
Questions would now be like this; "Is your
number between a and b, b and c, or c and d?"
This would make the game go much more
quickly, since each turn reduces the number of
2
remaining numbers by - .
To guess a number from 1 to 1000, the equation
that would model the situation would now be
„.iooo(l
. I substituted n = 1 into my eguation
1 = 10002
1
fo;
1000
(2)
1=10001.3
1000
q .log.,
c| = 9.965...
You would need to ask 10 questions to find a
number from 1 to 1000. This answer matches my
expenmental answer.
. To determine the number of questions needed
to guess a number from 1 to one billion, I
changed 1000 to 1 000 000 000 in my eguation.
n = 1 000 000 000 I ^
Foundations of Mathematics 12 Solutions Manual
1
1000
g = 6.287,.
C h a p t e r S e l f - T e s t , p a g e 501
1. To match the functions with the graphs, look at
the X- or yontercepts and directions of the
functions and of the graphs. This is because
each function and each graph has a unique
y-intercept.
a) It is a decreasing exponential function with
y-intercept of 0.2 so it must match with ii.
b) It is a decreasing logarithmic function, so it
must match with iv.
c) It is an increasing exponential function with
y-intercept of 2. so it must match with iii.
d) It is an increasing loganthmic function, so it
must match with i.
7-21
2. a | The exponential regression function is
y= 1 2 J 2 0 . . . ( 1 0 9 4 . , . f . where y is the debt in
billions of dollars and x is the number of years
after 1955.
N« t o d,-.,it r)..bt
4, xontercept: 1
yontercept: none
End behaviour; Ql to QIV
Domain: {x | x > 0, x o R}
Range; {y | y e R)
Function: decreasing
S'jil
5 a) I .>i ^ rf.pic:.. Ill !he energy released. Let y
H o'>-><-" Iho r.oigtiiPjde o f t h e earthquake.
I O ' - icgo- .-.ion cgoolion is
1 I'-o
t I) /;>o ..In X .
h)
' : h..'
/
1 hiO
> :i .'P.o ..in X
' i'^'^
'
..In (1.1 • 10^®)
•/ "• 10
:h , -o^'-.i-....,.,rfh,,iMke had a magnitude of
about 9.5.
',(HI
2 r.ti
,.
CO
•
1
too
.''XI
/0(,
Z
150nc
tq±t±t
I'' M;
Years since 1955
-a
.|C.
-ih
b| 1 9 8 8 - 1 9 5 5 = 33
y = 12.620...(1.094...f^
y = 12.620,.,(19.592...)
y = $247.282... billion
The net federal debt was $247.28 billion.
c) Assuming the same growth rate and that these
figures were taken at the end of the year, the net
fodoral dob! shculd have roordied $600 billion in
1998.
3. Number of xontercepts; 0
yontercept; y = 6
Domain; { x | x e R}
Range: {y | y > 0, y e R}
Or>o Bohavioor- OH lo Ql
C h a p t e r R e v i e w , p a g e 504
1. a) e.g., The end behaviours of exponential
functions extend from Qll to Ql, Polynomials
have varying end behaviours,
b) e.g.. The domain is always {x | x e R), the
range is always {y | y > 0. y o R). and they all
extend from Qll to Ql. a is the y-intercept.
2 a) domain: {x | x o R); range: {y | y > 0, y e R}
y-intercept: y = 9; end behaviour: Qll to Ql
This is a decreasing function.
b) To make the function into an increasing
exponential function, the b should be changed so
Its value is greater than 1.
3. a) i) number of x-intercepts: 0
ii) y-intercept: 125
iii) end behaviour: Qll to Ql
iv) domain: {x | x e R); range: {y | y > 0, y o R)
v) The function decreases.
b) i) number of x-intercepts: 0
ii) y-intercept: 0.12
iii) end behaviour: Qll to Ql
iv) domain: { x | x e R); range: { y | y > 0, y e R)
V ) The function decreases.
c) i) number of x-intercepts: 0
i i | y-intercept: 1
iii) end behaviour: Qll to Ql
i¥) domain: { x | x e R); range: {y | y > 0, y e R}
v) The function increases.
d) i) number of x-intercepts: 0
ii) y-intercept: 0.85
iii) end behaviour: Qll to Ql
iv) domain: {x | x c- R); range:
V ) The function increases.
4. To match the functions with the graphs, look at
the y-intercepts of the functions and of the
graphs. This is because each function and each
graph has a unique y-intercept.
a) It is an increasing exponential function with
y-intercept of 5, so it must match with i.
b) It is a decreasing exponential function with
y-intercept of 2 so it must match with ii.
7-22
Chapter r
fcxponential
and Logarithmic Functions
5. a) Let x be the number of months since
April 1, 1896.
Let y be the population of Dawson City,
a = 1000
l3 = 3
Tripling time; 3 months
An exponential equation that models the
population grov/th is ^ =1000 (3) , where y is the
population and x is the number of quarters after
April 1, 1896.
b) Domain; {x | 0 < x < 3, x e N} (quarters)
Range; {y | 1000 < y < 27 000, y e N) (population)
c) The population of Dawson City in mid-May of
1896 was 1732. and the population of Dawson
City in mid^August of 1896 was 5196. To
determine the answer for mid-May. I substituted
in 1.5 for X for my equation. To determine the
answer for mid-August, I substituted in 4.5 for x
for my eguation. These two points arc midpoints
of the first two quarters, and so correspond to
X = 0.5 and x = 1.5.
A(8000) = 1 0 0 ' ^ | ) ' ' "
/l(8000) = 100
/l(8000) = 100(0.379...)
A(8000)= 39.993...%
38% of the initial carbon-14 would be present in
the tools.
I>) I he ago (4 ihn iools wt;ro 7400 years old.
8. x-intercept; 1
Number of yontercepts; none
End Behaviour; QIV to Ql
Domain; {x | x > 0, x e R)
Range: {y | y e R}
6. a)
.oof
350
300
9. a) x-intercept: 1
Number of y-intercepts: none
End Behaviour: QIV to Ql
Domain: {x | x > 0, x
R)
Rang^.r {y \ y R]
250
200
150
100
SO
Ml
0
i
;
, 4 \: (
Years after 2005
b) Using the graphing calculator, the exponential
regression function is y = 200.798...(1.088...)".
c) y = 400.798.,.(1.088...)'°
y = 400.798...(2.330...)
y = 467.971...
The deer population 10 years after 2005 will be 468.
d) 200 798 .(2) = 402 (rounded)
Assuming the deer population numbers were
recorded at the beginning of the year, I would
expect the deer population to have doubled in the
year 2014.
Foundations of Mathematics 12 Solutions Manual
b) x-interccpt: 1
Number of y-intercopts: none
End Behaviour: Ql to QIV
Domain: (x | x > 0, x e R)
Range: {y | y e R}
7-23
c) x-intercept: 1
Number of y-intercepts: none
End Behaviour: QIV to Ql
Domain: {x | x > 0, x G R }
Range: {y | y e R}
ii) b, e.g., no x-intercept, y-intercept of 1,
increasing. Thus, the function is exponential so b
and c are the only possible choices. Since d > 1,
the function is increasing and therefore the
correct choice is b.
iii) a, e.g., x-intercept of 1, no y-intercept, QIV to
Ql. Thus, the function is logarithmic so a and d
are the only possible choices. The function has a
positive leading coefficient which means that the
function is an increasing function and therefore
the correct choice is a.
iv) c, e.g., no x-intercept, y-intercept of 6,
decreasing. Thus, the function is exponential so b
and c are the only possible choices. Since d < 1,
the function is decreasing and therefore the
correct choice is c.
11. The logarithmic regression equation of this
data is y = 151.211... - 32.836... (In X).
y = 151.211... - 3 2 . 8 3 6 . . . ( I n x )
200 = 1 5 1 . 2 1 1 . . . - 3 2 . 8 3 6 . . . ( I n x )
48.788... = - 3 2 . 8 3 6 . . . ( I n x )
-1.485... = l n x
X = e^^-'^^-X =0.226...
Therefore about 0.23% of the sunlight penetrates
water to a depth of 200 m.
Chapter Task, page 507
A. I found that the amount of caffeine in one
espresso is about 100 mg. So, a tnple espresso
will have 300 mg of caffeine.
d) x-intercept: 1
Number of y-intercepts; none
End Behaviour; Ql to QIV
Domain; {x | x > 0, x e R}
Range; {y | y e R}
B. I should use time as the independent vanable
because it is the change in time that causes the
change in the amount of caffeine.
If time is the independent vanable, then the
function is an exponential function.
C. I chose to use 6 h as my estimate because
that is the maximum possible time to reduce the
amount of caffeine to a half.
1 iow^ ih)
Caffeine (m g)
0
6
12
18
300
150
75
37.5
D. Using a graphing calculator, the exponential
regression function for the data is
y = 300(0.890...)^.
<£_2 | T T ]
1^ *ma»i12c06..10c ^
IgQ
10. i) d, e.g., x-intercept of 1, no y-intercept, Ql to
QIV. Thus, the function is logarithmic so a and d
are the only possible choices. The function has a
negative leading coefficient, which means that
the function is a decreasing function and
therefore the correct choice is d.
7-24
Chapter 7: Exponential and Logarithmic Functions
E. To determine the amount of caffeine in Maria's
body 24 h after her first triple espresso, I
substituted t = 24 into my exponential equation
from prompt D.
Ait) =300(0.890,..)'
Ai24) = 300(0.890...)^'^
4(24) =300(0.062...)^'*
4(24) = 18.749... mg
Maria will have 18.75 mg of caffeine in her
system 24 h after drinking her first triple
espresso.
Chapter 7 Diagnostic Test, T R page 462
i o| r = 64
F. I chose to use a spreadsheet to calculate the
amount of caffeine in Maria's system during the
d) (5)
Week
Starting
. .H
loi^^ilh^
Week
Starting
amount of
caffeine (mg)
Amount of
olleine after
.' I h
I) • / of the
•.O.sok
.05 irti
.Cl ,ount of
' cdfeine (mg)
,-\o ount of
rolteine after
L24h
i
f
, -
c) (121)2 =V121
(121)2-11
1
|./..,.
Tue.
(5)'"* =0.0016
300 OOOO
318.7500
2. a) 3 • 7 7 = 3(7)^
(I
o-r-.•...! of
cr fl.../.c, after
24 h
f
r
-zz
—j
(1
to
b)
19.9219
500
c)
.
d)
1
1
319.S
1951
n.
,.0 0997
19.9
Sa
a
a
(1
b =
a
a b
. . .y.y =
X X X
x-^
3. a) 27 = 3-*
b) 64 = 8^
64 = 4^
64 = 2"
c) 625 = 5'^
d) 169 = 13'
4. a) 6 ^ - 6 - " . 6 = - | ^
20.000fs
b)
20.0000
1 1
2 2
1.2500
On Saturday morning, Mana will have about
20.00 mg of caffeine in her system. If she does
not drink coffee (or any other caffeinated dnnk)
on Saturday, she will have only 1.25 mg of
caffeine in her system by Saturday evening.
1 f i f fiT'
3 K3J K3J
c) 2'-22 •2'-' =
n
13
1
2^-22
2^-22-2^'^ ^ 2 ^ . 2 " '
25.2^2.2^-2 =2^2
5J
V
1
52.5^.5
5/
5
- I -5^-5 = 5^
5
Foundations of Mathematics 12 Solutions Manual
7-25
RS
f1
0 .,1
IS
c)
5 x + T ^ = 8x
5x-8x = ^ ^
3
3
( 1]
RS
8x
0 , ^ 2 1 1 . 4 ' 33
4*-3""'
4 ^
4'*-3-'
= 432
i 9 J' 3
UJ
8
."••1
0
6
6^-2^
8
6 ^ . 6 - = 6''.2^^
9
0
9
LS = RS
6-^-2^
6
^
= 10 368
dl
'
1 I
/ , a| 6 - - 3 x = 8
I
-
!
3f - y
« 4
1 \
—
X -
3
-3x = 8 - 6
-3x = 2
13
K
2
x= —
3
6-3x
X
RS
8
LS
4
RS
3x
-(-II
3(4)
6 +2
8
-111
LS = RS
13
12
1
12
LS = RS
7-26
Chapter 7: Exponential and Logarithmic Functions
xontercept:
y=4x^5
8. a) No, a constant multiple was not used to
generate the numbers in this set. This can be
seen in the ratio between each successive
number.
Nurnhct
"
I "
0 = 4x^^5
5
Ratio
4x
5
X
'
7
I
b| I uc, u woclc^ot lOJdipio oi 2 -A'ac used Lo
generate the numbers in this set. This can be seen
I .P'. h»*f-.'iOHO --^cxh r y r r o c s i v o r-ismbor.
riortiber
•
R.itio
^
e) This would match with iii) because for y = x^,
the function increases when x is greater than
zero.
f) It would match with i) because the height of a
person over time is continuous not discrete. It is a
gradual increase. If the height of a person were
graphed, it would be a continuous line without a
break
g) This matches with ii) because the function is
linear since x has an exponent of 1.
c) Yes, a constant multiple of — was used to
generate the numbers in this set. This can be
seen in the ratio between each successive
number.
Ratio
nber
m
'
1
12
3
1
A
2. a) The domain of this function is
{f I f > 0 , f G R}. This is because t represents time
and the ball cannot have negative seconds after
it falls off a balcony,
b) The domain of this function is
{b\b>0,bG
R). This is because b represents
batches of cookies and it is impossible to bake
negative batches of cookies. It is possible to bake
zero or more batch of cookies.
1
1
c
d) No, a constant multiple was not used to
generate the numbers in this set. This can be
seen in the ratio between each successive
number.
r
_jNymt£er
V
4 1
^ _^
10^
23
3f.
Z
^
• I
Ratio
^71
'
^
'
'2 25
_ 1 77 7 .
I
~ J 562
„ 1 44
R e v i e w of T e r m s a n d C o n n e c t i o n s ,
T R page 464
1. a) It would match with iv) because the two
coordinates are the y-intercept and x-intercept of
y = 4x 5.
y-intercept:
y = 4(0)-5
y=
4
b) This would match with vi) because the number
of books collected in a year is a discrete number.
The number of books cannot be continuous; it
has to ; . o df Crete unit in which each unit is
separate. For example, it is impossible to have
0.5 books. It is possible to have 2 books. If this
were to be graphed, it would look like a staircase
because the line is not a continuous whole.
c) This matches with vii) because the domain of
y = x^ is all rea! numbers.
d) It would match with v) because the range of
y = x^' IS all values that are equal and greater than
zero.
3. a) The range of this function is
{11(f) I 0 < /7(f) < 3, h(t) G R} because if the ball is
falling off the balcony 3 m up, that must be its
maximum height. The minimum height would be
0 m, where the ball hits the ground,
b) The range of this function is
{n{b) I 0 < n{b), nib) e W} because it is
impossible to have a negative number of cups of
flour. The number of cups of flour must be zero
and greater to be able to bake b batches of
cookies.
4. a ) y = 3 x + 6
y-intercept:
y = 3x + 6
y=3(0) + 6
y =6
5
Foundations of Mathematics 12 Solutions Manual
7-27
x-intercept:
y = 3x + 6
(0) = 3x + 6
-3x = 6
x = -2
5. a) The domain is {x | x g R } and the range is
{y I y e R } because it is a linear function.
b) The domain is {x | x e R } and the range is
{y I y < 8, y G R } because it is a quadratic function.
c) The domain is {x | x € R } and the range is
{y I y G R } because it is a cubic function.
d) The domain is {x | x G R } and the range is
{y I y e R } because it is a cubic function.
The x-intercept is - 2 and the y-intercept is 6.
b)y = / ^ 4
y-intercept:
y = / ^ 4
y=(0f-4
x-intercepts:
y=X^-4
(0) = / - 4
4 = /
-2, 2 = X
6. a) The function is increasing because the
slope is positive.
b) The function is decreasing when x > 0 and
increasing when x < 0.
c) The function is decreasing because the slope
is negative.
d) The function is increasing because the slope is
positive.
7. a) y = -2/ + 8
y-intercept:
y =
+8
-2x^
x-intercept;
y =
+8
y = ^2(0)^+8
(0) = ^2/ + 8
-8
4 = /
±2 = x
The y-intercept is 8 and the x-intercepts are -2
and
2.
Oa.o)
•2"" * a'o)""^
The x-intercepts are - 2 and 2 while the
y-intercept is 4.
c)y=2x^
y-intercept:
y = 2x^
y=2(0f
y=0
x-intercept:
y = 2x^
(0) = 2x^
0 = x
6. $7'
b)y = x'
y-intercept;
y = x.
y - x \
(0) = x^
y=(Of
0 = x
y =0
x-intercept:
The y-intercept is 0 and the x-intercept is 0.
c)
y-intercept;
1
1
x-intercept;
(0)=lx-2
y = - x ^ 2
y == 2l ( 0 ) ^ 2
4
8=x
The y-intercept is 2 and the x-intercept is 8.
d)
y-intercept;
y = ^x + 3
x-intercept;
y = -4(0) + 3
-3 = ^ x
y = 3
0 = -4x + 3
3
— = X
to,o)
-s.s?
The x-intercept is 0 and the y-intercept is 0.
7-28
=-2/
y =8
I
1
-2x^
4
The y-intercept is 3 and the x-intercept is
8. Answers will vary, e.g., y = - 2 x + 2
For a linear function, the y-intercept will give a
value for b in equations such as y = mx + jb. From
this, the coordinates of the x-intercept can be
substituted in to calculate m.
Chapter 7: Exponential and Logarithmic Functions
C:f>;iptot / test
ri it) I t , i , p..I . I -p.- ,-:i'c .H. ( ..J ,= ».'ii;i,.l h.notion
!>-•.- . O ' 1 * tl.I'. ! > " o /•<.•!•* ; t'-.ij fi
I .' 'M M pair
f.i •• I;. . . ,-.fi ! • \
!Riwyt:4/|
I
M.
y iiifiircopt
j
t iici Bcli.ivitjur
I
V
R Stic
:
• j i l i.,
h'. (!. j •
{y I K > 0,
tjic.-c'j .:;t:) nr
,
b) This data docs not represent an exponential
function because it does not have a constant
ratio for each pair of successful y-values.
x-iiYii'fi
rt:'.
4 ' 4 )
I -'
Increasing Oi
Decreasing?
;
OlihiO!
y 1 / "^R.
ng
2. The y4nter«
The graph extends from guadrant II to guadrant I.
The domain is {x | x o R) and the range is
iy\y>0,ye
c) This data does represent an exponential
function because it has a constant ratio of 3 for
'•> h '
I
>
V
iir ( i~- ••il /- •.,nki
I
Ratio
I
R}.
The function is increasing.
3. The xontercept is 1.
There is no j/-intercept.
The graph extends from guadrant IV to guadrant I.
The domain is {x | x > 0, x e R} and the range is
{ y | y e R}.
The function is increasing.
4. a) This function matches with ii) because it has
a y-intercept of (0,4) and it is increasing.
b) Function a) is increasing, because a > T The
only graph that shows an increasing loganthmic
function is iv). So. function a) matches graph ii).
c) The graph for function i) extends from guadrant
to quadrant IV and has an xontercept but no
y-intercepts. These are charactenstics of a
loganthmic function. Function b) matches graph i).
d) This matches with iii) because it is a
decreasing function with a y-intercept of (0, 4).
Foundations of Mathematics 12 Solutions Manual
7. a) Using a graphing calculator, the eguation of
the exponential regression function that models
the data is y = 2.305...(1.181...)^ The regression
equation appears to fit the data very well. The
algebraic model would be y - 2.305. .(1.181...)',
where y represents the number of dandelions
and X represents the date in May.
b) Assuming that the population growth continues
at the same rate. 401 dandelions will be growing
in the yard on May 3 1 .
y = 2.305...(1.181...)''
y = 2.305..,(1.181...)^^
y = 401.425...
7-29
8 . Itjr- fM|u;il)<)n i)f t t m l ( K | a i ' f h n i i i : r e g r e s s i o n is
y = 1.Hiy
» 0 ? B 9 , . (In x)
w l i i ' H j \- !o ihs^ «Mit,igv lolf.'osoMi in k J , a n d y is t h e
m a < j n i l u d o ul i h o o o d h q i i o k o
'/ - 1 1;-)!)
» 0.280 (III x )
'» f< - 1 109
+ 0,200 (In X)
10.100...
0 . 2 0 0 . . . ( I n X)
3 5 . 2 3 2 . . . - In X
x = e3^^-=^32
x = 2 . 0 0 0 . . . X 10^^
A b o u t 2 . 0 X 10^*^ kJ o f e n e r g y w a s r e l e a s e d
d u r i n g the e a r t h q u a k e that o c c u r r e d off the c o a s t
of J a p a n in March 2 0 1 1 .
7-30
Chaptoi
Exponential and Logarithmic Functions
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