Download Chapter Review 151

Chapter Review
7. c. When x is 0 y is 1400, so Yo= 1400. The value
.
1
1
of y is halved every 3 years, so b = - and k =- .
2
3
Substituting these values into the general
exponential function y = yobkxwe have
</3
1 (1/3)<
1
Review Chapter 10
1. a.
x
y
0
2
8
y=1400
2
32
3
128
2. Yo= 200, b = 7; substituting into y = yob' we have
y = 200(7)'.
3. a. Ao= 2000, r = .068, n = 12; substituting into
= .40(1 +~r we have
121
=2000
( )
( )
1+ .~~8
A(7)
= 2000
=1400
()
"2
years,thereis
1 + .~~8
l3.. which is between
6.8
10 and 11 years.
5. Double the initial $2000 investment is $4000.
068
of the initialvalueor
1
c. Take - of the amount left after 245,000 years
2
to fmd the amount left after 490,000 years;
1 15 15
-*-==3.75g.
2 2
4
.!.
of the amountin (c) will givethe amount
2
ofU-234 left in 735,000 years;
1 15 15
-*-== 1.875 g.
2 4
8
Time (years)
Amount U-234 (g)
0
15
245,000
7.5
490,000
3.75
735,000
1.875
121
(
Y1 = 4000,Y2 = 2000(1+ .068/12)"(12X)
Solve 4000 = 2000 1+
2
1
15
-*15 =- = 7.5 g left after 245,000 years.
2
2
d.
Evaluate on the calculator by typing
2000(1+.068/12)"(12*7).
A(7)::::3214.92
In 7 years the IRA is worth $3214.92
4. The IRA doubles in about
.!.
.
12'7
b.
"2
b. Because the half-life for U-234 is 245,000
c. The outputs are increasing by a factor of 4;
b=4.
A(t)
()
8. a. Half-life is the amount of time it takes for half
of a substance to decay.
b. Find the y value when x = 0; Yo= 2.
A(t)
151
~)
Windowsmay vary: Xmin= 0, Xmax= 18.8,
Ymin= 0, Ymax= 5000.
e. The initial value is Yo= 15, the decay factor is
b = .!. , and the time scale is k =
!
2
245,000
Substituting into the general exponential
function A(t) = yobkt,we have
"245.000
1
A(t) = 15 2
.
()
9. The initial amount is Yo= 12. b = 1 - .08 = .92.
The time scale is k = .!.. Substituting into
5
Il'IhrsIQ:cti(t(J
~=U.;::;::2:195_ '1'='1(1(1(1
____
y = yobkx gives
y = 12(.92)"5
The IRA doubles in approximately 10.2 years.
6. Using the rule of 72, the population of the United
States will double in about E. = 60 years.
1.2
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10. a. Substituting 15 for t
y = 1.2(.98)'
= 1.2(.98)15
".9
In 15 years the snow line is about .9 miles
from the top of the mountain.
152
Chapter 10: Exponential and Logarithmic Functions
b. Substituting.6 for y
y =1.2(.98)'
.6 = 1.2(.98)'
Yl = .6, Y2 = 1.2(.98)"X
Windows may vary: Xmin = 0, Xmax= 47,
Ymin = 0, Ymax = 2.
15. a. log (.2) ::::'.699
b. 10g(%) "".222
c. log (78)::::1.892
16.
=6
=3
210g(x)
log(x)
IOlog(.) = 103
X = 1000
17. a. 410g(x)= 5
5
log(x)=-4
Il'Ihr5tC~i(a1'l
M=3Lt.30!:1618
_'1'=.6
10Iog(.) = lOs14
In approximately 34.3 years the snow line will
be .6 miles from the top of the mountain.
X = IOsI4
b. x = 17.78
11. Using the definition of negative integer exponents
from Section 6.1 we have 6'. = ~ and since I to
6'
any power
is I we may write
= 101.2S
18. a. 10' = 100;
x = 2 because102= 100.
b.
(i J . Therefore,
2(10)'IS= 20
10'1S= 10
log(10'ls) = log(10)
~=I
5
x=5
y=20(6)" =20(iJ
12. a. yo=8000, b=I-.!.=~,k=l;substitutefor
3 3
C.
Yo, b, and k into y = yoblrJ,we have
104.-1= 1000
10g(104.-1) = log(1000)
4x - 1= 3
4x=4
x=l
y = 8000(~}.
b. One-half of the initial amount is $4000, so we
19.
need to solve 4000 = 8000( ~ )'
YI = 4000,Y2 = 8000(2/3)"X
Windowsmay vary:Xmin= 0,Xmax=4.7,
Ymin= 0, Ymax= 10000.
1= 100(lOf.13
!
=IO'./3
100
10 '2 = 10 './3
10g(1O '2) = log(10 './3)
-2= ~
3
6=x
I.I.r"Cli~
M=1.709S:U3 _ 'I'=&tOOO
__
The piece of equipment is worth half of its
initial value in approximately 1.7 years. This
means the half-life is 1.7 years.
13. Domain: {xl 'oo<x<oo}; range: {yIO<y<oo}
14. a. log (100) = 2 because 100 = 102
b. log (10,000) = 4 because 10,000 = 104
c. 10g ~
( )=
10
-1 because ~=10'1
10
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Chapter Test
20. a.
25. The difference between the two sounds is
100 - 20 = 80 dB or 8 B. Because each bel
represents multiplication by 10, 8 B represents
multiplication by 108or 100,000,000. Thus the
100 dB sound is 100,000,000 times more intense
than the 20 decibel sound.
y
, I
I
,
IVi= 110 XI
x
i
153
i
i
;
b. Domain:{xIO<x<oo};
range: tv I-00< y < oo}. The graph does not
appear to cross the y-axis, this is because the
logarithm of 0 or a negative number does not
exist. Although it is hard to tell from the
graph, as x gets close to zero, y decreases
without bound. As we move away from the
y-axis the graph appears to flatten out, but y
does continue to increase without bound.
26. The difference in sound intensities is
90 - 45 = 45 dB or 4.5 B. This means the
recommended maximum sound from a nightclub
is 104.5:::::
31,623 or about 30,000 times more
intense than the recommended maximum sound
outdoors in a residential area at nighttime.
27. Double the initial investment is $2000. We need
to solve
2000 = 1000e.06t
Y1 = 2000,Y2 = 1000e"(.06X)
Windowsmay vary:Xmin= 0,Xmax= 18.8,
Ymin= 0, Ymax= 3000.
21. a. M(A) = 10g(A)
M(ll)
=log(ll)
:::::1.04
b. M(1412)
= 10g(l412)
:::::3.15
c.
M(53, 764, 720) = log(53, 764, 720)
:::::
7.73
28. a. In (e6)= 6
22. The Afghanistan earthquake is 7.4 - 4.4 = 3 units
higher on the Richter scale. Because each unit
higher on the scale means that the earthquake is
10 times as strong, the Afghanistan earthquake is
10*10*10 or 1000 times stronger than the
Peruvian earthquake.
23. a.
pH(1.0
pH(W)=-log(W)
* 10 -14) = -log(1.0
= -log(l
b. In(e-9)= -9
c. In (e1.8)= 1.8
d. In (e) = In (el) = 1
e. In (40):::::3.69
* 10 -14)
0-14)
= -n4)
=14
b. pH(1.0*10-3)= -log(1.0*10-3)
= -log(10-3)
=T3)
=3
c. Evaluate on a calculator because the log is not
of a familiar power of 10
pH(1.3 * 10 -7)
Il'IhrslQ:cti(lr.
H=11.552:~53 _'1'=2(1(1(1
---.
The investmentdoublesin approximately11.55
years.
= -log(1.3 *10-7)
:::::6.9
24. Each time the hydrogen ion concentration
increases by a factor of 10, the pH decreases by 1.
Therefore the pH decreases by 1.
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Chapter
10 Test
1. a. This function is of the formy = yoblcr..In this
case k= 1
Yo = 1,200,000
b. b = 1.052; this means the population in any
given year is 1.052 times the previous year's
population. The population increases by
1.052 - 1 = .052 = 5.2%. Thus the population
grows by 5.2% per year.
c. Substituting 20 for t
y = 1,200,000(1.052lo
y:::::3,307,470.98
We predict the population to be 3,307,471 in
the year 2010.
154
Chapter 10: Exponential and Logarithmic Functions
d. Using the rule of 72 gives
~5.2 '" 13.8 years.
6. a. Yo = 3000, b = 1-.!. = ~ , k = 1; substituting
4 4
Solve2,400,000= 1,200,000(1.052)'by
graphingto finda closerapproximation.
Yl = 2400000,Y2 = 1200000(1.052)AX
Windowsmayvary:Xmin= 0, Xmax= 18.8,
Ymin= 0, Ymax= 3000000.
into A(t)
= yobkr gives
= 3000(~)'
A(t)
b. A(2) = 3000( ~ J
= 1687.5
In 2 years the piece of hardware is worth
$1687.50.
c. One-half the initial value is $1500. This gives
Il'1hr5.;:c~i(lf1
~=U.&7ngg
_'I'=21t00001).....
A closerapproximationby graphingis 13.7
years.
07
the equation 1500 =3000(~)' .
To solve by graphing enter Y1 = 1500,
Y2 = 3000(3/4)AX
Windows may vary: Xmin = 0, Xmax= 4.7,
Ymin= 0, Ymax =4000.
3651
( )
( )
2. a. A(t) = 10000 1+':'365
365'6
A(6)=10000
1+~
365
Enter 10000(1 + .07/365)A(365*6) on the
calculator.
A(6)::: 15219.00; in 6 years the investment is
worth $15,219.00.
Il'1hrs.;:c~i(lf1
~=2.1t09't20B _'1'=1500 ___
b. Using the rule of 72, the investment doubles in
about 72 '" I0 years. Therule of 72 is
7
probably accurate enough because we want
our answer to the nearest year.
The piece of hardware is worth half its initial
value in approximately 2.4 years.
7. a. log (10)= 1
b. log (1) = 0
3. Using the rule of72, the colony doubles in size in
about 72 = 18 minutes. We don't have an initial
4
value, so we can't solve this problem by
graphing.
4. It takes 1600 years for the radium 226 to decay to
half its initial amount. The answer to this
problem is given in the half-life; every 1600 years
the radium 226 is halved. Thus if we start with
50 g, in 1600 years there will be 25 g.
c. 10g(1O-3) = -3
(~ )
d. 10g
100
=-2
8. a. 310g(x)= 2
2
log(x)=3
1010g(x) = 102/3
x=102l3
5. a. Yo= 8600, b = 1-.13 = .87, k = 1; substituting
into A(t)
= yobkr , we
b. 102/3:::4.64
have
A(t) = 8600(.87Y
9.
b. A(5) = 8600(.87)5
A(5) ::: 4286
In 5 years there are about 4286 cases of this
disease.
105>= 10000
log(l05» = 10g(lOOOO)
5x=4
4
x=5
10.
M(A) = 10g(A)
M(25270) = log(25270
",4.4
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-
Chapter Test
11. The earthquake that hit Taiwan was 100 or 102
times more intense than the Mexican quake, so it
was 2 higher on the Richter scale. Thus the
Taiwan earthquake measured 5.1 + 2 or 7.1 on the
Richter scale.
12. The difference in sound intensities is
100 - 70 = 30 dB or 3 B. Because each bel
represents multiplication by 10, 3 B represents
multiplication by 103or 1000. Therefore the
noise from a power lawn mower is 1000 times
louder than the sound from a vacuum cleaner.
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155