Download 3. a. FindM(A)ifA=Inun b. M(lOO)= 10g(100) = log(102) =2 c. M(lOO

Section
10.4
149
7. The relationship used to complete the table in
Exercise 6 works the same way in reverse. If
log (x) increases by 1, then x increases by a factor
of 10. Or iflog (x) decreases by 1, then x
decreases by a factor of 10.
x
log(x)
.058
-1.2
.58
-.2
5.8
.8
58
1.8
580
2.8
9. It is hard to distinguish between amplitudes that
vary so much. On the other hand, Richter
magnitudes usually vary between 1 and 9 making
it easier to see differences in earthquake intensity.
Section 10.4
1. The domain off(x) = log (x) is the set of positive
numbers. There is no number that we can put in
the exponent of 10 that will give 0 or a negative
number.
3. a. FindM(A)if A = I nun
M(A)
M(l)
=10g(A)
=log(l)
= log(l 0°)
=0
b. M(lOO)= 10g(100)
= log(102)
=2
11. The lower the pH, the more acidic the substance.
For example, many plants prefer soil that is near
neutral (pH = 7). When soils become acidic
(pH < 7), gardeners apply lime (sodium
hydroxide, pH = 12.4) to raise the pH of the soil.
13. a. pH and hydrogen ion concentration move in
opposite directions. Lemon juice has a lower
pH so it has the larger hydrogen ion
concentration.
b. The pH of lemon juice is 11 - 2 = 9 units
lower, so its hydrogen ion concentration is 109
or 1,000,000,000 time larger.
15. The difference in intensities is 50 - 30 = 20 dB or
2 B. Because each bel represents multiplication
by 10, 2 B represents multiplication by 102or
100. Thus, the noise from light traffic is 100
times louder than the sound from a soft whisper.
c. M(lOO,OO)
= 10g(100,000)
= 10g(lOs)
=5
Skills and Review 10.4
5. a. M(12,843)= log (12,843).. 4.1
17. a. 4x2+ 7 =43
b. M(240)= log (240).. 2.4
4x2 =36
c. M(351,060,000)= log (351,060,000)..8.5
N
x2 =9
=:119
x=:13
Check
4x2 + 7 = 43
?
4C3)2 + 7=43
?
4*9+7=43
?
36+7=43
43 = 43
@ Houghton MifflinCompany. All rights reserved.
?
4(3)2 + 7 =43
?
4*9+7=43
?
36+7=43
43 = 43
150
Chapter 10: Exponential and Logarithmic Functions
b. 4-h -1 =11
4-h =12
-h=3
c. Again we could use the formula from part (a)
to find A(2400) or recognize that 2400 years
(-hf =32
x=9
Check
4-h-l=11
amount is .!.Ao=.!. * 3 kg = ~ kg or .375 kg.
8
8
8
4.;9-1~11
is three half-lives. .!.*.!. *.!. of the initial
222
23. fix) and g(x) are inverses of each other because
f{g(x)) = x and g(f{x)) = x.
fix) =x + 2 andg(x) =x - 2
f(g(x)) =f(x + 2)
=(x-2)+2
Substitute 9 for x
g(f(x)) = g(x + 2)
=(x+2)-2
=x+2-2
=x
?
12 -1=11
11=11
19. a.
3(1Wx = 3000
(1WX = 1000
log(102X)= log(1000)
25. a. 2X2+ 6x
2x(x + 3)
10g(102x) 10g(103)
b. x2y -16y
y(x2 -16)
y(x - 4)(x + 4)
2x=3
3
x = - =1.5
2
Check
3(lOh)
=3000
?
3(102'15)';'3000
Substitute 1.5 for x
to
middle
-4xsplit
thatthe
sums
to 3xterm,
is -Ix3x.
andThe
4x.factor pair of
x2 -lx+4x-4
(x2 -lx)+(4x-4)
x(x-l)+4(x-l)
(x-I)(x+4) or (x+4)(x-l)
?
3(103)';'3000
?
3 * 1000=3000
3000 = 3000
b. 410g(x) = - 4
10108(x) = 10-1
1
x=lO -1 =-=.1
10
Check
410g(x) =? -1
410g(1O -1)';' - 4
?
4*Cl)=-4
c. ~ + 3x - 4
There is no GCF of all three terms and we
can't use the difference of two squares, so try
Substitute 10 -1 for x
-4= -4
21. a. Ao= 3, b =.!., k = ~ ; substitutinginto A(t)
2
800
1
= Ao(b)''',we have A(t) = 3 2"
,/800
()
.
b. We could substitute 800 for t and find A(800)
from the formula in (a), or recognize that 800
years is one half-life. In one half-life we have
one-half the initial amount
1
1
3
- Au =- * 3 kg =- kg or 1.5 kg.
2
2
2