Bleach:
p(xB) = –log xB
13 = –log xB
–13 = log xB
-13
xB = 10
−5
xC 10
=
xB 10 −13
⎛ 1⎞
7. a) ⎜ ⎟ log3 36 − log3 2 = log3 6 − log3 2
⎝ 2⎠
⎛ 6⎞
= log3 ⎜ ⎟
⎝ 2⎠
= log3 3
=1
b) log 625 + 2 log 4 = log 625 + log 16
= log (625 16)
= log 10 000
=4
⎛ 40 ⎞
c) log2 40 − log2 2.5 = log2 ⎜
⎝ 2.5 ⎟⎠
xC
1
=
xB 10 −8
xC
= 108
xB
xC
= 100 000 000
xB
Therefore black coffee is 100 000 000 times more
acidic than bleach.
Lesson 7.3: Laws of Logarithms, page 446
= log2 16
=4
⎛ 1⎞
d) log3 27 − ⎜ ⎟ log3 27 = log3 27 − log3 3
⎝ 3⎠
⎛ 27 ⎞
= log3 ⎜ ⎟
⎝ 3⎠
1. a) log5 (8 6) = log5 8 + log5 6
b) logz (xy) = logz x + logz y
= log3 9
2. a) log3 7 + log3 5 = log3 (7 5)
b) logb s + logb r = logb (sr)
⎛ 36 ⎞
3. a) log2 ⎜ ⎟ = log2 36 − log2 11
⎝ 11 ⎠
⎛ x⎞
b) logz ⎜ ⎟ = logz x − logz y
⎝y⎠
⎛ 15 ⎞
4. a) log3 15 − log3 11= log3 ⎜ ⎟
⎝ 11⎠
⎛ s⎞
b) logb s − logb r = logb ⎜ ⎟
⎝r⎠
3
5. a) log6 7 = 3 log6 7
y
b) logz x = y logz x
6. a) log6 2 + log6 108 = log6 (2 108)
= log6 216
=3
810
b) log3 810 – log3 20 = log3
10
= log3 81
=4
c) log 5 + log 0.002 = log (5 0.002)
1
= log
100
= –2
d)
16
2
= log2 8
=3
log2 16 – log2 2 = log2
Principles of Mathematics 12 Solutions Manual
=2
8. a) e.g., log2 32 = log2 8 + log2 4
log2 32 = log2 64 – log2 2
b) e.g., log 125 = log 25 + log 5
log 125 = log 250 – log 2
9. a) e.g., log2 32 = 5 log2 2
b) log 125 = 3 log 5
y 3
10. e.g., Daniel made an error when he simplified (2 )
y+3
to 2 . The last three lines should be:
3y
3
2 = 5 I used the exponent law for the power of a
power.
3
3y = log2 5 I wrote the equation in logarithmic form.
3
3 log2 5 = log2 5 I replaced y with the logarithm it
represented.
11. a)log3 2 + log3 4.5 = log3 (2 4.5)
= log3 9
=2
b)
c)
⎛ 162 ⎞
log3 162 – log3 2 = log3 ⎜
⎝ 2 ⎟⎠
= log3 81
=4
3
log2 16 = 3 log2 16
=34
= 12
⎛ 10 ⎞
d) log5 10 – log5 1250 = log5 ⎜
⎝ 1250 ⎟⎠
⎛ 1 ⎞
= log5 ⎜
⎝ 125 ⎟⎠
= –3
7-11
⎛ 40 ⎞
12. a) log 40 – log 8 = log ⎜ ⎟
⎝ 8⎠
= log 5
= 0.698…
It is approximately 0.70.
b) log 5 + 2 log 2 = log 5 + log 4
= log (5 4)
= log 20
= 1.301…
It is approximately 1.30.
⎛ 1⎞
13. 3 log8 2 − ⎜ ⎟ log8 64 = log8 8 − log8 8
⎝ 2⎠
=0
14. e.g., Wayne made an error when he simplified
log3 27
log3 27 – log3 9 as
. The solution should be
log3 9
⎛ 27 ⎞
log3 27 – log3 9 = log3 ⎜ ⎟
⎝ 9⎠
= log3 3
=1
⎛ 2⎞
15. log5 53 + ⎜ ⎟ log5 125 = log5 53 + log5 5 2
⎝ 3⎠
= 3+2
=5
2
16. 4 log2 4 + log2 16 = 4(2) + 2 log2 16
= 8 + 2(4)
= 16
17. e.g., Because exponents and logarithms are two
ways to express the same meaning, the laws are
related. They follow the same pattern.
The product law of logarithms is related to the product
law for exponents.
xy
x y
loga mn = loga m + loga n → a = a a ; multiply two
powers, add the exponents
log (100 10) = log 100 + log 10, which corresponds
2
1
2+1
to 10 10 = 10
The quotient law of logarithms is related to the
quotient law for exponents.
ax
m
x–y
= loga m – loga n → y = a ; divide two
loga
n
a
powers, subtract the exponents
100
= log 100 – log 10, which corresponds to
log
10
18. e.g., agree:
2
log (xy) = 2 log(xy)
= 2(log x + log y)
= 2 log x + 2 log y
(
)
(
19. log x 2 − 4x + 4 − log x 2 − 4
)
⎛ x − 4x + 4 ⎞
= log ⎜
⎟
2
⎝ x −4 ⎠
2
⎛ (x − 2)(x − 2) ⎞
= log ⎜
⎝ (x − 2)(x + 2) ⎟⎠
⎛ x − 2⎞
= log ⎜
⎝ x + 2 ⎟⎠
= log(x − 2) − log(x + 2)
Or log ( x 2 − 4x + 4 ) − log ( x 2 − 4 )
= log ( x − 2) − log (( x − 2)( x + 2))
2
= 2 log ( x − 2) − log ( x − 2) − log ( x + 2)
= log ( x − 2) − log ( x + 2)
20. log x x m−1 = ( m − 1)log x x
= mlog x x − 1log x x
= m −1
Math in Action, page 448
1
. This
3
result explains the position of log 2 on the scale,
because log 10 = 1 and log 2 is about one-third of
log 10.
1
ii) log 3 = 0.477…, which is a little less than . This
2
result explains the position of log 3 on the scale,
because log 3 is about one-half of log 10.
• The scale starts at 1 because log 1 = 0.
• log 0 is undefined, since no number raised to an
exponent yields zero.
1
• log 1.5 = 0.176…, which is slightly less than . This
5
1
means that log 1.5 should be placed about
of the
5
way between 1 and 10.
i) log 2 = 0.301…, which is a little less than
10 2
= 10 2−1
101
The power law of logarithms is related to the power
law for exponents.
n
x y
xy
loga m = n loga m → (a ) = a ; power of a power,
3
multiply the exponents log2 16 = 3 log2 16, which
4 3
43
corresponds to (2 ) = 2
7-12
Chapter 7: Logarithmic Functions
x
log x
Distance from 1 on a 13 cm Scale
(cm)
1
0.00
0.0
2
0.30
3.0
3
0.48
4.8
4
0.60
6.0
5
0.70
7.0
6
0.78
7.8
7
0.85
8.5
8
0.90
9.0
9
0.95
9.5
10
1.00
10.0
11
1.04
10.4
12
1.08
10.8
13
1.11
11.1
14
1.15
11.5
15
1.18
11.8
16
1.20
12.0
17
1.23
12.3
18
1.26
12.6
19
1.28
12.8
20
1.30
13.0
• To model division, align the dividend on the log
scale at the top (top row) with the divisor on the log
scale at the bottom (bottom row). The 1 of the bottom
row will indicate the quotient on the top row.
Lesson 7.4: Solving Exponential Equations
Using Logarithms, page 455
1. a) e.g., I estimate x = 2.1
x
10 = 3
x
log 10 = log 3
log 10 = x log 3
log10
=x
log3
2.095… = x
2.096 = x
b) e.g., I estimate x = 0.5
x
10 = 3
x
log 10 = log 3
x log 10 = log 3
log3
x=
log10
x = 0.477…
x = 0.477
c) e.g., I estimate x = 2.3
x
27 = 4
x
log 27 = log 4
log 27 = x log 4
log27
=x
log4
2.377… = x
2.377 = x
Principles of Mathematics 12 Solutions Manual
d) e.g., I estimate x = 2.6
x
78 = 5
x
log 78 = log 5
log 78 = x log 5
log78
=x
log5
2.706… = x
2.707 = x
e) e.g., I estimate x = 2.2
x
4.5 = 30
x
log 4.5 = log 30
x log 4.5 = log 30
log30
x=
log4.5
x = 2.261…
x = 2.261
f) e.g., I estimate x = –0.9
x
e = 0.4
x
ln e = ln 0.4
x ln e = ln 0.4
x = ln 0.4
x = –0.916…
x = –0.916…
2. a) e.g., I estimate x = 1.7
x
35 = 5(3 )
x
7=3
x
log 7 = log 3
log 7 = x log 3
log7
=x
log3
1.771… = x
1.771 = x
b) e.g., I estimate x = 5.3
x
117 = 3(2 )
x
39 = 2
x
log 39 = log 2
log 39 = x log 2
log39
=x
log2
5.285… = x
5.285 = x
c) e.g., I estimate x = 9.1
x
1000 = 500(1 + 0.08)
x
2 = 1.08
x
log 2 = log 1.08
log 2 = x log 1.08
log2
=x
log1.08
9.006… = x
9.006 = x
7-13