Chapter 8

)
c)
d)
8
6
4
2
–4 –2
0
–2
–4
–6
–8
function: y 522 log 5 3x
D 5 5xPR 0 x . 06
R 5 5 yPR6
asymptote: x 5 0
inverse: y 5 1 3 Q52 2R
x
D 5 5xPR 6
R 5 5yPR 0 y . 06
asymptote: y 5 0
(x22)
y510 3
function: y 5 2 1 3 log x
D 5 5xPR 0 x . 06
R 5 5 yPR6
asymptote: x 5 0
(x 2 2)
3
inverse: y 5 10
D 5 5xPR 6
R 5 5yPR 0 y . 06
asymptote: y 5 0
function: y 5 20(8) x
D 5 5xPR6
R 5 5 yPR0 y . 06
asymptote: y 5 0
inverse: y 5 log 8 Q x
20 R
D 5 5xPR0 x . 0 6
R 5 5yPR6
asymptote: x 5 0
y
6
4
y520(8) x
2
x
–6 –4 –2
0
2 4 6
–2
–4
y5log x 8 20 +
–6
2
10
y
8
6
4
2
–8 –6 –4 –2
0
2 4 6 8
–2
–4
y5213 log x
–6
–8
1 2– x y 5
3
5 2
4 6 8 10 12
y522 log 5 3x
y
(
(
+
x
x
e)
f)
16
12
8
y52(3) x+2 4
x
–8 –4
0
4 8 12 16
–4
y 5 log
x
3
– 2
2+
–8
function: y 5 2(3) x12
D 5 5xPR6
R 5 5yPR 0 y . 06
asymptote: y 5 0
inverse: y 5 log 3 Q x 2 R 2 2
D 5 5xPR0 x . 06
R 5 5yPR6
asymptote: x 5 0
–10 –8 –6 –4 –2
0
2
–2
y 5 25 x 2 3
y 5 log 5 (2x 2 3)
function: y 525 x 2 3
D 5 5xPR6
R 5 5yPR 0 y ,236
asymptote: y 523
inverse: y 5 log 5 (2x 2 3)
D 5 5xPR0 x ,236
R 5 5yPR6
asymptote: x 523
23. Given the constraints, two integer values are
possible for y, either 1 or 2. If y 5 3, then x
must be 1000, which is not permitted.
Lesson 8.4, pp. 475–476
1. a) log 45 1 log 68
b) log m p 1 log m q
c) log 123 2 log 31
d) log m p 2 log m q
e) log 2 14 1 log 2 9
f) log 4 81 2 log 4 30
2. a) log 35 d) log x y
b) log 3 2 e) log 6 504
c) log m ab f) log 4 6
3.
1
a) d) log 45
3
b) 21 log 7
1
e)
2 log 736
c) q log m p f) 1 5 log 5125
y
2
–4
–6
–8
–10
–12
–14
+
y
x
log 32;
log 4
4. a) log 3 27; 3 d) 7 log 4 4; 7
b) log 5 25; 2 e) 5 2
c) 1 1
2 f) log 10;
2 2
5. y 5 log 2 (4x) 5 log 2 x 1 log 2 4
5 log 2 x 1 2, so y 5 log 2 (4x) vertically
shifts y 5 log 2 x up 2 units;
y 5 log 2 (8x) 5 log 2 x 1 log 2 8
5 log 2 x 1 3, so y 5 log 2 (4x) vertically
shifts y 5 log 2 x up 3 units;
y 5 log 2 a x 5 log 2 x 2 log 2 2
2 b
5 log 2 x 2 1, so y 5 log 2 (4x) vertically
shifts y 5 log 2 x down 1 unit
6. a) 1.5 d) 20.5
b) 2 e) 4
c) 1.5 f) 2
7. a) log b x 1 log b y 1 log b z
b) log b z 2 (log b x 1 log b y)
c) 2 log b x 1 3 log b y
1
d)
2 (5 log b x 1 log b y 1 3 log b z)
8.
1
log means 5 x 5 3 5 3 and log 5 means
3
1
5 y 5 1 since 5 y 5 5 x(21) ;
3 5 321 ,
3 ; 1
therefore log 5 3 1 log 5 3 5 x 1 x(21) 5 0
9. a) log 5 56 d) log 3 4
b) log 3 2 e) log 4 (3!2)
c) log 2 45 f ) log 16
10. a) log 2 x 5 log 2 245; x 5 245
b) log x 5 log 432; x 5 432
c) log 4 x 5 log 7 5; x 5 5
d) log 7 x 5 log 7 5; x 5 5
e) log 3 x 5 log 3 4; x 5 4
f) log 5 x 5 log 5 384; x 5 384
11. a) log 2 xyz d) log 2 xy
uw
b) log 5
v
e) log 3 3x 2
a
c) log 6
bc
f)
x 5
v
12.
!x !y
log a
! 4 z 3
13. vertical stretch by a factor of 3, and vertical
shift 3 units up
14. Answers may vary. For example,
f (x) 5 2 log x 2 log 12
g(x) 5 log x 2
12
2 log x 2 log 12 5 log x 2 2 log 12
5 log x 2
12
15. Answers may vary. For example, any number
can be written as a power with a given base.
The base of the logarithm is 3. Write each
term in the quotient as a power of 3. The
laws of logarithms make it possible to
evaluate the expression by simplifying the
quotient and noting the exponent.
16. log x x m21 1 1 5 m 2 1 1 1 5 m
670 Answers
NEL