MHR • Advanced Functions 12 Solutions 585 585
Chapter 6 Exponential and Logarithmic Functions Chapter 6 Prerequisite Skills Chapter 6 Prerequisite Skills Question 1 Page 308 a)
b) i) Domain:
x !R{ }
ii) Range:
y !R, y > 0{ }
iii) y = 0 Chapter 6 Prerequisite Skills Question 2 Page 308 a) P(0) = 300(2)0 = 300(1) = 300 b) P(3) = 300(2)3 = 300(8) = 2400 c) i) ii)
t &= 2.74 days approximately 38 bacteria Chapter 6 Prerequisite Skills Question 3 Page 308 a) x7 b) m3 c) k6 d) –8x12
e)
!2b
a f)
2
x2
g)
1
u
MHR • Advanced Functions 12 Solutions 586 586
Chapter 6 Prerequisite Skills Question 4 Page 308
a) 22= 4 b)
103
102= 10
c)
�
1
3
!
" #
$
% &
6
1
3
9!
"
# #
$
%
& &
=1
3
!
" #
$
% &
'3
= 33
= 27
d)
�
36!3
5
37
+1=3
11
37
+1
= 34
+1
= 81+1
= 82
Chapter 6 Prerequisite Skills Question 5 Page 308 a)
b) Domain:
x !R, x " 0{ } , Range:
y !R, y " 0{ }
c) d)
e) Yes. f –1 is a function since it passes the vertical line test. f) Domain:
x !R, x " 0{ } , Range:
y !R, y " 0{ }
MHR • Advanced Functions 12 Solutions 587 587
Chapter 6 Prerequisite Skills Question 6 Page 308 a)
b) Domain:
x !R{ } , Range:
y !R, y " 4{ }
c)
d)
e) No. f –1 is not a function since it does not pass the vertical line test. f) Domain:
x !R, x " 4{ } , Range:
y !R{ }
Chapter 6 Prerequisite Skills Question 7 Page 308 Yes. Each curve is a reflection of the other in the line y = x. Chapter 6 Prerequisite Skills Question 8 Page 309 a) Translation of 3 units to the right and 1 unit up. b) Reflection in the x-axis and a vertical stretch of factor 2.
MHR • Advanced Functions 12 Solutions 588 588
Chapter 6 Prerequisite Skills Question 9 Page 309 a) A vertical stretch of factor 3 and vertical translation of 4 units down. b) The period is doubled followed by a reflection in the y-axis. Chapter 6 Prerequisite Skills Question 10 Page 309
Apply a horizontal translation of 2 units to the right and a vertical translation of 7 units down. Chapter 6 Prerequisite Skills Question 11 Page 309
Apply a vertical stretch by a factor of 2 and a reflection in the x-axis. Chapter 6 Prerequisite Skills Question 12 Page 309
Apply a vertical stretch by a factor of 3 and a reflection in the line y = x.
MHR • Advanced Functions 12 Solutions 589 589
Chapter 6 Section 1 The Exponential Function and Its Inverse Chapter 6 Section 1 Question 1 Page 318 C, D First Differences A B
C D
If you look at the first differences, y is increasing at a rate proportional to the functions for C and D so they are exponential.
MHR • Advanced Functions 12 Solutions 590 590
Chapter 6 Section 1 Question 2 Page 318 C:
y = bx
1= b0 This statement is true for any value of b.
3= b1
b = 3 The only valid value for b is 3.
Check:
9 = 32
Since b = 3, y = 3x.
D:
y = bx
1= b0 This statement is true for any value of b.
1
3= b1
b =1
3 The only valid value for b is 3.
Check:
9 =1
3
!"#
$%&
'2
Since b =1
3, y =
1
3
!"#
$%&
x
.
MHR • Advanced Functions 12 Solutions 591 591
Chapter 6 Section 1 Question 3 Page 318 a)
b) i)
m12=
1.52!1.5
1
2 !1
= 0.75
ii)
m23=
1.53!1.5
2
3! 2
&= 1.13
iii)
m34=
1.54!1.5
3
4 ! 3
&= 1.69
iv)
m45=
1.55!1.5
4
5! 4
&= 2.53
c)
m1=
1.51!1.5
0.999
1! 0.999
&= 0.6
m2=
1.52!1.5
1.999
2 !1.999
&= 0.9
m3=
1.53!1.5
2.999
3! 2.999
&= 1.4
m4=
1.54!1.5
3.999
4 ! 3.999
&= 2.1
m5=
1.55!1.5
4.999
5! 4.999
&= 3.1
d) The rates are increasing.
MHR • Advanced Functions 12 Solutions 592 592
Chapter 6 Section 1 Question 4 Page 319 a)
b) i)
�
m!3!2
=0.5
!2!0.5
!3
!2 + 3
= !4
ii)
�
m!2!1
=0.5
!1!0.5
!2
!1+ 2
= !2
iii)
�
m!10
=0.5
0!0.5
!1
0 +1
= !1
iv)
�
m01
=0.5
1!0.5
0
1!0
= !0.5
c)
m!3=
0.5!3! 0.5
!2.999
!3+ 2.999
&= !5.5
m!2=
0.5!2! 0.5
!1.999
!2 +1.999
&= !2.8
m!1=
0.5!1! 0.5
!0.999
!1+ 0.999
&= !1.4
m0=
0.50! 0.5
0.0001
0 ! 0.0001
&= !0.7
m1=
0.51! 0.5
0.999
1! 0.999
&= !0.35
d) The rates are increasing.
MHR • Advanced Functions 12 Solutions 593 593
Chapter 6 Section 1 Question 5 Page 319 a)
b) y = 3x c), d)
Chapter 6 Section 1 Question 6 Page 319 a)
b)
y =1
3
!"#
$%&
x
c), d)
MHR • Advanced Functions 12 Solutions 594 594
Chapter 6 Section 1 Question 7 Page 319 a) iii) b) i) c) ii) d) iv) Chapter 6 Section 1 Question 8 Page 320
a) Graph is the inverse of 7d): y =
1
5
!"#
$%&
x
b) Graph is the inverse of 7b): y = 5x
c) Graph is the inverse of 7c): y =
1
2
!"#
$%&
x
d) Graph is the inverse of 7a): y = 2x
Chapter 6 Section 1 Question 9 Page 320 a)
MHR • Advanced Functions 12 Solutions 595 595
b)
c) All three functions have the same domain and are increasing. d) h(x) = 3x is different than the other two functions for range, x-intercept, y-intercept, y = 0 asymptote and positive/negative intervals. e) For f(x) = 3x, the instantaneous rate of change is constant. For g(x) = x3, the instantaneous rate of change is decreasing then increasing. For h(x) = 3x, the instantaneous rate of change is increasing.
MHR • Advanced Functions 12 Solutions 596 596
Chapter 6 Section 1 Question 10 Page 320 a) i) N(0) = 10(2)0 ii) N(1) = 10(2)1 = 10 = 20 iii) N(2) = 10(2)2 iv) N(3) = 10(2)3 = 40 = 80 b) Yes, the function appears to be exponential since the rate of change is increasing.
c)
m12=
10(2)2!10(2)1
2 !1
= 20
d) i)
m1=
10(2)1!10(2)0.999
1! 0.999
&= 13.9
ii)
m2=
10(2)2!10(2)1.999
2 !1.999
&= 27.7
e) Answers may vary. A sample solution is shown. c) is an average rate of change between two
points. d) is an instantaneous rate at one point. If you average the two instantaneous rates in d) (20.8), it is close to the value in c) (20).
Chapter 6 Section 1 Question 11 Page 320 a) b), c)
MHR • Advanced Functions 12 Solutions 597 597
Chapter 6 Section 1 Question 12 Page 320 a) Domain:
x !R{ } , Range:
y !R, y > 0{ }
b) No x-intercept. c) The y-intercept is 1. d) The function is never negative. The function is positive for all intervals. e) The function is increasing for all intervals. f) The horizontal asymptote is y = 0. Chapter 6 Section 1 Question 13 Page 321 a) Domain:
x !R, x > 0{ } , Range:
y !R{ }
b) The x-intercept is 1. c) No y-intercept. d) For 0 < x < 1, the function is negative. For x > 1, the function is positive. e) The function is increasing for x > 0. f) The vertical asymptote is x = 0. Chapter 6 Section 1 Question 14 Page 321 a) b), c)
MHR • Advanced Functions 12 Solutions 598 598
Chapter 6 Section 1 Question 15 Page 321 a) Domain:
x !R{ } , Range:
y !R, y > 0{ }
b) No x-intercept. c) The y-intercept is 1. d) The function is positive for all intervals. e) The function is decreasing for all intervals. f) The horizontal asymptote is y = 0. Chapter 6 Section 1 Question 16 Page 321 a) Domain:
x !R, x > 0{ } , Range:
y !R{ }
b) The x-intercept is 1. c) No y-intercept. d) For 0 < x < 1, the function is positive. For x > 1, the function is negative. e) The function is decreasing for x > 0. f) The vertical asymptote is x = 0. Chapter 6 Section 1 Question 17 Page 321 a) The domain, range, y-intercept, and the horizontal asymptote are the same. They are positive
for the same interval.
b) y = 4x increases and
y =1
2
!"#
$%&
x
decreases.
Chapter 6 Section 1 Question 18 Page 321 a) The domain, range, x-intercept, and vertical asymptote are the same. b) f –1 and g–1 are positive and negative for different intervals. f –1 is increasing for all intervals
and g–1 is decreasing for all intervals.
MHR • Advanced Functions 12 Solutions 599 599
Chapter 6 Section 1 Question 19 Page 321 a) b), c)
d) y = 6x Chapter 6 Section 1 Question 20 Page 321
y =1
10
!"#
$%&
x
Chapter 6 Section 1 Question 21 Page 321 a)
x y 0 1 1 –2 2 4 3 –8 4 16
b)
c) No, the points do not form a smooth curve, they form a v-shape.
MHR • Advanced Functions 12 Solutions 600 600
d) i) ii)
The values are undefined because they are not real (square root of a negative value). e) Answers may vary. A sample solution is shown. Exponential functions are defined only for functions with positive bases since they are real. Exponential functions with negative bases are not real.
Chapter 6 Section 1 Question 22 Page 321
a)
m12=
1.42!1.4
1
2 !1
= 0.56
m34=
1.44!1.4
3
4 ! 3
&= 1.10
The ship’s average velocity between 1 s and 2 s is 56 km/s. The ship’s average velocity between 3 s and 4 s is approximately 110 km/s.
b)
m3=
1.43!1.4
2.999
3! 2.999
&= 0.92
m4=
1.44!1.4
3.999
4 ! 3.999
&= 1.29
The ship’s instantaneous velocity is 92 km/s at 3 s and 129 km/s at 4 s. Chapter 6 Section 1 Question 23 Page 322 a) For 0 < b < 1, f and f –1 have equal x- and y-coordinates at the point where they intersect the
line y = x. b) Yes; when b > 1, the graphs do not intersect the line y = x.
MHR • Advanced Functions 12 Solutions 601 601
Chapter 6 Section 1 Question 24 Page 322 a), b)
c) i) f is a function for b > 0. ii) f is undefined for b < 0. d) i) f –1 is a function for b > 0. ii) f –1 is undefined for b < 0. e) Yes, it seems that if f is a function then f –1 is a function. If f is undefined, f –1 is undefined.
MHR • Advanced Functions 12 Solutions 602 602
Chapter 6 Section 2 Logarithms Chapter 6 Section 2 Question 1 Page 328
a) log
464 = 3 b)
log
2128 = 7 c)
log5
1
25
!"#
$%&= '2 d)
log1
2
0.25 = 2
e) log
6y = x f)
log
10100 000 = 5 g)
log3
1
27
!"#
$%&= '3 h)
log
bv = u
Chapter 6 Section 2 Question 2 Page 328
a)
y = log2
64
2 y= 64
2 y= 26
y = 6
b)
y = log3
27
3y= 27
3y= 33
y = 3
c)
y = log2
1
4
!"#
$%&
2 y=
1
4
2 y=
1
22
2 y= 2'2
y = '2
d)
y = log4
1
64
!"#
$%&
4 y=
1
64
4 y=
1
43
4 y= 4'3
y = '3
e)
y = log5125
5y= 125
5y= 53
y = 3
f)
y = log21024
2 y= 1024
2 y= 210
y = 10
g)
y = log6
363
6 y= 363
6 y= 66
y = 6
h)
y = log381
3y= 81
3y= 34
y = 4
Chapter 6 Section 2 Question 3 Page 328
a)
y = log10
1000
10 y= 1000
10 y= 103
y = 3
b)
y = log10
1
10
!"#
$%&
10 y=
1
10
10 y= 10'1
y = '1
c)
y = log10
1
10 y= 1
10 y= 100
y = 0
d)
y = log10
0.001
10 y= 0.001
10 y= 10!3
y = !3
MHR • Advanced Functions 12 Solutions 603 603
e)
y = log10
10!4
10 y= 10!4
y = !4
f)
y = log10
1 000 000
10 y= 1 000 000
10 y= 106
y = 6
g)
y = log10
1
100
!"#
$%&
10 y=
1
100
10 y= 10'2
y = '2
h)
y = log10
10 000
10 y= 10 000
10 y= 104
y = 4
Chapter 6 Section 2 Question 4 Page 328 a) 72 = 49 b) 25 = 32 c) 104 = 10 000 d) bw = z
e) 23 = 8 f) 54 = 625 g)
10!2=
1
100 h) 72y = x
Chapter 6 Section 2 Question 5 Page 328 a) y = 2x x = 2y or y = log2 x
b) y = 4x x = 4y or y = log4 x
MHR • Advanced Functions 12 Solutions 604 604
Chapter 6 Section 2 Question 6 Page 328 a)
2
x= 6; y
1= 2
x, y
2= 6 b)
4
x= 180; y
1= 4
x, y
2= 180
log
26 &= 2.6
log
4180 &= 3.7
c)
3
x= 900; y
1= 3
x, y
2= 900 d)
9
x= 0.035; y
1= 9
x, y
2= 0.035
log
3900 &= 6.2
log
90.035 &= !1.5
Chapter 6 Section 2 Question 7 Page 328 Answers may vary. A sample solution is shown. d)
9x= 0.035
9x! 0.035 = 0
Solve by graphing y = 9
x! 0.035 and finding the zero.
log
90.035 &= !1.5
MHR • Advanced Functions 12 Solutions 605 605
Chapter 6 Section 2 Question 8 Page 328 Evaluate each expression using the log function on a calculator. An example is shown for part a). a) 2.63
b) –4.43 c) 0.95 d) –0.70 e) 1.23 f) 2.00 g) 2.26 h) 3.00 Chapter 6 Section 2 Question 9 Page 329 a) y = 10x
b)
Chapter 6 Section 2 Question 10 Page 329
a)
y = log33
3y= 3
3y= 31
y = 1
b)
y = log2
2
2 y= 2
2 y= 21
y = 1
c)
y = log12
12
12 y= 12
12 y= 121
y = 1
d)
y = log1
2
1
2
1
2
y
=1
2
1
2
y
=1
2
1
y = 1
MHR • Advanced Functions 12 Solutions 606 606
Chapter 6 Section 2 Question 11 Page 329 a) logx x = 1 for x > 0, x ≠ 1 b) Answers may vary. A sample solution is shown.
log0.1
0.1= 1
log1
3
1
3= 1
log11
11= 1
c)
log
xx = 1, x > 0, x ! 1
L.S. R.S.
y = logx
x y = 1
xy= x
xy= x
1
y = 1
Since L.S. = R.S., log
xx = 1, x > 0, x ! 1 is true.
Chapter 6 Section 2 Question 12 Page 329 a) Answers may vary. A sample solution is shown. y = log x
y = 10
x
The logarithmic function has decreasing slope (the rate of change is decreasing) and the exponential function has increasing slope (the rate of change is increasing).
b) Answers may vary. A sample solution is shown.
The average rate of change of a logarithmic function is the reciprocal of the average rate of change of its inverse, which is an exponential function.
log(100)! log(10)
100 !10=
1
90
102!10
1
2 !1= 90
MHR • Advanced Functions 12 Solutions 607 607
Chapter 6 Section 2 Question 13 Page 329 a)
It takes approximately 6.3 days for the number of visitors to the Web site to reach 1000. b)
It takes approximately 12.6 days for the number of visitors to the Web site to reach 1 000 000. Chapter 6 Section 2 Question 14 Page 329 a)
The car is approximately 66.5 m away.
MHR • Advanced Functions 12 Solutions 608 608
b)
No, if the headlight intensity doubles the car is approximately 16.2 m away, which is closer than half.
c) Answers may vary. A sample solution is shown. Drive slower. Chapter 6 Section 2 Question 15 Page 329 a)
The thickness of the hull needs to be at least 5.72 cm. b)
The thickness of the hull needs to be at least 7.15 cm, so it needs to be increased by approximately 1.43 cm.
MHR • Advanced Functions 12 Solutions 609 609
Chapter 6 Section 2 Question 16 Page 329 Solutions to Achievement Check questions are provided in the Teacher’s Resource. Chapter 6 Section 2 Question 17 Page 330 Answers may vary. A sample solution is shown.
a) Approximately 3 integer values can be viewed at one time. b) It is difficult to view a broad range of this function since y is an integer when x is a power of
10. For example: x y 10 1
100 2 1000 3
Chapter 6 Section 2 Question 18 Page 330 a) Answers may vary. A sample solution is shown.
x y 100 0 101 1 102 2 103 3 104 4 105 5 106 6
MHR • Advanced Functions 12 Solutions 610 610
b)
The graph is linear. The semi-log grid has turned each power into the exponent value. c) An advantage is to plot a greater range of values. Chapter 6 Section 2 Question 19 Page 330 a)
1 0 10 1
100 2 1000 3
10000 4 100000 5
1000000 6 b)
The graph is a curve with positive, decreasing slope.
MHR • Advanced Functions 12 Solutions 611 611
c) 1 0 0
10 1 1 100 2 2
1000 3 3 10000 4 4
100000 5 5 1000000 6 6
d)
The graph is linear with positive, constant slope. e) Answers may vary. A sample solution is shown. You can graph a greater range of values. Chapter 6 Section 2 Question 20 Page 330
�
D;10
20061+10
2( )2!10
2007=
1+100
2!10
=101
20
= 5.05
MHR • Advanced Functions 12 Solutions 612 612
Chapter 6 Section 2 Question 21 Page 330 Method 1:
m!1
m= 1 n !
1
n= !1
m2! m!1= 0 n2
+ n !1= 0
m =1± (!1)2
! 4(1)(!1)
2(1) n =
!1± (1)2! 4(1)(!1)
2(1)
m =1+ 5
2 n =
!1+ 5
2 m, n are positive
m+ n =1+ 5
2+!1+ 5
2
=2 5
2
= 5
Method 2:
m!1
m= 1
m2! m!1= 0
m =1± (!1)2
! 4(1)(!1)
2(1)
m =1+ 5
2
m+1
m=
1+ 5
2+
2
1+ 5
=(1+ 5)(1+ 5)+ 2(2)
2(1+ 5)
=1+ 2 5 + 5+ 4
2(1+ 5)
=5+ 5
1+ 5
=5+ 5
1+ 5"
1! 5
1! 5
=5! 4 5 ! 5
1! 5
=!4 5
!4
= 5
MHR • Advanced Functions 12 Solutions 613 613
Chapter 6 Section 3 Transformations of Logarithmic Functions Chapter 6 Section 3 Question 1 Page 338 a) iv; translated right 3 units b) ii; translated down 3 units c) i; translated left 3 units d) iii; translated up 3 units Chapter 6 Section 3 Question 2 Page 338 a) translate right 2 units b) translate left 5 units and down 4 units c) translate up 1 unit d) translate left 4 units and down 6 units Chapter 6 Section 3 Question 3 Page 338 a)
Domain:
x !R, x > 0{ } ; Range:
y !R{ }
x-intercept, let y = 0 y-intercept, let x = 0 0 = log x + 2 y = log 0 + 2 –2 = log x 10–2 = x x = 0.01 The x-intercept is 0.01. Since log 0 is undefined, there is no y-intercept. The vertical asymptote is x = 0.
MHR • Advanced Functions 12 Solutions 614 614
b)
Domain:
x !R, x > 3{ } ; Range:
y !R{ }
x-intercept, let y = 0 y-intercept, let x = 0 0 = log(x – 3) y = log(0 – 3) 100 = x – 3 y = log –3 x = 1 + 3 x = 4 The x-intercept is 4. Since log –3 is undefined, there is no y-intercept. The vertical asymptote occurs when x – 3 = 0. The vertical asymptote is x = 3. c)
Domain:
x !R, x > 3{ } ; Range:
y !R{ }
x-intercept, let y = 0 y-intercept, let x = 0 0 = log(x – 3) + 4 y = log(0 – 3) + 4 –4 = log(x – 3) y = log –3 + 4 10–4 = x – 3 x = 0.0001 + 3 x = 3.0001 The x-intercept is 3.0001. Since log –3 is undefined, there is no y-intercept. The vertical asymptote occurs when x – 3 = 0. The vertical asymptote is x = 3.
MHR • Advanced Functions 12 Solutions 615 615
d)
Domain:
x !R, x > "5{ } ; Range:
y !R{ }
x-intercept, let y = 0 y-intercept, let x = 0 0 = log(x + 5) – 1 y = log(0 + 5) – 1 1 = log(x + 5) y = log 5 – 1 101 = x + 5 y &= –0.301 x = 10 – 5 x = 5 The x-intercept is 5. The y-intercept is approximately –0.301. The vertical asymptote occurs when x + 5 = 0. The vertical asymptote is x = –5. Chapter 6 Section 3 Question 4 Page 338 a) y = 2log x; vertical stretch by a factor of 2
b) y = log(2x); horizontal compression by a factor of
1
2
c) y = log
1
2x
!"#
$%&
; horizontal stretch by a factor of 2
d) y =
1
2log x; vertical compression by a factor of
1
2
Chapter 6 Section 3 Question 5 Page 339
a) vertical compression by a factor of
1
2
b) horizontal compression by a factor of
1
5 and a reflection in the y-axis
c) horizontal stretch by a factor of 2 and a reflection in the y-axis d) vertical stretch by a factor of 5 and a reflection in the x-axis
MHR • Advanced Functions 12 Solutions 616 616
Chapter 6 Section 3 Question 6 Page 339 a) b)
c) d)
Chapter 6 Section 3 Question 7 Page 339 a) b)
MHR • Advanced Functions 12 Solutions 617 617
Chapter 6 Section 3 Question 8 Page 339 For y = –log(x – 4) For y = log(–x) – 3 a) Domain:
x !R, x > 4{ } a) Domain:
x !R, x < 0{ }
b) Range:
y !R{ } b) Range:
y !R{ }
c) x = 4 c) x = 0 Chapter 6 Section 3 Question 9 Page 339 a)
b) Range:
y !R, 2 " y " 6{ }
MHR • Advanced Functions 12 Solutions 618 618
Chapter 6 Section 3 Question 10 Page 339 Answers may vary. A sample solution is shown. The domain and range remain the same under a vertical reflection. y = log x and y = –log x y = log(x – 2) and y = –log(x – 2)
Domain:
x !R, x > 0{ } ; Range:
y !R{ } Domain:
x !R, x > 2{ } ; Range:
y !R{ }
y = 3 log x and y = –3 log x y = log 4x and y = –log 4x
Domain:
x !R, x > 0{ } ; Range:
y !R{ } Domain:
x !R, x > 0{ } ; Range:
y !R{ }
y = log x – 2 and y = –log x – 2
Domain:
x !R, x > 0{ } ; Range:
y !R{ }
MHR • Advanced Functions 12 Solutions 619 619
Chapter 6 Section 3 Question 11 Page 339 Answers may vary. A sample solution is shown. No, the same thing does not happen under a horizontal reflection. The domains are different, but the ranges are the same. y = log x and y = log(–x)
y = log x; Domain:
x !R, x > 0{ } ; Range:
y !R{ }
y = log(–x); Domain:
x !R, x < 0{ } ; Range:
y !R{ } y = 2 log x and y = 2 log(–x)
y = 2 log x; Domain:
x !R, x > 0{ } ; Range:
y !R{ }
y = 2 log(–x); Domain:
x !R, x < 0{ } ; Range:
y !R{ } y = log x – 2 and y = log(–x) – 2
y = log x – 2; Domain:
x !R, x > 0{ } ; Range:
y !R{ }
y = log(–x) – 2; Domain:
x !R, x < 0{ } ; Range:
y !R{ }
MHR • Advanced Functions 12 Solutions 620 620
Chapter 6 Section 3 Question 12 Page 340 Answers may vary. A sample solution is shown. The domains are different, but the ranges are the same. y = log x and y = –log(–x)
y = log x; Domain:
x !R, x > 0{ } ; Range:
y !R{ }
y = –log(–x); Domain:
x !R, x < 0{ } ; Range:
y !R{ } y = 2 log(x) – 1 and y = –2 log(–x) – 1
y = 2 log(x) – 1; Domain:
x !R, x > 0{ } ; Range:
y !R{ }
y = –2 log(–x) – 1; Domain:
x !R, x < 0{ } ; Range:
y !R{ }
MHR • Advanced Functions 12 Solutions 621 621
Chapter 6 Section 3 Questions 13 and 14 Page 340 a) b)
vertical stretch by a factor of 2 reflection in the y-axis
horizontal compression by a factor of
1
3
translation 4 units to the left. translation 2 units down c) d)
vertical stretch by a factor of 5 vertical stretch by a factor of 2
horizontal compression by a factor of
1
2 reflection in the x-axis
translation 2 units to the right horizontal stretch by a factor of 2 translation 6 units down translation 6 units to the left translation 3 units down
MHR • Advanced Functions 12 Solutions 622 622
Chapter 6 Section 3 Question 15 Page 340 a)
b)
Vo = 6 Vo &= 6.3
For an input voltage of 10 V, the output is 6 V. For an input voltage of 20 V, the output is approximately 6.3 V.
c)
Vi = 1020 The input voltage is 1020 V when the output is 25 V. d) The domain is the input voltage, {Vi ∈ R, Vi > 0}; the range is the output voltage, Vo, which
can be any real number: {Vo ∈ R}. Chapter 6 Section 3 Question 16 Page 340 Solutions to Achievement Check questions are provided in the Teacher’s Resource.
MHR • Advanced Functions 12 Solutions 623 623
Chapter 6 Section 3 Question 17 Page 340 a) b)
c)
d) i) Domain:
x !R{ }
ii) Range:
y !R, y > 0{ } iii) y = 0 e) y = 102x + 8 Chapter 6 Section 3 Question 18 Page 340 a) b)
MHR • Advanced Functions 12 Solutions 624 624
Chapter 6 Section 3 Question 19 Page 340 a) Domain:
t !R, 2n" < t < (2n +1)" , n !Z{ } ; Range:
V
o!R, V
o" 5{ }
Domain: The range of y = sin t is
y !R, "1# y #1{ } , but for y = log x,
x !R, x > 0{ } , so sin t > 0. t > 0 or t > 2π or t > 4π or t > 2nπ, n ∈ Z For 0 < sin t ≤ 1, look at the graph y = sin t.
We can see from the graph that 0 < sin t ≤ 1 when 0 < t ≤ π or –2π < t ≤ –π or 2nπ < t ≤ (2n + 1)π, n ∈ Z. The range is found by substituting the largest value for sin t = 1, Vo = 5. b)
c) Answers may vary. A sample solution is shown. Because the pattern is a beat, like a pulse. Chapter 6 Section 3 Question 20 Page 340
x = 64
MHR • Advanced Functions 12 Solutions 625 625
Chapter 6 Section 4 Power Law of Logarithms Chapter 6 Section 4 Question 1 Page 347 a)
log2(24 )3
= log2
212
= 12
b) log
464 = 3
c)
log 102( )!4
= log10!8
= !8
d)
1
2log(10!1) =
1
2(!1)
= !1
2
Chapter 6 Section 4 Question 2 Page 347
a)
log2(23)
1
2 =1
2log
223
=1
2(3)
=3
2
b)
log3(35)
1
2 =1
2(5)
=5
2
c)
log3
(34 )
1
3!
"#
$
%&
6
= log338
= 8
d)
log4
(42 )
1
5!
"#
$
%&
15
= log4
46
= 6
Chapter 6 Section 4 Question 3 Page 347
a)
log10 = log 4t
1= t log 4
1
log 4= t
t &= 1.66
b)
log5t= log 250
t log5 = log 250
t =log 250
log5
t &= 3.43
c)
log 2 = log1.08t
log 2 = t log1.08
log 2
log1.08= t
t &= 9.01
d)
500
100= 1.06t
log5 = log1.06t
log5 = t log1.06
log5
log1.06= t
t &= 27.62
MHR • Advanced Functions 12 Solutions 626 626
Chapter 6 Section 4 Question 4 Page 347 a) i)
A(0) = 500(1.07)0
= 500
The value of the investment is initially $500. ii)
A(2) = 500(1.07)2
= 572.45
The value of the investment after 2 years is $572.45. iii)
A(4) = 500(1.07)4
&= 655.40
The value of the investment after 4 years is approximately $655.40 b) i)
2 = 1.07t
log 2 = t log1.07
log 2
log1.07= t
t &= 10.2
It will take approximately 10.2 years for the investment to double in value. ii)
3= 1.07t
log3= t log1.07
log3
log1.07= t
t &= 16.2
It will take approximately 16.2 years for the investment to triple in value. Chapter 6 Section 4 Question 5 Page 347
a)
log 23
log3&= 2.854 b)
log 20
log6&= 1.672
c)
log 2
log7&= 0.356 d)
!log 4
log12&= !0.558
e)
log30
log1
2
&= !4.907 f)
log8
log3
4
&= !7.228
Chapter 6 Section 4 Question 6 Page 347
a) log
58 b)
log
917 c)
log2
3
1
2 d)
log
x!1( )x +1( )
MHR • Advanced Functions 12 Solutions 627 627
Chapter 6 Section 4 Question 7 Page 347 a) b)
Chapter 6 Section 4 Question 8 Page 347
Chapter 6 Section 4 Question 9 Page 347
a)
2 = x log3
2
log3= x
x &= 4.192
b)
100
10= x log1000
10 = x(3)
x &= 3.333
c)
4 = xlog15
log3
!"#
$%&
x =4
log15
log3
!"#
$%&
x &= 1.623
d)
12
2= x
log3
log5
!"#
$%&
x =6
log3
log5
!"#
$%&
x &= 8.790
Chapter 6 Section 4 Question 10 Page 347 a)
A(0) = 400(1.09)0
= 400
The initial value of the investment is $400, when t = 0. b)
2 = 1.09t
log 2 = t log1.09
t =log 2
log1.09
t &= 8
The investment will double in approximately 8 years.
MHR • Advanced Functions 12 Solutions 628 628
Chapter 6 Section 4 Question 11 Page 347
MHR • Advanced Functions 12 Solutions 629 629
log(mx) = m log x
log(mx) = log xm
mx = xm
xm! mx = 0
x(xm!1
! m) = 0
so x = 0, but x ≠ 0 since log 0 is undefined or xm – 1 = m, true for m = 1 log(mx) = m log x only when m = 1.
MHR • Advanced Functions 12 Solutions 630 630
Chapter 6 Section 4 Question 12 Page 347
log(xn ) = (log x)n only when n = 1 and/or when x = 1.
Chapter 6 Section 4 Question 13 Page 348
a)
log285
= log2
23( )5
= log2
215
= 15
b)
log2
85= 5log
28
= 5log2
23
= 5(3)
= 15
MHR • Advanced Functions 12 Solutions 631 631
c) Answers may vary. A sample solution is shown. I prefer the power law method since it involves simpler calculations. Chapter 6 Section 4 Question 14 Page 348 Answers may vary. A sample solution is shown.
log3
27
1
3 = log3
33( )1
3
= log331
= 1
log3
27
1
3 =1
3log
333( )
=1
3(3)
= 1
Chapter 6 Section 4 Question 15 Page 348
a)
t = log0.5
10
1000
!"#
$%&
t = log0.5
0.01
t &= 6.6
; t &= 6.6 h
The docking procedures should begin approximately 6.6 h after the reverse thrusters are first fired.
b) Domain:
d !R, 0 < d "1000{ } ; Range:
t !R, t " 0{ }
The domain represents the distance from the space station in kilometres. The range represents the time, in hours, required to reach a distance from the space station. Chapter 6 Section 4 Question 16 Page 348 Solutions to Achievement Check questions are provided in the Teacher’s Resource. Chapter 6 Section 4 Question 17 Page 348
Let w = logb
x
x = bw write in exponential form
logk
x = logk
bw take log
kof both sides
logk
x = w logk
b apply the power law of logarithms
w =log
kx
logk
b
logb
x =log
kx
logk
b express in terms of x
MHR • Advanced Functions 12 Solutions 632 632
Chapter 6 Section 4 Question 18 Page 348
a)
log2
9
log2
3 b)
log2
25
log210
Chapter 6 Section 4 Question 19 Page 348
log
2210( )
64
= 640
Chapter 6 Section 4 Question 20 Page 348 a)
A(t) = P(1+ r)i
�
r =0.035
4
= 0.008 75 interest rate each quarter
i = 4t number of pay periods each year
A(t) = P 1.008 75( )4t
b) i)
2 = (1.008 75)4t
log 2 = 4t log1.008 75
4t =log 2
log1.008 75
t =log 2
4 log1.008 75
t &= 19.9
It will take approximately 19.9 years for the investment to double. ii)
3= (1.008 75)4t
log3= 4t log1.008 75
4t =log3
log1.008 75
t =log3
4 log1.008 75
t &= 31.5
It will take approximately 31.5 years for the investment to triple.
MHR • Advanced Functions 12 Solutions 633 633
Chapter 6 Section 4 Question 21 Page 348
f (x) =1
1! x
f (a) =1
1! a substitute a in for x
1
1! a= 2 given: f (a) = 2
1
2= 1! a
1
4= 1! a square both sides
a =3
4
f (1! a) = f1
4
"#$
%&'
=1
1!1
4
=1
3
4
=1
3
2
=2
3
MHR • Advanced Functions 12 Solutions 634 634
Chapter 6 Section 4 Question 22 Page 348
Graph y = x x x and view the table for when y is an integer.
So e = 256 and f = 128. Check
�
f = 256 256 256
= 256 256!16
= 256 4096
= 256!64
= 16 384
= 128
e+ f = 256 +128
= 384
MHR • Advanced Functions 12 Solutions 635 635
Chapter 6 Section 5 Making Connections: Logarithmic Scales in the Physical Sciences
Chapter 6 Section 5 Question 1 Page 353 a)
pH = ! log H+"
#$%
= ! log(0.01)
= !(!2)
= 2
b)
pH = ! log H+"
#$%
= ! log(0.000 034)
&= !(!4.5)
&= 4.5
c)
pH = ! log H+"
#$%
= ! log(10!9 )
= !(!9)
= 9
d)
pH = ! log H+"
#$%
= ! log(1.5&10!10 )
&= !(!9.8)
&= 9.8
Chapter 6 Section 5 Question 2 Page 353 a) 10!pH
= H+"# $%
H+= 10
!11
b)
10!pH
= H+"
#$%
10!3= H
+"#
$%
H+= 0.001
c)
10!pH
= H+"
#$%
10!8.5
= H+"
#$%
H+
&= 3.2 &10!9
d)
10!pH
= H+"
#$%
10!4.4
= H+"
#$%
H+
&= 0.000 039 8
Chapter 6 Section 5 Question 3 Page 353 a) Answers may vary. A sample solution is shown.
The pH scale varies over several powers of 10. b) Answers may vary. A sample solution is shown. To ensure that most pH measurements are positive. Chapter 6 Section 5 Question 4 Page 353 a)
pH = ! log H+"
#$%
= ! log(0.000 01)
= !(!5)
= 5
b)
pH = ! log H+"
#$%
= ! log(2.5&10!11)
&= !(!10.6)
&= 10.6
MHR • Advanced Functions 12 Solutions 636 636
Chapter 6 Section 5 Question 5 Page 353 a) Answers may vary. A sample solution is shown.
b)
The pH level of acetic acid is approximately 2.7. c)
pH = ! log H+"
#$%
= ! log(1.9 &10!3)
&= !(!2.7)
&= 2.7
MHR • Advanced Functions 12 Solutions 637 637
Chapter 6 Section 5 Question 6 Page 354
a)
!2" !
1= 10 log
I2
I1
#
$%&
'(
80 ! 60 = 10 logI
2
I1
"
#$%
&'
20 = 10 logI
2
I1
"
#$%
&'
2 = logI
2
I1
"
#$%
&'
102=
I2
I1
100 =I
2
I1
The ratio of intensities is 100. A shout is 100 times more intense than a conversation.
b)
!2" !
1= 10 log
I2
I1
#
$%&
'(
80 ! 30 = 10 logI
2
I1
"
#$%
&'
50 = 10 logI
2
I1
"
#$%
&'
5 = logI
2
I1
"
#$%
&'
105=
I2
I1
100 000 =I
2
I1
The ratio of intensities is 100 000. A shout is 100 000 times more intense than a whisper.
MHR • Advanced Functions 12 Solutions 638 638
Chapter 6 Section 5 Question 7 Page 354
!2" !
1= 10 log
I2
I1
#
$%&
'(
150 "110 = 10 logI
2
I1
#
$%&
'(
40 = 10 logI
2
I1
#
$%&
'(
4 = logI
2
I1
#
$%&
'(
104=
I2
I1
10 000 =I
2
I1
A rock concert speaker sounds 10 000 times as intense as a loud car stereo. Chapter 6 Section 5 Question 8 Page 354
!2" !
1= 10 log
I2
I1
#
$%&
'(
60 " !1= 10 log
30 000
1
#$%
&'(
60 " !1
&= 45
!1
&= 15
The decibel level of a pin drop is approximately 15 dB. Chapter 6 Section 5 Question 9 Page 354
M = logI
I0
!
"#$
%&
M = log 10 000( )= 4
The earthquake in North Bay, Ontario, measured 4 on the Richter scale.
MHR • Advanced Functions 12 Solutions 639 639
Chapter 6 Section 5 Question 10 Page 354
a)
M = logI
I0
!
"#$
%&
2.3= logI
I0
!
"#$
%&
102.3=
I
I0
I
I0
&= 199.53
The earthquake in Welland, Ontario, was approximately 199.53 times as intense as a standard low-level earthquake.
b)
M = logI
I0
!
"#$
%&
1.1= logI
I0
!
"#$
%&
101.1=
I
I0
I
I0
&= 12.59
199.53
12.59&= 15.85
The Welland earthquake was approximately 15.85 times as intense as the St. Catharines earthquake.
MHR • Advanced Functions 12 Solutions 640 640
Chapter 6 Section 5 Question 11 Page 354
M = logI
I0
!
"#$
%&
8.9 = logI
I0
!
"#$
%&
108.9=
I
I0
I
I0
&= 794 328 235
The most intense earthquakes are approximately 794 328 235 times as intense as standard, low-level earthquakes. Chapter 6 Section 5 Question 12 Page 354
a)
m2! m
1= log
b1
b2
"
#$%
&'
0.12 ! (!1.5) = logb
1
b2
"
#$%
&'
1.62 = logb
1
b2
"
#$%
&'
101.62=
b1
b2
b1
b2
&= 41.69
Sirius appears to be approximately 41.69 times brighter than Betelgeuse.
b)
m2! m
1= log
b1
b2
"
#$%
&'
(!1.5)! m1= log 1.3(1010( )
m1= !1.5! log 1.3(1010( )
m1
&= !11.61
The apparent magnitude of the Sun is approximately –11.61.
MHR • Advanced Functions 12 Solutions 641 641
Chapter 6 Section 5 Question 13 Page 354 Solutions to Achievement Check questions are provided in the Teacher’s Resource. Chapter 6 Section 5 Question 14 Page 354 Answers may vary. A sample solution is shown. I think that astronomers would be more interested in the absolute magnitude since it takes distance into consideration. Chapter 6 Section 5 Question 15 Page 355 a) i)
M2! M
1= log
b1
b2
"
#$%
&'
0.4 ! (!1.8) = logb
1
b2
"
#$%
&'
2.2 = logb
1
b2
"
#$%
&'
b1
b2
= 102.2
b1
b2
&= 158.5
The brightest star is approximately 158.5 times brighter than the second brightest, in absolute terms.
ii)
M2! M
1= log
b1
b2
"
#$%
&'
3.2 ! (!1.8) = logb
1
b2
"
#$%
&'
5 = logb
1
b2
"
#$%
&'
b1
b2
= 105
b1
b2
= 100 000
The brightest star is 100 000 times brighter than the least brightest, in absolute terms. b) The closest star is Biffidus-V, next is Cheryl-XI, and farthest away is Roccolus-III.
MHR • Advanced Functions 12 Solutions 642 642
Chapter 6 Section 5 Question 16 Page 355 Answers may vary. A sample solution is shown. There have been several major earthquakes around the Pacific Ocean. The largest known earthquakes are: • Chile, 1960 – magnitude 9.5 • Indonesia, 2004 – magnitude 9.3 (Boxing Day tsunami) • Russia, 1737 – estimated magnitude 9.3 • Alaska, 1964 – magnitude 9.2 • Russia, 1952 –magnitude 9.0 There have been minor earthquakes (tremors) where I live, but nothing above 4.0. Chapter 6 Section 5 Question 17 Page 355 Answers may vary. A sample solution is shown. There is a logarithmic equation relating the apparent magnitude (how bright we see a star), m, and the star’s distance from earth, d, to the absolute magnitude (how bright the star would appear from 10 parsecs, or about 32.6 light years, away), M:
M &= m ! 5 logd
3.26
"#$
%&'!1
(
)*
+
,-
Chapter 6 Section 5 Question 18 Page 355 Answers may vary. A sample solution is shown. The eye sees luminosity on a logarithmic scale, so people can look at one object 1 000 000 times brighter than another, but they do not appear that different.
MHR • Advanced Functions 12 Solutions 643 643
Chapter 6 Section 5 Question 19 Page 355 B;
logsin x
cos x =1
2
log cos x( )log sin x( )
=1
2
2 log(cos x) = log(sin x)
log cos2x( ) = log sin x( )
cos2x = sin x
cos2x ! sin x = 0
(1! sin2x)! sin x = 0
! sin2x ! sin x +1= 0
sin x =1± (!1)2
! 4(!1)(1)
2(!1)
sin x =1± 5
!2
sin x =!1+ 5
2, since sin x cannot be less than !1
MHR • Advanced Functions 12 Solutions 644 644
Chapter 6 Section 5 Question 20 Page 355
The base of the triangle
= s2+ s
2
= 2s2
= 2s
Area of triangle
=1
2bh
=1
22s( )h
Area of square = s2
Area square = Area triangle
s2=
1
22s( )h
2s2= 2sh
h =2s
2
2s
!2
2
divide the s 's
multiply the numerator and denominator by 2 to get the square root out of the denominator
h =2 2s
2
h = 2s
s
MHR • Advanced Functions 12 Solutions 645 645
Chapter 6 Section 5 Question 21 Page 355 B;
a + b = 3
(a + b)2= 32
a2+ 2ab+ b
2= 9
a2+ b
2( ) + 2ab = 9
7 + 2ab = 9 substitute a2+ b
2= 7
2ab = 2
ab = 1
a =1
b
a2+ b
2( )2
= 72
a4+ 2a
2b
2+ b
4= 49
a4+ 2
1
b
!"#
$%&
2
b2+ b
4= 49
a4+ 2
b2
b2
!
"#
$
%& + b
4= 49
a4+ 2 + b
4= 49
a4+ b
4= 47
Chapter 6 Section 5 Question 22 Page 355 D;
logb(a
2 ) = 3
b3= a
2
b3( )
2
3 = a2( )
2
3
b2= a
4
3
loga
b2= log
aa
4
3
=4
3
MHR • Advanced Functions 12 Solutions 646 646
Chapter 6 Review Chapter 6 Review Question 1 Page 356 a)
i) Domain:
x !R{ } ; Range:
y !R, y > 0{ }
ii) Let y = 0.
�
0 = 3x
log 0 = x log 3 log 0 is undefined
No x-intercept. iii) Let x = 0.
y = 30
y = 1
The y-intercept is 1. iv) positive for all intervals v) increasing for all intervals vi) The horizontal asymptote is y = 0. b), c)
Domain:
x !R, x > 0{ } ; Range:
y !R{ } ; x-intercept is 1; no y-intercept;
f –1 is positive for x > 1 and negative for 0 < x < 1; f –1 is increasing for all intervals; vertical asymptote is x = 0
MHR • Advanced Functions 12 Solutions 647 647
Chapter 6 Review Question 2 Page 356 a)
i) Domain:
x !R{ } ; Range:
y !R, y > 0{ }
ii) Let y = 0.
�
0 =1
4
x
log 0 = x log1
4log 0 is undefined
No x-intercept. iii) Let x = 0.
y =1
4
0
y = 1
The y-intercept is 1. iv) positive for all intervals v) decreasing for all intervals vi) The horizontal asymptote is y = 0. b), c)
Domain:
x !R, x > 0{ } ; Range:
y !R{ } ; x-intercept is 1; no y-intercept;
g–1 is positive for 0 < x < 1 and negative for x > 1; g–1 is decreasing for all intervals; vertical asymptote is x = 0
MHR • Advanced Functions 12 Solutions 648 648
Chapter 6 Review Question 3 Page 356 a)
log
464 = 3 b)
log
328 = x c)
log
6y = 3 d)
log
2512 = 9
Chapter 6 Review Question 4 Page 356 a) 27 = 128 b) bx = n c) 35 = 243 d) b19 = 4 Chapter 6 Review Question 5 Page 356
log
250 &= 5.6
log
232 = 5 and
log
264 = 6 , so
log
250 must be between 5 and 6. Find the approximate exponent
to which 2 must be raised to give 50.
Estimate Check Analysis 5.5
25.5
&= 45.3 Too low. Try a higher value.
5.7 2
5.7&= 52.0 High, but close.
5.6 2
5.6&= 48.5 This is a good estimate.
log
250 &= 5.6
Chapter 6 Review Question 6 Page 356 a)
log216 = log
224
= 4
b)
log381= log
334
= 4
c)
log4
1
16
!"#
$%&= log
4
1
42
!"#
$%&
= log4
4'2
= '2
d)
log0.000 001= log10!6
= !6
MHR • Advanced Functions 12 Solutions 649 649
Chapter 6 Review Question 7 Page 356
log
xx = 1, x > 0, x ! 1
logx
x = y
xy= x
xy! x = 0
x(xy!1
!1) = 0
x = 0, not a solution, since log0
0 is not defined
xy!1
= 1
( y !1) log x = log1
( y !1) log x = 0
y !1= 0, y = 1
Therefore logx
x = 1 for x > 0, x " 1.
Chapter 6 Review Question 8 Page 356 a)
b) i) Domain:
x !R, x > 5{ }
ii) Range:
y !R{ }
iii) The vertical asymptote occurs when x – 5 = 0. The vertical asymptote is x = 5.
MHR • Advanced Functions 12 Solutions 650 650
Chapter 6 Review Question 9 Page 356 a)
b) i) Domain:
x !R, x < 0{ }
ii) Range:
y !R{ }
iii) The vertical asymptote occurs when –x = 0. The vertical asymptote is x = 0. Chapter 6 Review Question 10 Page 356
Chapter 6 Review Question 11 Page 356 a) Answers may vary. A sample solution is shown. Horizontal stretch by a factor of 2. Translation 8 units left. Translation 3 units down. Reflection in the x-axis.
y =
! log1
2x + 4
"#$
%&'! 3 or
y = ! log1
2(x + 8)
"
#$
%
&' ! 3
b) Answers may vary. A sample solution is shown.
MHR • Advanced Functions 12 Solutions 651 651
c) i) Domain:
x !R, x > "8{ }
ii) Range:
y !R{ }
iii) The vertical asymptote occurs when
1
2x + 4 = 0 . The vertical asymptote is x = –8.
Chapter 6 Review Question 12 Page 356 a)
log2
323= 3log
225( )
= 3(5)
= 15
b)
log1000!2= !2 log103
= !2 3( )= !6
c)
log0.001!1= ! log10!3
= ! !3( )= 3
d)
log1
4
1
16
!"#
$%&
4
= 4 log1
4
1
4
!"#
$%&
2
= 4(2)
= 8
Chapter 6 Review Question 13 Page 356
a)
x =log17
log3
x &= 2.579
b)
x =log0.35
log 2
x &= !1.515
c)
x log 4 = log10
x =1
log 4
x &= 1.661
d)
80
100=
1
2
!"#
$%&
x
0.8 =1
2
!"#
$%&
x
log0.8 = x log1
2
x =log0.8
log1
2
x &= 0.322
MHR • Advanced Functions 12 Solutions 652 652
Chapter 6 Review Question 14 Page 357
0.75 = (0.8)x
log0.75 = x log0.8
x =log0.75
log0.8
x &= 1.3
The glass should be approximately 1.3 mm thick. Chapter 6 Review Question 15 Page 357 a) Draw a graph of the inverse function y = 3x, then reflect it in the line y = x.
b)
MHR • Advanced Functions 12 Solutions 653 653
Chapter 6 Review Question 16 Page 357 Chemical A Chemical B
pH = ! log H+"
#$%
10!pH= H
+"#
$%
10!6= H
+
pH = ! log H+"
#$%
10!pH= H
+"#
$%
10!8= H
+
There is less hydronium ion concentration in Chemical B. Chapter 6 Review Question 17 Page 357 great light
M = logI
I0
!
"#$
%&
8 = logI
I0
!
"#$
%&
108=
I
I0
M = logI
I0
!
"#$
%&
4 = logI
I0
!
"#$
%&
104=
I
I0
No, a great earthquake is not twice as intense as a light earthquake, it is 10 000 times as intense as a light earthquake.
Chapter 6 Review Question 18 Page 357
a)
n = 1!7 logT
3
n = 1!7 log
1
83
n &= 3.1
The shade number that the welder should use is 3.
b)
n = 1!7 logT
3
14 = 1!7 logT
3
13= !7 logT
3
!39
7= logT
10!
39
7 = T
T &= 2.683"10!6
These glasses transmit a fraction of approximately
1
372 717 of light.
MHR • Advanced Functions 12 Solutions 654 654
c)
n = 1!7 logT
3
2 = 1!7 logT
3
1= !7 logT
3
!3
7= logT
10!
3
7 = T
T &= 0.373
0.373
2.683!10"6
&= 138 950
The #2 welding glasses transmit approximately 138 950 times as much visible light as the #14 welding glasses.
Chapter 6 Review Question 19 Page 357 a)
b) Domain:
T !R, 0 < T "1{ } ; Range:
n !R, n "1{ }
Since T is the fraction of light, it can be between 0 and including 1. T cannot be 0 since log T is undefined. From the graph we can see that see that n is greater than or equal to 1.
Chapter Problem Wrap-Up Chapter Problem Wrap-Up solutions are provided in the 12 Teacher’s Resource.
MHR • Advanced Functions 12 Solutions 655 655
Chapter 6 Practice Test Chapter 6 Practice Test Question 1 Page 358 The correct solution is D.
Chapter 6 Practice Test Question 2 Page 358 The correct solution is A.
log5
1
125
!"#
$%&= log
5
1
53
!"#
$%&
= log55'3
= '3
Chapter 6 Practice Test Question 3 Page 358 The correct solution is B.
log416!3
= !3log4
42
= !3 2( )= !6
Chapter 6 Practice Test Question 4 Page 358 The correct solution is C. Graph is y = log x translated 3 units to the left and 5 units down. Chapter 6 Practice Test Question 5 Page 359
a)
log6
27 =log 27
log6
&= 1.839
b)
log12
63= 3log
126
=3log6
log12
&= 2.163
MHR • Advanced Functions 12 Solutions 656 656
Chapter 6 Practice Test Question 6 Page 359
a) g(x) =
1
2log !(x + 3)"# $% + 6
Compress vertically by a factor of
1
2, translate 3 units left, translate 6 units up, and reflect in
the y-axis. b)
c) i) Domain:
x !R, x < "3{ }
ii) Range:
y !R{ } iii) The vertical asymptote occurs when –x – 3 = 0. The vertical asymptote is x = –3. d)
The inverse is an exponential function. Chapter 6 Practice Test Question 7 Page 359
MHR • Advanced Functions 12 Solutions 657 657
Chapter 6 Practice Test Question 8 Page 359
A(t) = P(1+ i)t
2 = (1.065)t
log 2 = t log1.065
t =log 2
log1.065
t &= 11
It will take approximately 11 years for the amount to double. Chapter 6 Practice Test Question 9 Page 359 a)
pH = ! log H+"
#$%
= ! log(3.2 &10!12 )
&= !(!11.5)
&= 11.5
The pH level of ammonia is approximately 11.5. b)
pH = ! log H+"
#$%
10!pH= H
+"#
$%
10!2= H
+"#
$% 10!3.5
= H+"
#$%
H+= 0.01 H +
&= 0.000 32
The range of the hydronium ion concentration in vinegar is approximately 0.000 32 < V < 0.01. c) Vinegar is acidic since its pH is less than 7. Ammonia is alkaline since its pH is greater than 7. Chapter 6 Practice Test Question 10 Page 359 a)
MHR • Advanced Functions 12 Solutions 658 658
b)
The atmospheric pressure is approximately 54.3 kPa at an altitude of 5 km. c)
The mountain climber is approximately 1.4 km above sea level. d) b)
P = 101.3(1.133)!d
= 101.3(1.133)!5
&= 54.3
c)
P = 101.3(1.133)!d
85 = 101.3(1.133)!d
85
101.3= 1.133!d
log85
101.3
"#$
%&'= !d log1.133
d = !log
85
101.3
"#$
%&'
log1.133
d &= 1.4
MHR • Advanced Functions 12 Solutions 659 659
Chapter 6 Practice Test Question 11 Page 359
a)
!2" !
1= 10 log
I2
I1
#
$%&
'(
!2" 85 = 10 log 315( )!
2= 10 log 315( ) + 85
!2
&= 110
The passing subway train has a level of approximately 110 dB.
b)
!2" !
1= 10 log
I2
I1
#
$%&
'(
118" 85 = 10 logI
2
I1
#
$%&
'(
33= 10 logI
2
I1
#
$%&
'(
3.3= logI
2
I1
#
$%&
'(
103.3=
I2
I1
I2
I1
&= 1995
The sound of the power saw is approximately 1995 times as intense as the sound of normal city traffic.
Chapter 6 Practice Test Question 12 Page 359 Examples may vary. A sample solution is shown. Graphing y = 2log x and y = log x2 results in the same graph when x > 0. The results are the same for y = 6log x and y = log x6 and y = 2.3log x and y = log x2.3.
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