MHR • Advanced Functions 12 Solutions 660

Chapter 7 Tools and Strategies for Solving Exponential and Logarithmic Equations 

Chapter 7 Prerequisite Skills 

Chapter 7 Prerequisite Skills Question 1 Page 362 

Using the product law: 

a) x 2 + 3 + 4 = x 9 b) 6a 5 b 4 c) x 4 y 5 d) 6r 2 s 

e) 5q 2 r f) v 3 w 4 g) 4ab 4 


Using the quotient law: 

a) k 6 – 2 = k 4 b) –6n c) 6x 3 y d) 2ab 2 

Using the product law then the quotient law: 

e) 

5q 2 r 

10r 2 = q2 

2r 

f) 

v 3 w 4 

vw 2 = v2 w 2 g) 

4ab 4 

4a 

= b4 


Using the power law: 

a) w 2 × 4 = w 8 b) 4u 2 v 6 c) a 6 b 3 d) x 6 y 6 

e) 8w 6 f) a 2 b 8 g) rs 6 


a) 

4a 3 

2ab = 2a2 , product and quotient laws 

b 

b) (–27k 3 m 9 )(4k 6 m 4 ) = –108k 9 m 13 , product and power laws 

c) 

d) 

(2x 3 y)(64x 2 y 4 ) 

= 32y 5 , product, quotient, and power laws 

4x 5 

(a 6 )a 3 b 6 

= a 4 b , product, quotient, and power laws 

a 5 b 5 

MHR • Advanced Functions 12 Solutions 660


a) (x + 6)(x ! 4) = 0 

x = !6 or x = 4 

b) 2m 2 ! 9m +10 = 0 c) (2a !1)(2a + 7) = 0 

(2m ! 5)(m ! 2) = 0 

m = 5 a = 1 

2 or m = 2 2 or a = ! 7 2 

d) q 2 ! q ! 20 = 0 

(q ! 5)(q + 4) = 0 

q = 5 or q = !4 

e) 9b 2 !1 = 0 

(3b !1)(3b +1) = 0 

b = 1 3 or b = ! 1 3 

f) (2y +1)(4y +1) = 0 

y = ! 1 2 or y = ! 1 4 

g) x 2 ! x ! 2 = 0 

(x ! 2)(x +1) = 0 

x = 2 or x = !1 

h) 2r 2 + 5r ! 3 = 0 i) (q + 4)(q +1) = 0 

(2r !1)(r + 3) = 0 

q = !4 or q = !1 

r = 1 2 or r = !3 


a) y = !6 ± 62 ! 4(1)(!5) 

2(1) 

y = 

y = 

y = 

!6 ± 56 

2 

!6 ± 14 " 4 

2 

!6 ± 2 14 

2 

y = !3± 14 

b) 4q 2 ! 2q !10 = 0 

q = !(!2) ± (!2)2 ! 4(4)(!10) 

2(4) 

q = 2 ± 164 

8 

q = 2 ± 41" 4 

8 

q = 2 ± 2 41 

8 

q = 

1± 41 

4 

MHR • Advanced Functions 12 Solutions 661

c) x 2 + x !1 = 0 

x = !1± 12 ! 4(1)(–1) 

2(1) 

x = !1± 5 

2 

e) x 2 ! x ! 7 = 0 

x = !(!1) ± (!1)2 ! 4(1)(!7) 

2(1) 

x = 

1± 29 

2 

d) a = !2 ± 22 ! 4(2)(!10) 

2(2) 

a = 

a = 

a = 

a = 

!2 ± 84 

4 

!2 ± 21" 4 

4 

!2 ± 2 21 

4 

!1± 21 

2 

f) r 2 ! 5r + 3 = 0 

r = !(!5) ± (!5)2 ! 4(1)(3) 

2(1) 

r = 5 ± 13 

2 

g) m = !6 ± 62 ! 4(2)(3) 

2(2) 

m = 

!6 ± 12 

4 

m = !6 ± 4 " 3 

4 

m = !6 ± 2 3 

4 

m = !3± 3 

2 

h) 3a 2 ! 3a ! 4 = 0 

a = !(!3) ± (!3)2 ! 4(3)(!4) 

2(3) 

a = 

3± 57 

6 



a) 4 ! 2 = 2 2 b) 2 4 ! 5 = 2 ! 2 5 

= 4 5 

c) 9 ! 2 = 3 2 

d) 3 9 ! 2 = 3! 3 2 

= 9 2 

e) 25! 3 = 5 3 

f) 

4 + 16 ! 3 

2 

= 4 + 4 3 

2 

= 2 + 2 3 

( ) 

= 2 3 +1 

g) 

2 ± 4 ! 6 

2 

= 2 ± 2 6 

2 

= 1± 6 


a) log 4 

(4 2 ) 3 = log 4 

4 6 

= 6 

1 

b) log 2 

(2 5 ) 4 

= log 2 

2 

= 5 4 

5 

4 

c) log 3 

(3 3 ) 2 = log 3 

3 6 

= 6 

d) log 5 

(5 2 ) 2 = log 5 

5 4 

= 4 

g) log 4 

(4 2 ) 2 = log 4 

4 4 

= 4 

e) log 2 

(2 5 ) 2 = log 2 

2 10 

= 10 

1 

h) log 2 

4 = log ! 1 $ 

2 

" 

# 

% 

& 

2 2 

= log 2 

2 '2 

= '2 

f) log 3 

3 4 = 4 


a) 

log10 

log3 &= 2.096 b) log 1 5 

log 1 2 

&= 2.322 c) 

log15 

log 2 

&= 3.907 

d) 

g) 

log17 

log 4 &= 2.044 e) log3 

log 2 &= 1.585 f) log8 

&= 1.893 

log3 

log 20 

log 4 &= 2.161 h) log19 

log 2 

&= 4.248 



a) t log5 = log 4 

t = log 4 

log5 

t &= 0.86 

b) 2t log1.2 = log1.6 

t = log1.6 

2log1.2 

t &= 1.29 

c) t log7 = log 2 

t = log 2 

log7 

t &= 0.36 

d) 3t log 1 4 = log 2 

t = log 2 

3log 1 4 

t &= !0.17 

g) t log 2 = log9 

t = log9 

log 2 

t &= 3.17 

e) t log3 = log 2 

t = log 2 

log3 

t &= 0.63 

h) 3t log 2 = log15 

t = log15 

3log 2 

t &= 1.30 

f) t log8 = log3 

t = log3 

log8 

t &= 0.53 


y = b x 

log y = log b x 

log y = x log b 

x = log y 

log b 


Chapter 7 Section 1 

Equivalent Forms of Exponential Equations 

Chapter 7 Section 1 Question 1 Page 368 

a) (2 2 ) 6 = 2 12 b) (2 3 ) 3 = 2 9 

c) 

! 1 $ 

" 

# 

% 

& 

2 3 

2 

= ( 2 ) '3 2 

= 2 '6 log 2 k = log14 

k log 2 = log14 

d) 2 k = 14 

k = log14 

log 2 

log14 

log 2 

14 = 2 


a) (3 3 ) 2 = 3 6 b) (3 –1 ) 4 = 3 –4 

c) 3 k = 10 

log3 k = log10 

k log3 = log10 

k = 1 

log3 

10 = 3 

1 

log 3 

d) 3 k = 1 2 

log3 k = log 1 2 

k log3 = log 1 2 

k = 

log 1 2 

log3 

log 1 2 

1 

2 = 3 log 3 


a) Answers may vary. A sample solution is 4 4 . 

b) Answers may vary. A sample solution is 16 2 . 

c) 



a) 4 3 b) 4 2 

( ) 1 2 

3 

= 4 

3 

c) 4 3 

( ) 1 " 

2 

! $ 4 

# 

7 

2 

% 

' 

& 

3 

4 

3 

= 4 

2 

! 4 

12 

= 4 

8 + 21 

8 

33 

= 4 

8 

21 

8 

! 

1 

$ 

d) 2 

2 

# & 

" % 

8 

! 

1 

$ 

' 23 

# & 

" % 

4 

4 

= 2 4 ' 2 

3 


= 2 

3 + 4 3 

16 

= 2 

3 

! 

1 

$ 

= 4 

2 

# & 

" % 

8 

= 43 

16 

3 


a) 2 4x = ( 2 ) 2 x+3 

2 4x = 2 2x+6 

4x = 2x + 6 

2x = 6 

x = 3 

b) ( 5 ) 2 x!1 = 5 3x 

5 2x!2 = 5 3x 

2x ! 2 = 3x 

!2 = 3x ! 2x 

x = !2 

c) 3 w+1 = ( 3 ) 2 w!1 

3 w+1 = 3 2w!2 

w +1 = 2w ! 2 

1+ 2 = 2w ! w 

w = 3 

d) ( 6 ) 2 3m!1 = 6 2m+5 

6 6m!2 = 6 2m+5 

6m ! 2 = 2m + 5 

6m ! 2m = 5+ 2 

4m = 7 

m = 7 4 

Check: 

a) 


) 

c) 

d) 


a) ( 2 ) 2 3x = ( 2 ) 3 x!3 

b) ( 3 ) 3 x = ( 3 ) 2 2x!3 

c) ( 5 ) 3 2 y!1 = ( 5 ) 2 y+4 

d) ( 2 ) 4 2k !3 = ( 2 ) 5 k +3 

2 6x = 2 3x!9 

3 3x = 3 4x!6 

5 6 y!3 = 5 2 y+8 

2 8k !12 5k +15 

= 2 

6x = 3x ! 9 

3x = 4x ! 6 

6y ! 3 = 2y + 8 

8k !12 = 5k +15 

3x = !9 

4x ! 3x = 6 

4y = 11 

8k ! 5k = 15+12 

Check: 

a) 

x = !3 

x = 6 

y = 11 

4 

3k = 27 

k = 9 


) 

c) 

d) 


a) 10 2x = ( 10 ) 2 2x!5 

10 2x = 10 4x!10 

2x = 4x !10 

4x ! 2x = 10 

2x = 10 

x = 5 

b) log10 2x = log100 2x!5 

2x log10 = (2x ! 5)log100 

2x(1) = (2x ! 5)(2) 

2x = 4x !10 

4x ! 2x = 10 

2x = 10 

x = 5 

c) Answers may vary. A sample solution is shown. 

I like using logarithms because I do not need to find a common base. 



a) Answers may vary. A sample solution is shown. 

n = 0 1 2 

b) Answers may vary. A sample solution is shown. 

n = 0 1 2 


Both models are the same, so they illustrate that (2 3 ) n = 8 n . 

d) L.S. = (2 3 ) n 

= 2 3n 

R.S. = 8 n 

= (2 3 ) n 

= 2 3n 

Since L.S. = R.S., f(n) = g(n), n ∈ R. 



f(n) = (3 2 ) n and g(n) = 9 n 


n = 0 1 2 

c) L.S. = ( 3 ) 2 n 

= 9 n 

R.S. = 9 n 

Since L.S. = R.S., f(n) = g(n), n ∈ R. 



! 

a) 2 3 

" 

# 

( ) 1 2 

$ 

% 

& 

x 

= 2 x'2 

3 

2 

2 x = 2 x'2 

3 

2 x = x ' 2 

3 

2 x ' x = '2 

Check: 

1 

2 x = '2 

x = '4 

b) 3 2 

" 

( ) k !1 = ( 3 3 

) 1 $ 2 

$ 

# 

3 2k !2 = 3 3k 

2k ! 2 = 3k 

3k ! 2k = !2 

k = !2 

Check: 

% 

' 

& 

2k 



a) log10 = log 2 x 

1 = x log 2 

x = 1 

log 2 

b) 10 = 2 x 

log10 = log 2 x 

1 = x log 2 

x = 1 

log 2 

Therefore 10 = 2 

1 

log 2 

c) 10 = 3 x 

log10 = log3 x 

1 = x log3 

x = 1 

log3 

Therefore 10 = 3 

1 

log 3 

Chapter 7 Section 1 Question 12 Page 368 

10 = b x 

log10 = log b x 

1 = x log b 

x = 1 

log b 

log 

Therefore 10 = b 

b , b > 0 

1 


log a 

log b 

a = b 

log a = log b 

log a 

log b 

log a = log a 

log b log b 

log a = log a 

log a 

log 

Since both sides are equal, a = b 

b for any a, b > 0. 


Solutions to Achievement Check questions are provided in the Teacher’s Resource. 



a) i) 2 3x > ( 2 ) 2 x+1 

2 3x > 2 2x+2 

3x > 2x + 2 

3x ! 2x > 2 

x > 2 

ii) ( 3 ) 4 x!2 < ( 3 ) 3 2x+1 

3 4x!8 < 3 6x+3 

4x ! 8 < 6x + 3 

!8 ! 3 < 6x ! 4x 

!11 < 2x 

x > ! 11 2 


Graph each side of the inequality as a separate function. Find their point of intersection. Test 

a point other than the point of intersection to ensure that the inequality is true. 

Check a) i): 

181.02 > 128 


Check a) ii): 

5.4 × 10 –16 < 1.8 × 10 –16 


8 x > 2 x + 1 

x > 1 2 



a) i) 

2 x > x 3 for 0 < x < 1.37; 2 x < x 3 for x > 1.37 (correct to 2 decimal places) 

ii) 

2 x > x 3 for x > 9.94 (correct to 2 decimal places) 

1.1 x > x 10 for –0.99 < x < 1.01 (correct to 2 decimal places) 


) 

3 x > x 4 for –0.80 < x < 1.52 (correct to 2 decimal places) 



Techniques for Solving Exponential 

Equations 


Substitute x = 4 into all the equations to find a point. 

Match the graphs to the equations using the point found. 

! 

i) y = 100 1 $ 

" 

# 2% 

& 

= 100 

16 

= 6.25 

4 

! 

ii) y = 50 1 $ 

" 

# 2% 

& 

= 25 

4 

4 

! 

iii) y = 50 1 $ 

" 

# 2% 

& 

= 50 

16 

= 3.125 

4 

! 

iv) y = 100 1 $ 

" 

# 2% 

& 

= 50 

4 

4 

a) iii) b) i) c) ii) d) iv) 


a) log 2 = log1.07 t 

log 2 = t log1.07 

t = log 2 

log1.07 

t &= 10.24 

b) log3 = log1.1 t 

log3 = t log1.1 

t = log3 

log1.1 

t &= 11.53 

c) 

100 

10 = 1.04t 

10 = 1.04 t 

log10 = log1.04 t 

1 = t log1.04 

t = 

1 

log1.04 

t &= 58.71 

d) log5 = log1.08 t +2 

log5 = (t + 2)log1.08 

t + 2 = 

t = 

log5 

log1.08 

log5 

log1.08 ! 2 

t &= 18.91 

e) log0.5 = log1.2 t !1 

log0.5 = (t !1)log1.2 

t !1 = log0.5 

log1.2 

t = log0.5 

log1.2 +1 

t &= !2.80 


! 

f) log10 = log 1 $ 

" 

# 4% 

& 

3t 

! 

g) log15 = log 1 $ 

" 

# 2% 

& 

t 

4 

h) 

100 

25 = ! 1 $ 

" 

# 2% 

& 

6 

t 

1 = 3t log 1 4 

1 

3t = 

log 1 4 

t = 

1 

log15 = t 4 log 1 2 

t 

4 = log15 

log 1 2 

t = 4log15 

! 

4 = 1 $ 

" 

# 2% 

& 

6 

t 

2 2 = ( 2 ) 6 

'1 t 

2 2 = 2 ' 6 t 

3log 1 4 

t &= '0.55 

log 1 2 

t &= '15.63 

2 = ' 6 t 

t = ' 6 2 

t = '3 


a) 

90 s 

= 1.5 min 

60 s 

1.5 

3.1 

! 

A( 1.5) = 50 1 $ 

# & 

" 2 % 

&= 35.75 

Approximately 35.75 mg of a 50 mg polomium-218 sample will remain after 90 s. 

! 

b) 5 = 50 1 $ 

" 

# 2% 

& 

! 

0.1 = 1 $ 

" 

# 2% 

& 

t 

3.1 

! 

log0.1 = log 1 $ 

" 

# 2% 

& 

t 

3.1 

t 

3.1 

log0.1 = t 

3.1 log 1 2 

t 

3.1 = log0.1 

log 1 2 

t = 3.1log0.1 

log 1 2 

t &= 10.30 

It will take approximately 10.3 min for the sample to decay to 10% of its initial amount. 



The answer to part b) would be the same for any initial sample, since 10% (0.1) is the ratio of 

A(t) to A 0 . So A(t) 

A 0 

= 0.1. 


a) log 2 x = log3 x!1 

x log 2 = (x !1)log3 

x log 2 = x log3! log3 

x log 2 ! x log3 = ! log3 

x(log 2 ! log3) = ! log3 

x = 

c) log8 x+1 = log3 x!1 

log3 

log3! log 2 

(x +1)log8 = (x !1)log3 

x log8 + log8 = x log3! log3 

log8 + log3 = x log3! x log8 

log8 + log3 = x(log3! log8) 

x = 

log8 + log3 

log3! log8 

b) log5 x!2 = log 4 x 

(x ! 2)log5 = x log 4 

x log5! 2log5 = x log 4 

x log5! x log 4 = 2log5 

x(log5! log 4) = 2log5 

x = 

2log5 

log5! log 4 

d) log7 2x+1 = log 4 x!2 

(2x +1)log7 = (x ! 2)log 4 

2x log7 + log7 = x log 4 ! 2log 4 

log7 + 2log 4 = x log 4 ! 2x log7 

log7 + 2log 4 = x(log 4 ! 2log7) 

x = 

log7 + 2log 4 

log 4 ! 2log7 


a) 2.710 b) 14.425 c) –3.240 d) –1.883 


a) ( 2 ) x 2 + 2 x ! 6 = 0 

a = 1, b = 1, c = !6 

b) 2 x = !1± 12 ! 4(1)(!6) 

2(1) 

2 x !1± 25 

= 

2 

2 x = !1± 5 

2 

2 x = 2 " x = 1 

or 

2 x = !3, not true 

c) 2 x = –3 



a) (8 x ) 2 ! 2(8 x ) ! 5 = 0 

a = 1, b = !2, c = !5 

b) 8 x = !(!2) ± (!2)2 ! 4(1)(!5) 

2(1) 

8 x = 2 ± 24 

2 

8 x = 2 ± 4 " 6 

2 

8 x = 2 ± 2 6 

2 

8 x = 1± 6 

8 x = 1+ 6 

log8 x = log(1+ 6) 

x log8 = log(1+ 6) 

x = 

or 

log(1+ 6) 

log8 

8 x = 1! 6, not true 

c) 8 x = 1! 6 



! 1$ 

a) A(t) = A 0 

" 

# 2% 

& 

! 

17 = 20 1 $ 

" 

# 2% 

& 

17 

20 = ! 1 $ 

" 

# 2% 

& 

5 

h 

t 

h 

5 

h 

! 

log0.85 = log 1 $ 

" 

# 2% 

& 

5 

h 

log0.85 = 5 h log 1 2 

5 

h = log0.85 

log 1 2 

h = 

5log 1 2 

log0.85 

h &= 21.3 

The half-life of thorium-233 is approximately 21.3 min. 

! 

b) 1 = 20 1 $ 

" 

# 2% 

& 

1 

20 = ! 1 $ 

" 

# 2% 

& 

t 

21.3 

! 

log0.05 = log 1 $ 

" 

# 2% 

& 

t 

21.3 

t 

21.3 

log0.05 = 

t 

21.3 log 1 2 

t 

21.3 = log0.05 

log 1 2 

t = 21.3log0.05 

log 1 2 

t &= 92.06 

It will take approximately 92.06 min for the sample to decay to 1 mg (92.17 min using the 

exact half-life). 



! 1$ 

a) A(t) = A 0 

" 

# 2% 

& 

! 

9 = 10 1 $ 

" 

# 2% 

& 

! 

0.9 = 1 $ 

" 

# 2% 

& 

3 

h 

t 

h 

3 

h 

! 

log0.9 = log 1 $ 

" 

# 2% 

& 

3 

h 

log0.9 = 3 h log 1 2 

3 

h = log0.9 

log 1 2 

b) 

h = 

3log 1 2 

log0.9 

h &= 19.7 

The half-life of bismuth-214 is approximately 19.7 min. 

c) i) A shorter half-life means less time for the substance to decay. 

The graph would decrease faster since it takes less time for the sample to decay. 

ii) A longer half-life means more time for the substance to decay. 

The graph would decrease at a slower rate since it takes more time for the sample to 

decay. 

d) i) A greater sample size means more substance to decay in the same amount of time. 

The graph would have a steeper slope (negative). 

ii) A smaller sample size means less substance to decay in the same amount of time. 

The graph would have a slope that isn’t as steep. 



(4 x ) 2 + 2(4 x ) + 3 = 0 

a = 1, b = 2, c = 3 

4 x = !2 ± 22 ! 4(1)(3) 

2(1) 

4 x !2 ± !8 

= 

2 

There are no real roots. 


a) (3 x ) 2 – 7(3 x ) + 12 = 0 

a = 1, b = –7, c = 12 

3 x = !(!7) ± (!7)2 ! 4(1)(12) 

2(1) 

3 x = 7 ± 1 

2 

3 x = 7 ±1 

2 

3 x = 8 2 

or 3 x = 6 2 

3 x = 4 3 x = 3 

x log3 = log 4 x = 1 

x = log 4 

log3 

x &= 1.26 


) (5 x ) 2 + (5 x ) – 3 = 0 

a = 1, b = 1, c = –3 

5 x = !1± 12 ! 4(1)(!3) 

2(1) 

5 x = 

!1± 13 

2 

5 x !1+ 13 

= 

2 

" !1+ 13 % 

x log5 = log $ ' 

# 2 & 

" !1+ 13 % 

log $ ' 

# 2 & 

x = 

log5 

x &= 0.16 

5 x = 

or 

c) ( 3) 2x+3 = 24 

8 

3 2x+3 = 3 1 

2x + 3 = 1 

2x = !2 

x = !1 

!1! 13 

2 

< 0, extraneous root 

d) 

18 

2 = 92x!1 

9 1 = 9 2x!1 

1 = 2x !1 

2x = 2 

x = 1 

e) 3 x +1+ 56(3 ! x ) = 0 

(3 x )(3 x ) + 3 x + 56(3 ! x )(3 x ) = 0(3 x ) multiply each term by 3 x 

( 3 ) x 2 + 3 x + 56 = 0 

a = 1,b = 1,c = 56 

3 x = !1± 12 ! 4(1)(56) 

2(1) 

3 x = 

!1± !223 

, no real roots 

2 


f) 4 x ! 3!18(4 ! x ) = 0 

(4 x ) 2 ! 3(4 x ) !18 = 0 multiply both sides by 4 x 

a = 1, b = !3, c = !18 

4 x = !(!3) ± (!3)2 ! 4(1)(!18) 

2(1) 

4 x = 

3± 81 

2 

4 x = 3+ 9 

2 

or 4 x = 3! 9 

2 

4 x = 6 4 x = !3 < 0, extraneous root 

x log 4 = log6 

x = log6 

log 4 

x &= 1.29 


a) 


) 

c) 


d) 

e) 

There is no zero so there is no solution. 

f) 



! 1$ 

a) A(t) = A 0 

" 

# 2% 

& 

! 

0.1 = 1 $ 

" 

# 2% 

& 

t 

20 

t 

h 

! 

log0.1 = log 1 $ 

" 

# 2% 

& 

t 

20 

log0.1 = t 

20 log 1 2 

t 

20 = log0.1 

log 1 2 

t = 20log0.1 

log 1 2 

t &= 66.4 

It would take approximately 66.4 h for 90% of platinum-197 to turn into gold. 


This is not a “get-rich quick” scheme since the cost of the nuclear process exceeds the price 

of the gold. 


1! 2 !0.05n > 0.2 

!2 !0.05n > 0.2 !1 

!2 !0.05n > !0.8 

2 !0.05n < 0.8 

log 2 !0.05n < log0.8 

!0.05nlog 2 < log0.8 

!0.05n < log0.8 

log 2 

n > 

log0.8 

!0.05log 2 

n > 6.44 

After approximately 6.44 days the probability is greater than 20%. 



! 

a) V (t) = 24 000 1 $ 

" 

# 2% 

& 

! 

0.25 = 1 $ 

" 

# 2% 

& 

t 

3 

t 

3 

log0.25 = t 3 log 1 2 

t 

3 = log0.25 

log 1 2 

t = 3log0.25 

log 1 2 

t = 6 

It would take 6 years for the minivan to depreciate to one quarter of its initial value. 

! 

b) 0.1 = 1 $ 

" 

# 2% 

& 

t 

3 

log0.1 = t 3 log 1 2 

t 

3 = log0.1 

log 1 2 

t = 3log0.1 

log 1 2 

t &= 10 

It would take approximately 10 years for the minivan to depreciate to 10% of its initial value. 



! 

a) i) 0.5 = 1 $ 

" 

# 2% 

& 

t 


log0.5 = t 

12 log 1 2 

t 

12 = log0.5 

log 1 2 

t 

12 = 1 

t = 12 

It will take 12 s for the note to decay to half of its initial value. 

! 

ii) 0.1 = 1 $ 

" 

# 2% 

& 

t 


log0.1 = t 

12 log 1 2 

t 

12 = log0.1 

log 1 2 

t = 12log0.1 

log 1 2 

t &= 40 

It will take approximately 40 s for the note to decay to 10% of its initial value. 

b) When I 0 = 1: 

c) i) Answers may vary. A sample solution is shown. 

The half-life would be longer, so the 12 in the exponent would increase. 

ii) Answers may vary. A sample solution is shown. 

The graph would not be as steep. It would decrease at a slower rate. 





" 1% 

a) If A 0 is the initial amount, y = A 0 

! A 0 

# 

$ 2& 

' 

t 

20 

b) Time starts at 0 s and continues forever. The amount starts at 0 g and continues until all the 

platinum-197 is gold-197. 

Domain: {t ∈ R, t ≥ 0}; Range: {y ∈ R, 0 < y ≤ A 0 } 


Answers may vary. A sample solution is shown. 

a) The carbon fusion cycle takes place in very hot stars. During the cycle, the addition of four 

hydrogen atoms changes a carbon atom into other elements before returning it back to carbon 

and releasing a helium atom and two positrons. At each of the six stages of the process, 

energy is released. 

Every time a carbon cycle takes place, two of the four protons are destroyed. 

b) Call the protons in hydrogen and helium atoms available protons. If there are 5 × 10 13 

available protons in a star, how many will there be after three complete carbon cycles? 

Solution: 5 × 10 13 (0.5) 3 = 6.25 × 10 12 

After how many carbon cycles will there be fewer than 10% of the original available protons? 

Solution: Find when 0.5 x is less than 0.1: (0.5) 3 = 0.125 and (0.5) 4 = 0.0625 

After four complete carbon cycles. 



D; 

4 x ! 4 x!1 = 24 

4 x!1 (4 !1) = 24 

4 x!1 (3) = 24 

4 x!1 = 24 3 

4 x!1 = 8 

( 2 ) 2 x!1 = 2 3 

2 2x!2 = 2 3 

2x ! 2 = 3 

2x = 5 

x = 5 2 

( 2x) x " " 

= 2 5 % % 

$ 

# 

$ 2& 

' ' 

# & 

5 

= 52 

= 5 5 

= 5 4 ( 5 

= 25 5 

5 

2 



C; 

2 a = 5 2 b = 3 

a log 2 = log5 

log 3 

(5! 2) = 

= 

= 

= 

log(5! 2) 

log3 

log5+ log 2 

log3 

a log 2 + log 2 

blog 2 

log 2(a +1) 

blog 2 

= a +1 

b 

blog 2 = log3 

product law 


h(2x +1) = 2h(x) +1 

h(2(0) +1) = 2h(0) +1 

h(1) = 2(2) +1 

h(1) = 5 

h(2(1) +1) = 2h(1) +1 

h(3) = 2(5) +1 

h(3) = 11 



Product and Quotient Laws of Logarithms 


! 

a) log(9 × 6) = log 54 b) log 48 $ 

" 

# 6 % 

& = log 8 

! 36$ 

c) log 3 (7 × 3) = log 3 21 d) log 5 

" 

# 18 % 

& = log 2 5 


screen shot for part c): 

a) 1.732 b) 0.903 c) 2.771 d) 0.431 


a) log(2xyz), x > 0, y > 0, z > 0 

! 3ab$ 

b) log 2 

" 

# 2c % 

& ,a > 0, b > 0, c > 0 

c) 

" 

log m 2 + log n 3 ! log y 4 = log m2 n 3 % 

$ 

# y 4 ' , m > 0, n > 0, y > 0 

& 

d) logu 2 + log v + log w = log( u 2 v w), u > 0, v > 0, w > 0 


a) log 6 

(18 ! 2) = log 6 

36 

= 2 

c) log 12 

(8 ! 2 ! 9) = log 12 

144 

= 2 

b) log(40 ! 2.5) = log100 

= 2 

d) log(5! 40 ! 5) = log1000 

= 3 



! 54$ 

a) log 3 

" 

# 2 % 

& = log 27 3 

= 3 

! 320$ 

c) log 4 

" 

# 5 % 

& = log 64 4 

= 3 

! 

b) log 

" 

# 

50 000 

5 

$ 

% 

& = log10 000 

= 4 

! 2 $ 

d) log 

" 

# 200% 

& = log0.01 

= '2 


a) log 16 

2 3 + log 16 

8 2 ! log 16 

2 = log 16 

8 + log 16 

64 ! log 16 

2 

# 8 " 64& 

= log 16 

$ 

% 2 ' 

( 

= log 16 

256 

= 2 

1 

b) log 20 + log 2 + log125 3 

= log 20 + log 2 + log5 

= log(20 ! 2 ! 5) 

= log 200 

&= 2.301 


a) log 7 c + log 7 d b) log 3 m – log 3 n 

1 

c) logu + log v 3 = logu + 3log v d) log a + log b 2 

! log c 2 = log a + 1 log b ! 2log c 

2 

e) log 2 

(5! 2) = log 2 

5+ log 2 

2 

= log 2 

5+1 

f) log 5 

(25! 2) = log 5 

25+ log 5 

2 

= 2 + log 5 

2 



log 5 

(5!10) = log 5 

5+ log 5 

10 

= 1+ log 5 

10 

Since log 5 10 can be simplified again using 10 = 5 × 2, the original solution is the simplest result. 



! 

a) log x2 $ 

# & 

" x % 

= log x2 ' log x 

1 

2 

= 2log x ' 1 2 log x 

= 4 2 log x ' 1 2 log x 

= 3 log x, x > 0 

2 

! 

b) log# 

" 

! 

1 

m $ 

m 3 & + log( m) 7 = log m $ 

2 

# & 

! 

1 

$ 

# 

% 

m 3 & 

+ log m2 

# & 

" 

# 

% 

& " % 

1 

= log m2 '3 + log m 

7 

2 

1 

= log m2 ' 6 2 

+ 7 2 log m 

7 

c) log k + log( k ) 3 + log k 2 

= log m ' 5 2 

+ 7 2 log m 

= ' 5 2 log m + 7 2 log m 

= log m,m > 0 

1 

3 

= log k 

! 

2 

+ log# 

k 

" 

1 

2 

$ 

& 

% 

3 

+ log( k ) 1 

2 3 

= 1 2 log k + 3 2 log k + 2 3 log k 

= 3 6 log k + 9 6 log k + 4 6 log k 

= 16 6 log k 

= 8 log k, k > 0 

3 


d) 2log w + 3log w + 1 2 log ! 

1 

$ 

w2 = 2log w + 3log w2 

# & + 1 

" % 2 (2)log w 

= 2log w + 3 log w + log w 

2 

= 4 2 log+ 3 2 log w + 2 2 log w 

= 9 log w, w > 0 

2 


" 

a) log x2 ! 4% 

" (x ! 2)(x + 2) % 

$ 

# x ! 2 

' = log 

& # 

$ x ! 2 & 

' 

= log( x + 2), x > 2 

! 

b) log x2 + 7x +12$ 

! (x + 4)(x + 3) $ 

# 

" x + 3 

& = log 

% " 

# x + 3 % 

& 

= log( x + 4), x > '3 

" 

c) log x2 ! x ! 6% 

" (x ! 3)(x + 2) % " 

$ 

# 2x ! 6 

' = log 

& # 

$ 2(x ! 3) & 

' d) log x2 + 7x +12% 

" (x + 4)(x + 3) % 

$ 

x 2 ' = log 

# ! 9 & # 

$ (x ! 3)(x + 3) & 

' 

" 

= log x + 2 % 

# 

$ 2 & 

' , x > 3 " 

= log x + 4 % 

# 

$ x ! 3& 

' , x > 3 


! 

a) V o 

= log V $ 

2 

# & 

" % 

V 1 

! 

b) i) V o 

= log 10V $ 

1 

# & 

" % 

= log10 

= 1 

V 1 

! 

ii) V o 

= log 100V $ 

1 

# & 

" % 

= log100 

= 2 

V 1 

! 

iii) V o 

= log V $ 

1 

# & 

" % 

= log1 

= 0 

V 1 


a) log(10nx) = log10 + log n + log x 

= 1+ log n + log x 

translate y = log x up (1 + log n) units, n > 0 


) Answers may vary. A sample solution is shown. 

log 20x = log(10 ! 2 ! x) 

= 1+ log 2 + log x 

( ) = log$ 

10 ! 1 2 ! x 

log 5x 

" 

# 

% 

' 

& 

= log10 + log 1 2 + log x 

= 1+ log 1 2 + log x 



This process cannot be applied when n is an integer since you cannot have a logarithm of a 

negative number. You can have positive integers only. 


a) 

b) 

c) 

d) p(x) = g(x) ; this demonstrates the power law. 


a) q(x) = f (x) + g(x) 

= 2log x + 4log x 

= 6log x 

= log x 6 


) 


The graph has values for x < 0 and x > 0 because x 6 is an even power: the graph has a 

discontinuity at x = 0. 



Y 1 is the difference of two functions f(x) and g(x). Y 2 is the simplified form once the quotient law 

is used. 

log(20x 2 ) ! log 2x 

f (x) = log(20x 2 ) and g(x) = log(2x) 

" 

$ 

# 

( ) = log 20x2 

2x 

= log(10x) 

% 

' 

& 



Using the product law: 

log10 2 = log(10 !10) 

= log10 + log10 

= 1+1 

= 2 

Using the power law: 

log10 2 = 2log10 

= 2(1) 

= 2 


a) A(t) = 40 000 + 2000t 

B(t) = 38 500(1.05) t 

b) i) 

It will take approximately 6.7 years for Renata to earn the same amount at either firm. 

Renata will earn more at firm B after 10 years. 


ii) 

Renata will earn more at firm B after 20 years. 


Does Renata plan to stay at the job for more than 6.7 years? If not, she would get more 

money taking Firm A’s offer. She should also consider the benefits and experiences the jobs 

offer. 




" 

log b 

# 

$ 

Let x = log b 

m and y = log b 

n 

b x = m and b y = n 

m 

n = bx 

b y 

m 

n = bx! y 

m% 

n & 

' = x ! y 

" m% 

log b 

# 

$ n & 

' = log m ! log n 

b b 

product law of exponents 

logarithm to base b of both sides 


a) Let x = log b 


n 


log b 

mn = log b 

m + log b 

n product law of logarithms 

log b 

mn = x + y 

mn = b x+ y 

b x b y = b x+ y 


) Let x = log b 


n 


! m$ 

log b 

" 

# n % 

& = log m ' log n quotient law of logarithms 

b b 

! m$ 

log b 

" 

# n % 

& = x ' y 

m 

n = bx' y 

b x 

b y = bx' y 



This will occur when the line y = –x + c is tangent to the circle x 2 + y 2 = 1 and has a y-intercept 

above the x-axis. 

c = 1.41 

f( x) = - x+ c 

AB: x2+ y2 = 12 

5 

4 

3 

2 

1 

-8 -6 -4 -2 A B 2 4 6 8 

-1 

y = –x + c intersects x 2 + y 2 = 1 at point P(x, y). 

The distance to P(x, y) is the radius of the circle r = 1 and 

x = y 

1 = x 2 + x 2 

1 = 2x 2 

1 

2 = x2 

-2 

x = 1 2 and y = 1 2 

y = !x + c 

1 

2 = ! 1 2 + c 

c = 1 2 + 1 2 

c = 2 2 

c = 2 



k = -1 

f( x) = x2 +k 

2 

1 

-6 -4 -2 2 4 6 

-1 

-2 

k = –1 



B; 

equation of a line with slope ! 2 

b = !2a + d 

!3 = !2r + d substitute the point (r,!3) 

d = !3+ 2r 

( ) 

b = !2a + 2r ! 3 1 

equation of a line with slope 1 2 

(perpendicular to ! 2) 

b = 1 2 a + c 

r = 1 (6) + c substitute the point ( 6,r) 

2 

c = r ! 3 

b = 1 2 a + r ! 3 ( 2 ) 

1 

a + r ! 3 = !2a + 2r ! 3 

2 

1 

a + 2a ! 3+ 3 = 2r ! r 

2 

1 

2 a + 4 2 a = r 

r = 5 2 a 

a = 2 5 r 


Pounds of coffee per k sacks = c k 

pounds of coffee per (k + n) sacks = 

The sacks hold c k ! 

c 

k + n 

c 

k + n 

fewer pounds. 



Techniques for Solving Logarithmic 

Equations 


a) log(x ! 2) = 1 

x ! 2 = 10 1 

x = 10 + 2 

x = 12 

d) 1! log(w ! 7) = 0 

1 = log(w ! 7) 

10 = w ! 7 

10 + 7 = w 

w = 17 

b) 2 = log(x + 25) 

10 2 = x + 25 

100 ! 25 = x 

x = 75 

e) log(k ! 8) = 2 

k ! 8 = 10 2 

k = 100 + 8 

k = 108 

c) 4 = 2log( p + 62) 

2 = log( p + 62) 

10 2 = p + 62 

100 ! 62 = p 

p = 38 

f) 6 ! 3log 2n = 0 

6 = 3log 2n 

2 = log 2n 

10 2 = 2n 

100 

2 = n 

n = 50 

Check: 

a) log(12 – 2) = 1 b) 2 = log(75 + 25) c) 4 = 2log(38 + 62) 

d) 1 – log (17 – 7) e) log(108 – 8) = 2 f) 6 – 3log 2(50) = 0 



a) log 3 

(x + 4) = 2 

log 3 

(x + 4) = log 3 

9 

x + 4 = 9 

x = 9 ! 4 

x = 5 

c) 2 = log 4 

(k !11) 

log 4 

16 = log 4 

(k !11) 

16 = k !11 

k = 16 +11 

k = 27 

e) log 8 

(t !1) = !1 

log 8 

(t !1) = log 8 

1 

8 

t !1 = 1 8 

t = 1 8 + 8 8 

t = 9 8 

b) 5 = log 2 

(2x !10) 

log 2 

32 = log 2 

(2x !10) 

32 = 2x !10 

2x = 32 +10 

2x = 42 

x = 21 

d) 9 = log 5 

(x +100) + 6 

9 ! 6 = log 5 

(x +100) 

3 = log 5 

(x +100) 

log 5 

125 = log 5 

(x +100) 

125 = x +100 

x = 125!100 

x = 25 

f) log 3 

(n 2 ! 3n + 5) = 2 

log 3 

(n 2 ! 3n + 5) = log 3 

9 

n 2 ! 3n + 5 = 9 

n 2 ! 3n + 5! 9 = 0 

n 2 ! 3n ! 4 = 0 

(n ! 4)(n +1) = 0 

n = 4 or n = !1 



a) log x + log(x ! 4) = 1 

Check: 

log " 

# x(x ! 4) $ 

% = log10 

x 2 ! 4x = 10 

x 2 ! 4x !10 = 0 

x = !(!4) ± (!4)2 ! 4(1)(!10) 

2(1) 

x = 4 ± 56 

2 

x = 4 ± 2 14 

2 

x = 2 ± 14 

x = 2 + 14 

2 ! 14 is an extraneous root since both log x and log(x ! 4) 

are undefined for this value. 


) log x 3 ! log 2 ! log(2x 2 ) = 0 

Check: 

" x 3 % 

log$ 

# 2(2x 2 ' 

)& 

= 0 

" 

log x % 

# 

$ 4& 

' = 0 

x 

4 = 100 

x 

4 = 1 

x = 4 

c) log(v !1) ! log(v !16) = 2 

" v !1 % 

log 

# 

$ v !16& 

' = 2 

v !1 

v !16 = 102 

v !1 = 100v !1600 

100v ! v = 1600 !1 

99v = 1599 

v = 1599 

99 

v = 533 

33 


Check: 

d) 1 = log( y + 9) ! log y 

" 

1 = log y + 9 % 

# 

$ y & 

' 

10 1 = y + 9 

y 

10y = y + 9 

9y = 9 

y = 1 

Check: 


e) log " 

#(k + 2)(k !1) $ 

% = 1 

Check: 

k 2 + k ! 2 = 10 1 

k 2 + k !12 = 0 

(k + 4)(k ! 3) = 0 

k = 3 

k = !4 is an extraneous root since both log(k + 2) and log(k !1) 

are undefined for this value. 


! 

f) log p + 5 $ 

" 

# p +1% 

& = 3 

p + 5 

p +1 = 103 

p + 5 = 1000 p +1000 

1000 p ' p = 5'1000 

Check: 

999 p = '995 

p = ' 995 

999 


Graph both sides of the equation and find the points of intersection. 

MHR • Advanced Functions 12 Solutions 712


1 

a) log(x 2 ! 3x) 2 

= 1 2 

1 

2 log(x2 ! 3x) = 1 2 

log(x 2 ! 3x) = 1 

x 2 ! 3x = 10 1 

x 2 ! 3x !10 = 0 

(x ! 5)(x + 2) = 0 

x = 5 or x = !2 

Check: 

1 

a) log(5 2 ! 3" 5) 2 

= log10 

= 1 2 

1 

log((!2) 2 ! 3" (!2)) 2 

= log10 

1 

2 

1 

2 

1 

b) log(x 2 + 48x) 2 

= 1 

( ) = 1 

1 

2 log x2 + 48x 

log(x 2 + 48x) = 2 

x 2 + 48x = 10 2 

x 2 + 48x !100 = 0 

(x + 50)(x ! 2) = 0 

x = !50 or x = 2 

1 

b) log((!50) 2 + 48 " (!50)) 2 

= log100 

= 1 

1 

log((2) 2 + 48 " 2) 2 

= log100 

= 1 

1 

2 

1 

2 

= 1 2 


! x + 5$ 

a) log 2 

" 

# 2x % 

& = log 256 2 

x + 5 

2x = 256 

x + 5 = 512x 

511x = 5 

x = 5 

511 

b) log(2k + 4) ! log k = 1 

" 

log 2k + 4 % 

# 

$ k & 

' = log10 

2k + 4 

= 10 

k 

2k + 4 = 10k 

10k ! 2k = 4 

8k = 4 

k = 1 2 



a) 

b) 

x &= 9.05 

x &= 2.16 


# 0.9 & 

a) L = 10log 

$ 

% 

1.0 !10 "12 ' 

( 

&= 119.54 

The concert is approximately 119.54 dB. 


# I & 

b) 20 = 10 log 

$ 

% 

1.0 !10 "12 ' 

( 

( ) = I 

100 1.0 !10 "12 

# I & 

2 = log 

$ 

% 

1.0 !10 "12 ' 

( 

10 2 I 

= 

1.0 !10 "12 

I = 1!10 "10 

The whisper’s intensity is 1 × 10 –10 W/m 2 . 

# I & 

c) 120 = 10 log 

$ 

% 

1.0 !10 "12 ' 

( 

10 12 1.0 !10 "12 

# I & 

12 = log 

$ 

% 

1.0 !10 "12 ' 

( 

10 12 I 

= 

1.0 !10 "12 

( ) = I 

I = 1 

The car stereo’s intensity is 1 W/m 2 . 


a) No, the statement is not true since log (–3) and log (–4) are undefined. 

b) Yes. 

L.S. = ! log3! log 4 

= !(log3+ log 4) 

= ! # 

$ log(3" 4) % 

& 

= ! log12 

R.S. = ! log12 

Since L.S. = R.S., –log 3 – log 4 = –log 12. 



! $ 

# 1 & 

a) n = log 2 # & 

# 

1 

& 

" 4 % 

n = log 2 

4 

n = 2 

! 1 $ 

b) 4 = log 2 

" 

# p % 

& 

! 1 $ 

log 2 

16 = log 2 

" 

# p % 

& 

16 = 1 p 

p = 1 

16 


a) 

w 2 !10w !100 = 0 

2 

3 = log 3 

w2 !10w 

( ) 1 3 

2 

3 = log w2 !10w 

2 

3 = 1 3 log ( w2 !10w) 

( ) 

2 = log w 2 !10w 

10 2 = w 2 !10w 

w = !(!10) ± (!10)2 ! 4(1)(!100) 

2(1) 

w = 

w = 

10 ± 500 

2 

10 ±10 5 

2 

w = 5 ± 5 5 


) 


Answers may vary. A sample answer is shown. 

Graph each side of the equation as a function and find the point of intersection. 

x = –1 



log b 

a = c and log y 

b = c 

b c = a and y c = b 

( y ) c c = a substitute y c = b 

y c2 

= a 

log a 

y c2 

= log a 

a logarithm of base a of both sides 

c 2 log a 

y = 1 power law of logarithms 

log a 

y = 1 c 2 

log a 

y = c !2 


1 × 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 44 



3 = m + 1 m 

3 m = m +1 

m 2 ! 7m +1 = 0 

9m = m 2 + 2m +1 

square both sides 

m = !(!7) ± (!7)2 ! 4(1)(1) 

2(1) 

m = 7 ± 45 

2 

m = 7 ± 3 5 

2 

substitute m = 7 + 3 5 

2 

m ! 1 m = 7 + 3 5 

2 

2 

! 

7 + 3 5 

(7 + 3 5)(7 + 3 5) 4 

= ! 

2(7 + 3 5) 2(7 + 3 5) 

49 + 42 5 + 45! 4 

= 

2(7 + 3 5) 

90 + 42 5 

= 

2(7 + 3 5) 

45+ 21 5 

= 

7 + 3 5 

= 

= 

45+ 21 5 

7 + 3 5 " 7 ! 3 5 

7 ! 3 5 

315!135 5 +147 5 ! 315 

49 ! 45 

= 12 5 

4 

= 3 5 

substitute m = 7 ! 3 5 

2 

m ! 1 m = 7 ! 3 5 

2 

2 

! 

7 ! 3 5 

(7 ! 3 5)(7 ! 3 5) ! 2(2) 

= 

2(7 ! 3 5) 

49 ! 42 5 + 45! 4 

= 

2(7 ! 3 5) 

90 ! 42 5 

= 

2(7 ! 3 5) 

= 

= 

45! 21 5 

7 ! 3 5 " 7 + 3 5 

7 + 3 5 

315!12 5 ! 315 

49 ! 45 

= !12 5 

4 

= !3 5 

m ! 1 m = ±3 5 



D; 

" 

# 

$ 

10 ! v 

1+ v 

u + uv + v = 10 

u(1+ v) = 10 ! v 

% 

& 

' 

2 

u = 10 ! v 

1+ v 

+ v 2 = 40 

" 100 ! 20v + v 2 % 

$ 

# 1+ 2v + v 2 ' = 40 ! v2 

& 

100 ! 20v + v 2 = 40 ! v 2 

v 4 + 2v 3 + v 2 ! 39v 2 ! 20v ! 80v +100 ! 40 = 0 

( )( 1+ 2v + v ) 

2 

100 ! 20v + v 2 = 40 + 80v + 40v 2 ! v 2 ! 2v 3 ! v 4 

v 4 + 2v 3 ! 38v 2 !100v + 60 = 0 

v &= 0.5 or v &= 6.3 since it is a positive real number. 

Case 1: v &= 0.5 

u = 

10 ! (0.5) 

1+ (0.5) 

&= 6.3 

Case 2: v &= 6.3 

u = 

10 ! (6.3) 

1+ (6.3) 

&= 0.5 

u + v &= 6.3 + 0.5 

&= 6.8 

The closest integer value is 7. 



Making Connections: Mathematical 

Modelling With Exponential and Logarithmic 

Equations 


P = 1006(1.016) t 

6500 = 1006(1.016) t 

6500 

1006 = 1.016t 

! 

log 6500 $ 

" 

# 1006 % 

& = t log1.016 

! 

log 6500 $ 

" 

# 1006 % 

& 

t = 

log1.016 

t &= 117.5 

1920 + 117.5 = 2037.5 

The off-ramp should be built in approximately June of 2037. 


1920 – 145.4 &= 1774.6 

In approximately the year 1774 the town’s population was only 100. 



a) i) P = 1006(1.016) t 

P = 1006(1.016) 100 

&= 4920 

The town’s population will be approximately 4920 in 100 years from 1920. 

P = 1000 ! 2 

t 

43.5 

100 

P = 1000 ! 2 

43.5 

&= 4921 

The town’s population will be approximately 4921 in 100 years from 1920. 

ii) 

P = 1006(1.016) t 

20 000 = 1006(1.016) t 

20 000 

1006 = 1.016t 

! 20 000$ 

log 

" 

# 1006 % 

& = t log1.016 

! 20 000$ 

log 

" 

# 1006 % 

& 

t = 

log1.016 

t &= 188.4 

It will take approximately 188.4 years for the town’s population to reach 20 000. 

P = 1000 ! 2 

20 000 = 1000 ! 2 

20 = 2 

t 

43.5 

log 20 = log 2 

t 

43.5 

t 

43.5 

t 

43.5 

log 20 = 

t 

43.5 log 2 

t log 20 

= 

43.5 log 2 

t = 

43.5log 20 

log 2 

t &= 188 

It will take approximately 188 years for the town’s population to reach 20 000. 


Quite close. The differences are due to rounding in each method. 




The town may decide that the Lakeland Savings Bond account is now more favourable since 

Northern Equity will not be as flexible and the principle will mature to $80 000 slightly faster. 



A = 50 000(1.0325) 2t 

80 000 = 50 000(1.0325) 2t 

8 

5 = 1.03252t 

log 8 = 2t log1.0325 

5 

2t = 

t = 

log 8 5 

log1.0325 

log 8 5 

2log1.0325 

A = 50 000(1.005) 12t 

80 000 = 50 000(1.005) 12t ! 0.01(50 000) 

805 

500 = (1.005)12t 

log 805 = 12t log1.005 

500 

log 805 

12t = 500 

log1.005 

t &= 7.35 

t &= 7.96 

Rural Ontario Investment Group takes a little less time to make $80 000 (7.35 years) so it could 

be considered. Muskoka Guaranteed Certificate takes approximately the same amount of time as 

the other options, so it doesn’t need to be considered. 

t = 

log 805 

500 

12log1.005 



increasing in a curved pattern 


) i) 

y &= 4.21x – 3.07 

ii) 

y &= 0.86x 2 – 0.93x + 1.2 

iii) 

y &= 1.37(1.65) x 


Quadratic, since the circle is increasing in two dimensions. 

d) Answers may vary. A sample solution is shown. 

i) y &= 1.37(1.65) 10 

&= 207.8 

The oil spill is approximately 207.8 m 2 after 10 min. 


ii) A = !r 2 

" 

= ! 30 % 

# 

$ 2 & 

' 

&= 706.9 

2 

y &= 1.37(1.65) t 

706.9 &= 1.37(1.65) t 

706.9 

&= 1.65t 

1.37 

! 

log 706.9 $ 

" 

# 1.37 % 

& &= t log1.65 

! 

log 706.9 $ 

" 

# 1.37 % 

& 

t &= 

log1.65 

t &= 12.44 

The spill will take approximately 12.44 min to reach a diameter of 30 m. 

e) Answers may vary. A sample solution is shown. 

The spill is in the shape of a circle since it is spreading equally in all directions. 


a) A = P(1+ i) n 

P = 1000 

i = 0.08 ÷ 4 

= 0.02 

n = 4t 

A = 1000(1.02) 4t 

b) A = 1000(1.02) 16 

&= 1372.79 

The value of the investment will be approximately $1372.79 in 4 years. 

c) 2 = (1.02) 4t 

log 2 = 4t log1.02 

4t = log 2 

log1.02 

t = 

log 2 

4log1.02 

t &= 8.75 

The value of the investment will take approximately 8.75 years to double. 



a) A = 1000(1.02) 4t !1000(0.05) 

A = 1000(1.02) 4t ! 50,for t < 4 

b) Translates the graph down 50 units. 



a, b) 

Time (min) Temperature (°C) 

0 100 

2 90 

4 85 

6 81 

8 78 

10 74 

12 71 

14 68 

16 66 

18 64 

20 62 

c) 

d) 

e) I would use the exponential model because the quadratic model may work well with these 

data, but it will eventually start increasing, which the data will not. 


f) 40 = 94.54(0.978) t 

log 40 

94.54 = log(0.978)t 

log 40 = t log 0.978 

94.54 

log 40 

t = 

94.54 

log 0.978 

t &= 38.7 

30 = 94.54(0.978) t 

log 30 

94.54 = log(0.978)t 

log 30 = t log 0.978 

94.54 

log 30 

t = 

94.54 

log 0.978 

t &= 51.6 

0 = 94.54(0.978) t 

0 

log 

94.54 = log(0.978)t 

0 

log = t log 0.978 

94.54 

0 

log 

t = 

94.54 

log 0.978 

t!is!undefined 

This makes sense because the water cannot cool less than room temperature. 


" 

a) 150 !120 = 10log I % 

2 

$ ' 

# & 

I 1 

" 

30 = 10log I % 

2 

$ ' 

# & 

I 1 

" 

3 = log I % 

2 

$ ' 

# & 

10 3 = I 2 

I 1 

I 1 

1000 = I 2 

I 1 

The ratio of the intensities is 1000. Rocco’s signal needs to be boosted to match Biff’s sound 

level. 

1 

b) 1000 ÷ = 1000 !10 000 

10 000 

= 10 000 000 

= 10 7 

The sound crew should reduce the drums by a factor of 10 7 . 






a), b) 

Data on carbon-dating that gives the approximate age of piece of organic material, and 

the ratio of the current amount of carbon-14 to the expected original amount. 

Age (years) Carbon-14 Ratio 

750 0.91 

1000 0.88 

2400 0.74 

2850 0.70 

4000 0.61 

7500 0.39 

8200 0.36 

c) 

d) Knowing that carbon dating involves a half-life, the exponential regression is the best. 


e) How old is bone that contains 10% of its original carbon-14? 

Solution: 

0.1 = 0.999!122(0.999!875) t 

0.1 

log = log 0.999!875t 

0.999!122 

0.1 

log = t log 0.999!875 


0.1 

log 

t = 


log 0.999!875 

t &= 18!412.5 

The bone is about 5545 years old. 

Using the formula from the exponential regression, determine the half-life of carbon-14. 

Solution: 

t 

! 1$ 

h 

" 

# 

2% 

& = 0.999!875 

t 

t 

! 

log 1 $ h 

" 

# 

2% 

& = log 0.999!875 

t 

t 

h log ! 1 $ 

" 

# 

2% 

& = t log 0.999!875 

! 

t log 1 $ 

" 

# 

2% 

& 

h = 

t log 0.999!875 

! 

log 1 $ 

" 

# 

2% 

& 

h = 

log 0.999!875 

h &= 5545 

The half-life is about 5545 years. 




a), b) 

Data on bacteria in a petri dish over several days. 

Day 

Bacteria (thousands) 

1 2.8 

2 7.2 

3 17.6 

4 40.0 

5 78.6 

6 125.2 

7 162.8 

8 183.8 

9 193.4 

10 197.4 

c) 

d) The logistic curve models the data almost perfectly. 


e) How many bacteria will there be after two weeks? 

Solution: 

200 

y = 

1+181e !0.953x 

200 

y = 

1+181e !0.953(14) 

y &= 199.9 

There will be about 199 900 bacteria. 

After how many days did the bacteria population reach 100 000? 

Solution: 

200 

100 = 

1+181e !0.953x 

1+181e !0.953x = 200 

100 

e !0.953x = 2 !1 

181 

log e !0.953x = log 1 

181 

!0.953x log e = log 1 

181 

log 1 

x = ! 

181 

0.953log e 

x &= 5.45 

The population reached 100 000 just before 5.5 days. 



The data is constantly curving, so a piecewise linear would not be the best option. 

If there were more data about the end behaviour, it might be possible to approximate the data 

using two almost horizontal function (especially since the data is levelling off towards y = 200), 

with a steep linear function in the middle. 



downhill rate = x uphill = 1 

16 h level = 1 24 h downhill = 1 x h average = 1 24 h 

! 1 

16 + 1 24 + 1 $ 

" 

# x % 

& ÷ 3 = 1 average 

24 

1 

48 + 1 

72 + 1 3x = 1 24 

1 

3x = 6 

144 ' 3 

144 ' 2 

144 

1 

3x = 1 

144 

3x = 144 

x = 48 

She would have to travel 48 km/h downhill in order for her average to be 24 km/h for the entire 

route. 


(–1, 1) 

(1, 1) 

A = 1 ( 

2 r2 ! " sin! ) 

A = 1 2 

2 

( ) 2 * $ # ' 

& 

% 2 ( 

)" sin $ 

& 

# '- 

, 

% 2 )/ 

+ 

(. 

A = 1 ( 

2 2 

* 

), 

# + 2 "1 

- 

/ 

. 

A = # 2 "1 



xy = z 4 

z = 4xy 

increase x by 50% (1 + 0.5) 

decrease y by 25% (1 – 0.25) 

z = 4(1.5x)(0.75y) 

z = 4(1.125xy) 

z is increased by 12.5%. 


Chapter 7 Review 

Chapter 7 Review Question 1 Page 408 

a) 4 3 b) 4 1 c) 

1 

4 = ! 

1 

$ 

2 4!2 d) (2) 5 = 4 

2 

# & 

" % 

5 

= 4 

2 

5 


a) 5 x = 20 

x log5 = log 20 

x = 

log 20 

log 

5 

5 = 20 

log 20 

log5 

b) 5 x = 0.8 

x log5 = log0.8 

x = log0.8 

log5 

log 0.8 

log 

5 

5 = 0.8 


a) 3 5x = ( 3 ) 3 x!1 

3 5x = 3 3x!3 

5x = 3x ! 3 

2x = !3 

b) ( 2 ) 3 2x+1 = ( 2 ) 5 x!3 

2 6x+3 = 2 5x!15 

6x + 3 = 5x !15 

x = !18 

x = ! 3 2 

Check a): 


Check b): 


! 

a) 40 = 50 1 $ 

" 

# 2% 

& 

4 

5 = ! 1 $ 

" 

# 2% 

& 

1.6 

h 

1.6 

h 

log 4 5 = 1.6 

h log 1 2 

1.6 

h = log 4 5 

log 1 2 

1.6log 1 

h = 2 

log 4 5 

h &= 5 

The half-life of cobalt-60 is approximately 5 min. 


! 

b) 0.05 = 1 $ 

" 

# 2% 

& 

t 

5 

log0.05 = t 5 log 1 2 

t 

5 = log0.05 

log 1 2 

t = 5log0.05 

log 1 2 

t &= 21.6 

It takes approximately 21.6 min for the sample to decay to 5% of its initial amount. 


a) log3 x!2 = log5 x 

(x ! 2)log3 = x log5 

x log3! 2log3 = x log5 

x log3! x log5 = 2log3 

x(log3! log5) = 2log3 

x = 

2log3 

log3! log5 

b) log 2 k !2 = log3 k +1 

(k ! 2)log 2 = (k +1)log3 

k log 2 ! 2log 2 = k log3+ log3 

k log 2 ! k log3 = 2log 2 + log3 

k(log 2 ! log3) = 2log 2 + log3 

k = 

2log 2 + log3 

log 2 ! log3 



a) x &= –4.301 k &= –6.129 

b) 3 x – 2 = 5 x 

2 k – 2 = 5 k + 1 



a) (4 x ) 2 – 4 x – 20 = 0 

a = 1, b = –1, c = –20 

4 x = !(!1) ± (!1)2 ! 4(1)(!20) 

2(1) 

4 x = 

1± 81 

2 

4 x = 1± 9 

2 

4 x = 5 or 4 x = !4 extraneous root since the base is negative 

x log 4 = log5 

x = log5 

log 4 

b) 2 x (2 x ) +12(2 ! x )(2 x ) ! 7(2 x ) = 0 

( 2 ) x 2 ! 7 2 x 

( ) +12 = 0 

(2 x ! 4)(2 x ! 3) = 0 

2 x = 4 or 2 x = 3 

2 x = 2 2 x log 2 = log3 

x = 2 

x = log3 

log 2 


! 

a) 1516 = 2000 1 $ 

" 

# 2% 

& 

0.758 = 1 2 

1 

h 

log0.758 = 1 h log 1 2 

1 

h = log0.758 

log 1 2 

h = 

log 1 2 

log0.758 

h &= 2.5 

The half-life is approximately 2.5 years. 

1 

h 

t 

2.5 

b) 0.10 &= 1 2 

log0.10 &= 

t 

2.5 log 1 2 

t log0.10 

&= 

2.5 

log 1 2 

t &= 2.5log0.10 

t &= 8.3 

log 1 2 

It will take approximately 8.3 years to 

depreciate to 10% of its purchase price. 



a) log 6 

(8 ! 27) = log 6 

216 

= 3 

c) log 2 2 + log5 2 = log 4 + log 25 

= log(4 ! 25) 

= log100 

= 2 

! 128$ 

b) log 4 

" 

# 8 % 

& = log 16 4 

= 2 

d) log3 2 + log12.5 = log9 + log12.5 

= log(9 !12.5) 

= log112.5 

&= 2.05 


" 8 ! 4% 

a) log 7 

# 

$ 16 & 

' = log 2 7 

! 

b) log a 2 + log(3b) – log c = log 3a2 b$ 

# & , a > 0, b > 0, c > 0 

" c % 


a) log a 2 + log b + log c = 2 log a + log b + log c, a > 0, b > 0, c > 0 

1 

b) log k ! log m = log k ! log m2 

= log k ! 1 log m, k > 0, m > 0 

2 

Chapter 7 Review Question 12 Page 408 

" 

a) log 2m + 6 % 

# 

$ 

m 2 ! 9 & 

' = log " 2(m + 3) % 

# 

$ (m ! 3)(m + 3) & 

' 

" 2 % 

= log 

# 

$ m ! 3& 

' ,m > 3 

" 

b) log x2 + 2x !15% 

" (x + 5)(x ! 3) % 

$ 

# x 2 ' = log 

! 7x +12& 

# 

$ (x ! 4)(x ! 3) & 

' 

" 

= log x + 5 % 

# 

$ x ! 4& 

' , x > 4 or x < !5 



a) log(2x +10) = 2 

2x +10 = 10 2 

2x = 100 !10 

2x = 90 

x = 45 

b) 1! log(2x) = 0 

1 = log(2x) 

10 1 = 2x 

x = 5 


log 2 

! 

" x(x + 2) # 

$ = log 8 2 

x 2 + 2x = 8 

x 2 + 2x % 8 = 0 

(x + 4)(x % 2) = 0 

x = 2 

x = %4 is an extraneous root since it would 

make log 2 

x and log 2 

(x + 2) undefined. 




! 

a) t = 5 log0.2 $ 

" 

# log0.5% 

& 

&= 11.6 

It takes approximately 11.6 h for the caffeine to drop to 20%. 

! 

b) 3 = 5 log P $ 

" 

# log0.5% 

& 

3 

5 = log P 

log0.5 

3 

log0.5 = log P 

5 

3 

log0.5 5 

= log P 

3 

P = 0.55 

P &= 0.66 

Approximately 66% of the caffeine will remain in the body at noon. 


a) P = 500, i = 0.066 , n = 2t 

2 

= 0.033 

A = P(1 + i) n 

= 500(1+ 0.033) 2t 

= 500(1.033) 2t 

b) A = 500(1.033) 10 

&= 691.79 

The investment is worth $691.79 after 5 years. 

c) 2 = (1.033) 2t 

log 2 = 2t log1.033 

2t = 

t = 

log 2 

log1.033 

log 2 

2log1.033 

t &= 10.67 

It will take approximately 10.67 years for the investment to double. 


d) i) A = 500(1.033) 2t + 500(0.01) 

A = 500(1.033) 2t + 5, t > 3 

The graph would be the same shape translated up 5 units. 

ii) A = 1000(1.033) 2t 

The graph would be above the original function and increasing at a faster rate. 

Chapter 7 Chapter Problem Wrap-Up 

Solutions to the Chapter Problem Wrap-Up are provided in the Teacher’s Resource. 


Chapter 7 Practice Test 

Chapter 7 Practice Test Question 1 Page 410 

The correct solution is C. 

( 2 ) 1 3 

3 2 

= 2 

2 


The correct solution is A. 

2 x+1 = ( 2 ) 2 x 

2 x+1 = 2 2x 

x +1 = 2x 

x = 1 


The correct solution is D. 

The sum of the logarithms equals the logarithm of the product. 


The correct solution is C. 

" 5!10% 

log 2 

# 

$ 25 & 

' = log 2 2 

= 1 


" % 

$ 6 ! 4' 

" 

log 3 $ ' = log 

$ 

8 

3 

24 ! 3 % 

' # 

$ 8& 

' 

# 3 & 

= log 3 

9 

= 2 



a) ( 2 ) 3 x+2 = ( 2 ) 2 2x!1 

2 3x+6 = 2 4x!2 

3x + 6 = 4x ! 2 

4x ! 3x = 6 + 2 

x = 8 

b) 3 x = ( 3 ) 4 x!4 

3 x = 3 4x!16 

x = 4x !16 

4x ! x = 16 

3x = 16 

x = 16 3 

c) 1 = log(x + 5) ! log 2 

" 

log10 = log x + 5 % 

# 

$ 2 & 

' 

10 = x + 5 

2 

20 = x + 5 

x = 15 

d) log8 ! log(x ! 2) = 1 

" 8 % 

log 

# 

$ x ! 2& 

' = log10 

8 

x ! 2 = 10 

8 = 10x ! 20 

10x = 28 

x = 2.8 


a) 


) 

c) 


d) 


a) log (x + 1) = 5 log x – 2 

b) x &= 3.37, from the graph 

c) log(x +1) = 5log x ! 2 

2 = 5log x ! log(x +1) 

log100 = log x 5 ! log(x +1) 

" x 5 % 

log100 = log$ 

# x +1 

' 

& 

100 = x5 

x +1 

100x +100 = x 5 

x 5 !100x !100 = 0 

Since x &= 3.37 is the only value where log (x + 1) and log x are defined, it is the solution. 

The other values are extraneous roots. 



! 

a) 41 = 50 1 $ 

" 

# 2% 

& 

0.82 = 1 2 

10 

h 

10 

h 

log0.82 = 10 h log 1 2 

10 

h = log0.82 

log 1 2 

h = 

h &= 35 

10log 1 2 

log0.82 

The half-life is approximately 35 min. 

! 

b) 0.01 = 1 $ 

" 

# 2% 

& 

t 

35 

log0.01 = t 

35 log 1 2 

t 

35 = log0.01 

log 1 2 

t = 35log0.01 

log 1 2 

t &= 232.5 

It will take approximately 232.5 min (approximately 3 h, 52 min, and 30 s) for the material to 

decay to 1% of its initial amount. 



! 

0.10 = 1 $ 

" 

# 2% 

& 

33 

h 

log0.10 = 33 

h log 1 2 

33 

h = log0.10 

log 1 2 

h = 

h &= 9.9 

33log 1 2 

log0.10 

The half-life is approximately 9.9 min. 


a) log(x !1) 2 = log(x +1) 

b) 

(x !1) 2 = x +1 

x 2 ! 2x +1! x !1 = 0 

x 2 ! 3x = 0 

x(x ! 3) = 0 

x = 3 

x = 0 is an extraneous root since log(x !1) 

would be undefined 


Chapter 7 Practice Test Question 12 Page 411 


Yes, a quadratic curve of best fit could be drawn through the points since the points are 

nearly the shape of a quadratic. 


Yes, an exponential curve of best fit could be drawn through the points since they follow the 

shape of an exponential equation. 


i) population of a bacteria culture over time 

ii) surface area affected by an oil spill over time 


a) 

b) 

c) 

y &= 0.17x 2 – 5.34x + 97.93 

y &= 95.67(0.96) x 

d) Answers may vary. A sample solution is shown. 

The exponential model is better for predicting values for x > 15.74. The quadratic model 

gives increasing values, which is not realistic. 


e) i) y &= 95.67(0.96) 15 

y &= 49.6 

The tea is approximately 49.6°C after 15 min. 

ii) 38 &= 95.67( 0.96) x 

0.40 &= 0.96 x 

log0.40 &= x log0.96 

x &= log0.40 

log0.96 

x &= 21.1 

It would take approximately 21.1 min for the tea to reach 38°C. 

f) Answers may vary. A sample solution is shown. 

The temperature of the room was constant.

MHR • Advanced Functions 12 Solutions 660

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