3
3A 3B 3C 3D 3E 3F
The index laws Logarithm laws Exponential equations Logarithmic equations using any base Exponential equations (base e) Equations with natural (base e) logarithms 3G Inverses 3H Literal equations 3I Exponential and logarithmic modelling
Exponential and logarithmic equations arEas oF sTudy
• Logarithm and exponent laws • Identification of appropriate solution processes for solving logarithmic and exponential equations • Numerical solutions of exponential and logarithmic equations • Solution of literal equations such as emx + n = k
• Application of algebraic and logarithmic properties to the simplification of expressions and the solution of equations • Finding the rule of an inverse function involving logarithmic and exponential equations • Application of logarithmic and exponential functions in modelling practical situations eBook plus
3a
Digital doc
The index laws
10 Quick Questions
A number in index form has two parts, the base and the index, power, exponent or logarithm. A number in index form is represented like this: Index, power, exponent or logarithm
ax Base
The index laws are summarised below. ax = x a
ax ÷ ay = ax - y
a y = ax =
(ax)y = axy
(ab)x = axbx
a0 = 1 (a ≠ 0)
ax a b = b x , b ≠ 0
a
132
1
ax × ay = ax + y
−x
=
x
1 (a ≠ 0) ax
maths Quest 12 mathematical methods Cas for the Casio Classpad
y
x
( a) y
x
Worked Example 1
Simplify
( 2 x 2 y3 ) 3 × 3( xy4 )2 . 6 x 4 × 2 xy4
Think
Write
1
Remove the brackets by multiplying the indices.
2
Add the indices of x and add the indices of y. Simplify 23 to 8 and multiply the whole numbers.
3
Subtract the indices of x and y. Divide 24 by 12.
(2 x 2 y 3 )3 × 3( xy 4 )2 23 x 6 y 9 × 3 x 2 y8 = 6 x 4 × 2 xy 4 12 x 5 y 4 =
24 x8 y17 12 x 5 y 4
= 2x3y13
For negative indices and fractional or decimal indices, the same rules apply.
Worked Example 2
Write in simplest form: 2
− 0.4
a 64 3 b 32 −2
c 125
3
using a CAS calculator.
THINK
WRITE/display x
a
b
2 y
1
Rewrite using the index law a y = a x .
2
Rewrite using a x = ( y a ) x .
= ( 3 64 )2
3
Simplify by taking the cube root of 64.
= 42
4
Square 4.
a 64 3 =
y
3
64 2
= 16 −0.4
1
Write as a fraction with a positive index.
2
Change 0.4 to 104 .
3
Simplify the fractional index.
4
Rewrite using the index law a y = a x .
5
Simplify by taking the 5th root of 32.
6
Square 2.
b 32
= = =
1 320.4 1 4
3210 1 2
32 5 x
y
=
1 5
( 32 )2 1 = 2 2 1 = 4
Chapter 3 Exponential and logarithmic equations
133
−2
c
1
c
To simplify 125 3 with a CAS calculator, open the Main page. Complete the entry line as: −2
125 3 Then press E.
2
−2 3
Write the answer.
125
1
= 25
Worked Example 3
Simplify, leaving your answer with positive indices: a
−
−
a 2 b4 × ( a 3 b 4 )
−1
1 −1 a2 b b − 1 2 3 b
−1
using a CAS calculator.
THINK a
1
Remove the brackets by multiplying the indices.
2
Add the indices of a and of b. Place a5 in the denominator with a positive index.
3
b
WRITE/display
1
In the Main page, complete the entry line as: −1 12 1 a b . 3 1 b2 Highlight the equation and tap: • Interactive • Transformation • simplify
-
- -1
a a 2b4 × (a3b 4)
-
-
= a 2b4 × a 3b4 -
= a 5b8 =
b8 a5
b
−
−
2
Write the answer. The answer is expressed with positive powers.
12 −1 a b −1 2 3 b
−1
=
b
3
3 a
If the expression contains different numbers that do not have the same base, write each number as a product of prime factors.
134
Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad
WorkEd ExamplE 4
Simplify
3 n × 6 n + 1 × 12 n − 1 : 32 n × 8 n
a showing working, and
Think a
WriTE/display
1
Write each number as the product of prime factors.
2
Remove the brackets.
3
In the numerator, add the indices of numbers with base 3 and add indices of numbers with base 2. Subtract the indices.
a
3n × 6 n + 1 × 12 n − 1 32 n × 8 n 3n × (3 × 2) n + 1 × (22 × 3) n − 1 = 32 n × ( 2 3 ) n 3n × 3n + 1 × 2 n + 1 × 2 2 n − 2 × 3n − 1 32 n × 2 3 n 3 n 3 n − 1 3 ×2 = 2n 3 × 23 n =
-1
5
Write the term with a negative index in the denominator with a positive index.
= 3n × 2 1 = 3n × 2
6
Simplify.
=
1
On the Main page, complete the entry line as: 3n × 6 n + 1 × 12 n − 1 simplify 32 n × 8 n
4
b
b with a CAS calculator.
3n 2
b
Highlight the equation and then tap: • Interactive • Transformation • simplify
2
Write the answer.
3n × 6 n + 1 × 12 n − 1 3n = 2 32 n × 8 n
WorkEd ExamplE 5
eBook plus
Simplify, leaving your answer with positive indices. 3 3 1 − a x 2− − b − + −1 2 1 x x +2 x −2 Think a
1
Tutorial
int-0528 Worked example 5
WriTE
Rewrite the expression with positive indices.
a x
3 − x 2 1 = 2 − 3x 2 x
−2
−
Chapter 3
Exponential and logarithmic equations
135
2
Find the lowest common denominator.
b
=
1 3x 4 − 2 x2 x
1 − 3x 4 x2 1 3 b + = 1 1 x + 2 x − 2
Simplify.
1
Rewrite the question using positive indices.
2
Find the common denominator for the terms in the brackets and simplify. Follow the process for division of fractions (change the division sign to a multiplication sign and invert the second fraction).
=
4
Find the common denominator.
5
Expand the brackets on the numerator.
6
Simplify the numerator by adding like terms.
Factorise the numerator and write the answer.
7
1 x2 − 3x 2 × 2 2 x x
=
3
3
=
3 x + 1 + 2x 1 − 2x x x
=
3x x + 1 + 2x 1 − 2x
=
3 x (1 − 2 x ) + x (1 + 2 x ) (1 + 2 x ) (1 − 2 x )
=
3x − 6 x 2 + x + 2 x 2 (1 + 2 x ) (1 − 2 x )
=
4x − 4x2 (1 + 2 x ) (1 − 2 x )
∴
4 x (1 − x ) (1 + 2 x ) (1 − 2 x )
REMEMBER
1. axay = ax + y ax 2. ax ÷ ay = ax − y or y = a x − y a 3. (ax)y = axy 4. a0 = 1, a ≠ 0 1 1 5. a x = x and − x = a x a a −
1
x
y
y
y
6. a y = a and a y = a x = ( a ) x Exercise
3A
The index laws 1 WE1 a
x3
Simplify: ×
x4
x7
( x 2 )3 × x 5 x e ( x 5 )2
136
d (x 3) 2 16
b x7 ÷ x 2 x5
c (x2) 5 x10
5 x 2 y 4 × 4 x 5 y 5x4y3 f 22 x 3 y 2
2 3 4 2 4 2 g (2 xy ) × 5( x y ) 10 x y 3 4 x 5 y 3 × 3x 2 y 3
Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad
x
2 WE2
Simplify:
2 a 27 3
−3
1 125
b 25 2
9
c 810.25 3
−3
2
27 3 d 64
243 5 e 32
9 16
8 27
3 Simplify: a
x4 × x5 x6 x3
3
−3
1
d 9 2 49 4 WE3
c 361.5 216
b 16 4 8 e 256 4 81
3 7
27 64
Simplify, leaving your answer with positive indices. 1
a
3x 3y2
-4
× (x2y)
−3 − 1 2 x 2y2
c
3 11 2 x y
b 5
5 5 3 2 × 9x 2 y2 6x 4 y 3
1 2 2x 2 y 3
3 2 a2b c d − 1 3a 2 bc 2
5 −1 −1 2 × 9 x 5 y 2 243x 2 y2
−2
−
−
−
5 Simplify: 4y-1
×
a x
c
1 16 2
(x 2y3) 1
−1 2 −1 2 x5y 4
x6 4 y
b
1 2 12 × 4x 5 y 2
−1 3 5x 3 y 4
2
1 2 −1 20 x × 83 x 3 y 2 1 y2
−3 3 a 2b4 d ab 2
3 8y 8
3
2 a 3 b3 3 ÷ 3 1 2 9 3 a c a b
−2
1
15 7
9a − 3 b 2 2a 2 b 2 ÷ 2 3 3 4a b
6 WE4 Simplify: a 2n × 4n + 1 × 8n - 1 26n − 1 b 3n × 9n - 1 × 27n + 1 36n + 1 c 2n × 3n + 1 × 9n 2n × 33n + 1 d
32 × 2
−3
3 92
× 16
7 Simplify: a
2n - 1
2 3
22n × 22n + 1
×
3n
× 6n + 1
b
52 × 3
−1
125 × 9
−2
÷
27 1 5
8 WE5 Simplify, writing your answer as a single fraction with positive indices. −
a x 1 + c
x
1 x
−1
1 + x2 x
b (x 1 + x 2) 2
1 1 2x + − + 1 x 1 − 1 1 − x2
-
( x + 1)2 x4 -
d 2x(x2 - y2) 1 - (x - y) 1
−1
1 x+y
−
9 MC 3 x + 3x is equal to: A 1
✔ B
1 + 32 x 3x
C 3
− x2
D 6
x E 1 + 3 3x
Chapter 3 Exponential and logarithmic equations
137
3B
Logarithm laws If a > 0, then N = ax ⇔ loga (N) = x. For example, an expression in index form can also be rewritten in logarithmic form. 8 = 23 ⇔ log 2 (8) = 3 • Since ≠ 0 then loga (0) is undefined. • a0 = 1 ∴ loga (1) = 0 • a1 = a ∴ loga (a) = 1 Let m = ax ⇔ loga (m) = x and n = ay ⇔ loga (n) = y. • mn = ax × ay ∴ loga (mn) = x + y = ax + y = loga (m) + loga (n) x m m a • = ∴ log a = x − y n n ay = ax - y = loga (m) − loga (n) • mp = (ax)p ∴ loga (mp) = px = axp = p loga (m) ax
Change-of-base rule Suppose b = ax, then loga (b) = x. Consider N = by, then logb (N) = y. But N = by = (ax)y = axy. Therefore, loga (N) = xy = loga (b) logb (N). Thus, log b ( N ) =
log a ( N ) . log a (b)
This is called the change-of-base rule.
Worked Example 6
Evaluate: a log2 (1) b log5 (5). Think
Write
a
Log of 1 to any base is equal to zero: loga (1) = 0.
a log2 (1) = 0
b
If the number and base are equal the answer is 1: loga (a) = 1.
b log5 (5) = 1
Worked Example 7
Write in index form: a log2 (8) = 3 b logx (81) = 4. Think
138
Write
a Use ax = y ⇒ loga (y) = x.
a log2 (8) = 3 ⇔ 23 = 8
b Use ax = y ⇒ loga (y) = x.
b logx (81) = 4 ⇔ x4 = 81
Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad
WorkEd ExamplE 8
Simplify: a log10 (5) + log10 (2)
b log4 (20) - log4 (5)
c log2 (16)
Think a
b
c
d
d log 5
( x ). 5
WriTE a log10 (5) + log10 (2) = log10 (5 × 2)
1
Rewrite using loga (mn) = loga (m) + loga n.
2
Simplify.
= log10 (10)
3
Simplify using loga (a) = 1. m Rewrite using log a = log a (m) − log a ( n). n
=1
1
20 5
b log 4 (20) − log 4 (5) = log 4
2
Simplify.
= log 4 (4)
3
Simplify using loga (a) = 1.
=1
1
Rewrite 16 as a number with base 2.
2
Rewrite using loga (mp) = p loga (m).
= 4 log2 (2)
3
Simplify using loga (a) = 1.
=4
1
Rewrite using y a = a y .
2
Rewrite using loga (mp) = p loga (m).
1
c log2 (16) = log2 (24)
d log 5
( x ) = log x 5
5
1 5
1
= 5 log 5 ( x )
WorkEd ExamplE 9
Simplify log3 (27) + log3 (9) - log3 (81). Think
WriTE/display
1
On the Main page, complete the entry line as: log3(27) + log3(9) + log3(81) Then press E.
2
Write the answer.
log3 (27) + log3 (9) +log3 (81) = 1
WorkEd ExamplE 10
eBook plus
Simplify: a 2 + log10 (3)
b 3 log3 (6) - 3 log3 (18)
log 3 ( 9 ) c . log 3 ( 27 )
Chapter 3
Tutorial
int-0529 Worked example 10
Exponential and logarithmic equations
139
Think a
b
1
c
Write
Write 2 as 2 log10 (10) because log10 (10) = 1. (mp)
a 2 + log10 (3) = 2 log10 (10) + log10 (3)
= log10 (102) + log10 (3)
= p loga (m).
2
Rewrite using loga
3
Rewrite using loga (mn) = loga (m) + loga (n).
= log10 (102 × 3)
4
Write 102 as 100.
= log10 (100 × 3)
5
Multiply the numbers in the brackets. (mp)
= log10 (300) b 3 log3 (6) − 3 log3 (18) = log3 (63) − log3 (183)
= p loga (m).
1
Rewrite using loga
2
Rewrite using m log a = log a (m) − log a ( n). n
3
Write 63 as 6 × 6 × 6 and 183 as 18 × 18 × 18.
4
Simplify.
1 = log3 3 3
5
Write the numbers with the base 3.
= log3 (3 3)
6
Rewrite using loga (mp) = p loga (m).
= −3 log3 (3)
7
Simplify using loga a = 1.
= −3 × 1 = −3
1
Write the numbers with the same base. It is not possible to cancel the 9 and the 27 because they cannot be separated from the log.
2
Rewrite using loga (mp) = p loga (m).
=
2 log3 (3) 3 log3 (3)
3
Cancel the logs because they are the same.
=
2 3
63 = log3 3 18 6×6×6 = log3 18 × 18 × 18
−
c
log3 (9) log3 (32 ) = log3 (27) log3 (33 )
Worked Example 11
Calculate the value of log2 (18), correct to 2 decimal places. Think
140
1
On the Main page, complete the entry line as: log2(18) Then press E. Highlight the answer and tap . to get a decimal approximation.
2
Write the answer.
Write/display
log2 (18) = 4.17, correct to 2 decimal places.
Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad
REMEMBER
For a > 0: 1. loga (1) = 0 2. loga (a) = 1 3. loga (0) is undefined. 4. loga (x) does not exist when x ≤ 0. 5. loga (mn) = loga (m) + loga (n) Exercise
3B
m 6. log a = log a (m) − log a ( n) n 7. loga (mp) = p loga (m) log a ( N ) 8. log b ( N ) = log a (b)
Logarithm laws 1 WE6 Evaluate the following. 0 a log3 (1) 0 b log5 (1)
c log2 (2) 1
d log6 (6) 1
2 WE7 Write the following in index form. a log2 (16) = 4 24 = 16 b logx (25) = 2 x2 = 25 d log3 (x) = 5 35 = x 3
1 − e log 5 5 = −1 5 1 =
1 5
Write the following in logarithmic form. a 23 = 8 log2 (8) = 3 b 34 = 81 log3 (81) = 4 − log5 (125) = x d 5x = 125 = 12 log2 ( 12 ) = −1 e 2 1
4 WE8
= x log4 (x) = 3 c 43 3 f x = 27 logx (27) = 3
Simplify:
a log6 (3) + log 6 (2) 1 d log3 (81) 4 5
c log5 (125) = x 5x = 125
b log2 (10) − log2 (5) 1 e
log 5 1 5
c log2 (32) 5 f log3
−
1
1 27
−3
Simplify: a log 2 ( x )
1 2
log2 (x)
b 3 log3 ( 3 x ) log3 (x)
6 WE9 Simplify: a log4 (10) + log4 (2) − log 4 (5) 1 c 1 log10 (16) + log10 (52 ) 2
x4 y 2
c log 2
x2 log2 y
b log5 (25) + log5 (125) − log5 (625) 1 d log3 (2) − log3 (10) + log3 (15) 1
2
e log2 (16) + log2 (8) + log2 (4) 9 7 WE10 Simplify: a 4 log2 (12) − 4 log2 (6) 4 d
log 2 (64) 2 log 2 (8)
b 2 + log5 (10) − log5 (2) 3 e
log a ( x ) log a ( x )
1 2
8 WE11 Evaluate correct to 3 decimal places. a log10 (3) 0.477 b log5 (4) 0.861 d log2 (0.8) −0.322 9
2.161 e log4 (20)
Simplify: a 5 log3 (x) + log3 (x2) − log3 (x7) 0
c 1 + log2 (5) log2 (10)
c log10 (0.5) −0.301 f log3 (60) 3.727
b 4 log2 (x) + log2 (x3) − log2 (x6 ) log2 (x)
c 3 log4 (x) − 5 log4 (x) + 2 log4 (x) d 4 log6 (x) − 5 log6 (x) + log6 (x) 0 0 e log10 (x2) + 3 log10 (x) − 2 log10 (x) 3 log10 (x) f 4 log10 (x) − log10 (x) + log10 (x2 ) 5 log10 (x) g log5 (x + 1) + log5 (x + 1)2
h log4 (x − 2)3 − 2 log4 (x − 2) log4 (x − 2)
log5 (x + 1)3 or 3 log5 (x + 1)
Chapter 3 Exponential and logarithmic equations
141
10 MC 2 log10 (5) − log10 (20) + log10 (8) is equal to: ✔ C 1 B −log10 (2) A log10 (2)
D −1
E log10 (4)
11 MC If loga (b) = 2, then b is equal to: ✔ E a2 A 0 B 1 C 2 D a 12 If y = a log10 (x), find x when a = 2 and y = 3. Give your answer correct to 3 decimal places. 31.623
3C
Exponential equations The equation ax = b is an example of a general exponential (or indicial) equation and 2x = 32 is an example of a more specific exponential equation. To solve one of these equations it is necessary to write both sides of the equation with the same base if the unknown is an index or with the same index if the unknown is the base.
Worked Example 12
Solve for x in each of the following. 1 a 2x = 32 b 3 x = c 2 × 3x = 162 d 2(1 − x) = 16 27 Think a
1
b
The indices are equal because the base is 2 on each side of the equation.
1
Write 27 with base 3.
3 1
d
Write 32 with base 2, the same as the left-hand side.
2
2
c
Write a 2x = 32
2x = 25 x=5
1 27 1 = 3 3 − 3x = 3 3
b 3x =
Write 1 as a number with base 3. 33 Equate the indices. Divide both sides by 2 to leave 3x on the left-hand side.
2
Write 81 as a number with base 3.
3
Equate the indices.
1
Write 16 with base 2.
2
Equate the indices.
3
Solve for x.
x = −3 c 2 × 3x = 162
3x = 81
3x = 34 x=4 d
= 16 21 − x = 24
21 − x
1−x=4 x = −3
Worked Example 13
Solve 5x × 252x − 3 = 625 for x: a using index laws b with a CAS calculator. Think a
1
Write/display
Write all numbers with the same base.
5x × 252x − 3 = 625 5x × (52)2x − 3 = 54
2
142
Simplify.
5x × 52(2x − 3) = 54
Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad
5x × 54x - 6 = 54
3
Remove the brackets in the index.
4
Add the indices on the left-hand side.
55x - 6 = 54
5
Equate the indices.
5x - 6 = 4
6
Solve the equation.
5x = 10 x=2
b
1
On the Main page, enter the equation as: 5x × 252x - 3 = 625 Highlight it and tap: • Interactive • Advanced • solve • OK
2
Write the solution.
Solving 5x × 25(2x - 3) = 625 for x gives x = 2.
Sometimes it is possible to use the methods for solving quadratic equations to help solve indicial equations. Remember that 22x = (2x)2. WorkEd ExamplE 14
eBook plus
Solve for x in the following. b 32x - 12 × 3x + 27 = 0
a (2x - 16)(2x + 4) = 0 Think a
b
1
c 4x - 2x + 3 + 16 = 0 WriTE
Use the Null Factor Law to solve by making each bracket equal to zero.
(2x - 16) = 0 or (2x + 4) = 0
Solve each equation.
2x = 16 or 2x = -4
3
Write 16 as a number with base 2 but -4 can not be written with base 2.
2x = 24 or no real solution
4
Solve by equating the indices.
1
Write
as
(3x)2.
int-0530 Worked example 14
a (2x - 16)(2x + 4) = 0
2
32x
Tutorial
x=4 b 32x - 12 × 3x + 27 = 0
(3x)2 - 12 × 3x + 27 = 0
2
Let 3x = a to make a simpler quadratic equation to solve.
a2 - 12a + 27 = 0, where a = 3x
3
Factorise.
(a - 3)(a - 9) = 0
4
Use the Null Factor Law by making each bracket equal to zero.
a - 3 = 0, a - 9 = 0
5
Solve for a.
6
Substitute back a = 3x.
3x = 3, 3x = 9
7
Write numbers with base 3.
3x = 31, 3x = 32
a = 3, a = 9
Chapter 3
Exponential and logarithmic equations
143
c
8
Equate the indices.
1
4x
Rewrite
as
(2x)2
x = 1, x = 2 and
2x + 3
as
2x
×
23.
4x – 2x + 3 + 16 = 0 − 2x × 23 + 16 = 0
c
(2x)2
(2x)2 − 2x × 8 + 16 = 0
2
Rewrite 23 as 8.
3
Let 2x = a to make a simpler quadratic equation to solve.
4
Replace a × 8 with 8a because the coefficient precedes the pronumeral.
5
Factorise.
(a − 4)(a − 4) = 0
6
Use the Null Factor Law.
a − 4 = 0, a − 4 = 0
7
Solve for a.
a = 4 and a = 4
8
The two factors are equal because a2 − 8x + 16 = 0 is a perfect square.
9
Substitute back a = 2x.
2x = 4
10
Write 4 as a number with base 2.
2x = 22
11
Solve by equating the indices.
a2 − a × 8 + 16 = 0 where a = 2x a2 − 8a + 16 = 0
x=2
Remember to always make the right-hand side equal to zero when solving quadratic equations. It is a good idea to substitute your answer back into the original equation to check the accuracy of your work. If the base is not the same and the numbers cannot be written with the same base, then logarithms can be used. It is possible to take the logarithm of both sides of an equation provided the same base is used.
Worked Example 15
Solve for x in the following. Give your answers in exact form using base 10 and correct to 3 decimal places. a 5x = 10 b 2(x + 1) = 12 Think a
b
144
Write a
5x = 10 log10 (5x) = log10 (10)
1
Take the logarithm of both sides to base 10.
2
Use loga (mp) = p loga (m) and loga (a) = 1.
x log10 (5) = 1
3
Divide both sides by log10 (5).
x=
4
Use a calculator to simplify.
x ≈ 1.431, correct to 3 decimal places.
1
Take the logarithm of both sides to base 10.
2
Use loga (mp) = p loga (m) to simplify.
3
Divide both sides by log10 (2).
b
1 (exact form) log10 (5)
2(x + 1) = 12 log10 (2(x + 1)) = log10 (12) (x + 1) log10 (2) = log10 (12)
Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad
( x + 1) =
log10 (12) log10 (2)
4
Use a calculator to simplify the right-hand side.
5
Solve for x.
∴ x=
log10 (12) − 1 (exact form) log10 (2)
x ≈ 2.585, correct to 3 decimal places.
Note: Logarithms in bases other than 10 may be used. Inequalities are worked in exactly the same way except that there is a change of sign when dividing or multiplying both sides of the inequality by a negative number. Worked Example 16
Solve the following equations for x, giving your answers both in exact form and correct to 3 decimal places. a 2x > 5 b 0.5x ≤ 1.4 Think
Write/display
1
On the Main page, enter the equations in Alg Standard mode as: 2x > 5 0.5x ≤ 1.4 Highlight each equation and tap: • Interactive • Advanced • solve • OK To get an answer in decimal form, either highlight it and tap . or repeat the above process in Alg Decimal mode.
2
Write the answers in exact form. The calculator defaults to base e when solving exponential equations in exact form. Note: ln (A) ⇔ loge (A). This will be discussed later in this chapter. Note: In part b , the inequality ≤ has been 1 changed to ≥ because log e 2 < 0.
a Solving 2x > 5 for x gives
Write the answers correct to 3 decimal places.
a Solving 2x > 5 for x gives x > 2.322,
3
x>
log e (5) . log e (2)
b Solving 0.5x ≤ 1.4 for x gives −
x≥
7 log e 5 log e (2)
.
correct to 3 decimal places.
b Solving 0.5x ≤ 1.4 for x gives x ≥ −0.485, correct
to 3 decimal places.
REMEMBER
1. The equations ax = y and 2x = 32 are exponential equations. 2. Write numbers with the same base to help simplify problems. The most commonly used bases are 2, 3 and 5. 3. If the base is the same, equate the indices. 4. If the indices are the same, equate the bases.
Chapter 3 Exponential and logarithmic equations
145
5. Use the Null Factor Law to solve quadratic equations. 6. A negative number cannot be expressed in index form, for example, -4 cannot be expressed with base 2. 7. a2x = (ax)2 8. Take the logarithm of both sides of an equation or inequation using the same base. 9. Change the sign of an inequality when multiplying or dividing by a negative number. 10. loga (x) > 0 if x > 1 11. loga (x) < 0 if 0 < x < 1 ExErCisE
3C
Exponential equations 1
a
eBook plus
Index form
3x
2
b 10-x = 1000
= 81 4
d 7x = 1 49
Digital doc
SkillSHEET 3.1
Solve for x in each of the following.
WE12
a 3 × 2x = 48 4 3
Digital doc
c 2x × 4x - 1 = 16 4
3x
×
Digital doc
SkillSHEET 3.3 Solving indicial equations by equating the bases
7
Digital doc Solving liner inequations
c 52 x − 1 = 1 125
-1
d 22x - 6 = 1 3
b 5x × 52x + 1 = 625 1
3
d
2
33 x + 1 = 81 9x − 2
-1
Solve for x in the following. a - 1) = 0 2, 0 b 22x - 6 × 2x + 8 = 0 2, 1 2x x c 6 - 7 × 6 + 6 = 0 1, 0 d 4x - 6 × 2x - 16 = 0 3 x x e 9 =2×3 +3 1 b 42x - 20 × 4x = -64 1, 2
WE15 Solve for x correct to 3 decimal places.
a 2x = 5 2.322
b (0.3)x - 1 = 10 -0.912 c (1.4)2 - x = 6
d 3 × 5x = 27 1.365
e 5 × 7x = 1
-0.827
x ≥ -0.356
e (0.4)x > 0.2
-3.325
f 2x × 3x + 1 = 10 0.672
WE16 Solve for x correct to 3 decimal places. a 3x > 5 x > 1.465 b 22x ≤ 7 x ≤ 1.404
c (0.2)x > 3
x < -0.683
x < 1.756
mC The value of x for which 5 × 2x = 1255, correct to 3 decimal places, is:
✔D
9
-5
9)(3x
7.972
B 897.750
C 897.749
E 2.059
mC The solution to the equation 102x = 3 × 10x + 4 is:
A log10 (-1), log10 (4) D 0, 0.602
146
243
A 7.971
eBook plus SkillSHEET 3.4
3x - 1 =
d 7x ≥ 0.5 8
= 32
1 5
b 6x - 2 = 216 5
5 Solve for x in the following. a 25x + 4 × 5x - 5 = 0 0 6
1 2x
WE14
(3x
eBook plus
c
Solve for x in each of the following.
WE13
a
SkillSHEET 3.2
3
Solve for x in each of the following.
eBook plus
Solving equations
e 243x = 3
-2
-
B ✔E
-1,
4
log10 (4)
maths Quest 12 mathematical methods Cas for the Casio Classpad
C 10x + 1, 10x - 1
3d
logarithmic equations using any base The equation loga (y) = x is an example of a general logarithmic equation. Laws of logarithms and indices are used to solve these equations.
WorkEd ExamplE 17
Solve for x in the following equations. a log2 (x) = 3 b log3 (x4) = -16
c log5 (x - 1) = 2
Think a
1
b
c
WriTE
Rewrite using ax = y ⇔ loga (y) = x.
23 = x
x=8
2
Rearrange and simplify.
1
Rewrite using loga
(mp)
2
Divide both sides by 4.
3
Rewrite using ax = y ⇔ loga (y) = x.
4
Rearrange and simplify.
1
Rewrite using ax = y ⇔ loga (y) = x.
2
a log2 (x) = 3
= p loga (m).
b
log3 (x4) = -16 4 log3 (x) = -16 log3 (x) = -4 -
3 4 = x 1 34 1 = 81
x=
c log5 (x - 1) = 2
52 = x - 1
x - 1 = 25 x = 26
Solve for x.
The base of a logarithmic function and the base of an exponential function must be a positive real number other than 1. In the expression loga (x), a ∈ R+\{1}.
WorkEd ExamplE 18
eBook plus
Solve for x in each of the following: a logx (4) = 2
without technology
Think a
1 125
b log x
= − 3 with a CAS calculator.
Tutorial
int-0793
Worked example 18
WriTE/display
1
Rewrite using ax = y ⇔ loga (y) = x.
2
Solve the quadratic equation.
logx (4) = 2 x2 = 4
a
x2 - 4 = 0 (x - 2)(x + 2) = 0 x - 2 = 0 or x + 2 = 0 x = ± 2
Chapter 3
Exponential and logarithmic equations
147
3
Check to see if solutions are valid.
x = 2 is the only solution.
This is the only solution. The solution x = −2 is not valid because the base of a logarithmic function must be a positive real number other than 1. b
1
On the Main page, enter the equation as: log x
1 125
= −3
Highlight it and tap: • Interactive • Advanced • solve • OK
2
Solving log x
Write the solution.
1 125
= − 3 for x gives x = 5.
Worked Example 19
Solve for x in the following. a log2 (16) = x b log 3 1 = x c log9 (3) = x 3
Think a
b
1
Write
Rewrite using
ax
= y ⇔ loga (y) = x.
2
Write 16 with base 2.
3
Equate the indices.
1
Rewrite using ax = y ⇔ loga (y) = x.
a log2 (16) = x
2x = 16 2x = 24 x=4
b log 3 1 = x 3
1 3 1 = 1 3
3x =
c
148
1
2
Write 3 with base 3.
3
Equate the indices.
1
Rewrite using ax = y ⇔ loga (y) = x.
−1
3x = 3
x = −1 c log9 (3) = x
9x = 3
(32)x = 3
2
Write 9 with base 3.
3
Remove the brackets.
32x = 31
4
Equate the indices.
2x = 1
5
Solve.
Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad
x=
1 2
WorkEd ExamplE 20
Solve for x in the following. a log2 (4) + log2 (x) - log2 (8) = 3
b log10 (x) + log10 (x - 3) = log10 (4)
Think a
b
1
WriTE a log2 (4) + log2 (x) - log2 (8) = 3
Simplify the left-hand side. Use loga (mn) = loga (m) + loga (n) and m log a = log a (m) − log a ( n) . n
2
Rewrite using ax = y ⇔ loga (y) = x.
3
Solve.
1
Simplify the left-hand side by using loga (mn) = loga (m) + loga (n).
4 × x log 2 =3 8 x log 2 = 3 2 23 =
x 2
x = 2 × 23 =2×8 = 16 b log10 (x) + log10 (x - 3) = log10 (4)
log10x (x - 3) = log10 (4) x(x - 3) = 4
2
Equate the logs.
3
Expand.
4
Solve the quadratic equation.
x2 - 3x - 4 = 0 (x - 4)(x + 1) = 0 x = 4 or x = -1
5
x > 0, x - 3 > 0 because it is not possible to take the logarithm of a negative number, x > 3.
x = 4 is the only solution.
x2 - 3x = 4
rEmEmBEr
1. In a logarithmic equation, the unknown can be: (a) the number, log2 (x) = 5 (b) the base, logx (8) = 3 (c) the logarithm, log2 (4) = x. 2. In the expression loga (x), a is a positive real number other than 1. 3. Use laws of logarithms and indices to solve the equations. 4. Check that all solutions are valid. ExErCisE
3d eBook plus Digital doc
WorkSHEET 3.1
logarithmic equations using any base 1
WE17 Solve for x in the following. i log5 (x) = 2 25 a
i i i log10 (x2) = 4 v i i log5 (1 - x) = 4
1 8
i v log3 (x + 1) = 3
100
v log4 (2x - 3) = 0
i i log2 (x) = -3
2
vi log2
(-x)
=
-5
26 −1 32
-624
Chapter 3
Exponential and logarithmic equations
149
i i log4 (x) = −2
i log3 (x) = 4 81
b
i i i log2 (x3) = 12 16
i v log5 (x − 2) = 3 127 vi log3 (−x) = −2
v log10 (2x + 1) = 0 0 v i i log10 (5 − 2x) = 1
−5 2
2 WE18 Solve for x in the following. a i logx (9) = 2 3
−1 9
i i log x (25) = 23 125
i i i log x 1 = − 3 2
i v logx (62) = 2 6
i logx (16) = 4 2
i i log x (125) =
8
b
1 16
i i i log x
1 64
3 4
625
i v logx (43) = 3 4
= −2 8
3 WE19 Solve for x in the following. a
i i log 5 1 = x −1
i log2 (8) = x 3 i i i log4 (2) = x
5
i v log6 (1) = x 0
1 2
v log 1 (2) = x −1 2 b
1 16
i log3 (9) = x 2 i i i log8 (2) = x v log 1 ( 9) = x
i i log 4
= x −2
i v log8 (1) = x 0
1 3 −2
3
4 WE20 Solve for x in the following. a
i log2 (x) + log2 (4) = log2 (20) 5
i i log5 (3) + log5 (x) = log5 (18) 6
i i i log3 (x) − log3 (2) = log3 (5) 10
i v log10 (x) − log10 (4) = log10 (2) 8
v log4 (8) − log4 (x) = log4 (2) 4 b
i log3 (10) − log3 (x) = log3 (5) 2 i i i log2 (x) + log2 (5) = 1
i i log6 (4) + log6 (x) = 2 9 i v 3 − log10 (x) = log10 (2) 500
2 5
v 5 − log4 (8) = log4 (x) 128 5
5 Solve for x in the following. i log2 (x) + log2 (6) − log2 (3) = log2 (10) i i log2 (x) + log2 (5) − log2 (10) = log2 (3) 6 a i i i log3 (5) − log3 (x) + log3 (2) = log3 (10) i v log5 (4) − log5 (x)+ log5 (3) = log5 (6) 2 v log5 (x) + log5 (x − 2) = log 5 (3) 3 b
i log3 (x) + log3 (x + 2) = log3 (8) 2 i i i log5 (x) + log5 (x + 20) = 3
5
1
i i log4 (x) + log4 (x − 6) = 2 8 i v log5 (x + 1) + log5 (x − 3) = 1 4
v log6 (x − 2) + log6 (x + 3) = 1 3
6 MC If loga (x) = 0.7, then loga (x2) is equal to: A 0.49 D 0.837
✔ B
1.4
C 0.35
E 0
7 MC If log10 (x) = (a), then (log10 x)2 + log10 (x) − 6 becomes: ✔
150
A (log10 (a))2 + log10 (a) − 6
B a2 + a + 6
D (a − 2)(a + 3)
E log10 (106x3)
Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad
C log10 (x3) − 6
a 10,
3E
1 100
8 Solve for x in the following. a (log10 (x))2 + log10 (x) − 2 = 0 (Hint: Let a = log10 (x)) c (log2 (x))2 − 2 log2 (x) = 8 16, 14 = 0 3, 27 e (log3 (x))2 − log3 (x4) + 3 g log2 (x4) = (log2 (x))2 1, 16 i log10 (x2 + 2x − 5) = 1 −5, 3 log10 ( x ) = 4, find x. 16 9 If log10 (2)
b 1000,
b d f h j
1 10
(log10 − 2 log10 (x) − 3 = 0 2 (log2 (x)) + 3 log2 (x) = 4 2, 161 (log5 (x))2 − log5 (x3) + 2 = 0 25, 5 log3 (x3) = (log3 (x))2 1, 27 log3 (x2 − 3x − 7) = 1 −2, 5 x)2
Exponential equations (base e)
Euler’s number e, named after an 18th century Swiss mathematician, is a very important number used in problems involving natural growth and natural decay. Like π, it is irrational and has to be approximated: e = 2.718 281 828 459 . . . The number e can be used to find the value of an investment after a period of time, or the temperature of a liquid after it has been cooling. n 1 To find the value of e, take the expression 1 + and evaluate it for increasing values of n. n n 1 1 1 n=1 1 + n = 1 + 1 = 2 n
2
n
3
n
5
n = 2
1 1 1 + n = 1 + 2 = 2.25
n = 3
1 1 1 + n = 1 + 3 = 2.370 37
n = 5
1 1 1 + n = 1 + 5 = 2.48832
n = 10
1 1 1 + n = 1 + 10
n = 100
1 1 1 + n = 1 + 100
n
n
n
10
= 2.59374 100
1 1 n = 1000 1 + = 1 + 1000 n n
= 2.70481 1000
1 1 n = 10 000 1 + = 1 + n 10 000
= 2.716 92 10 000
= 2.71815
n
1 As n increases, 1 + becomes closer and closer to 2.718 281 or e, or n n 1 e = lim 1 + n→ ∞ n An answer given in terms of e is an exact answer. The laws of indices apply in the same way if e is the base, that is: • ex × ey = ex + y • ex ÷ ey = ex − y • (ex)y = exy • e0 = 1 1 −x • e = x e •
x ey
y
= ex
Chapter 3 Exponential and logarithmic equations
151
Worked Example 21
Solve for x in e3x = e. Think
Write
1
Write the equation.
e3x = e
2
Write e with a power of 1.
e3x = e1
3
Equate the indices.
4
Solve for x.
3x = 1 1 x= 3
CAS calculators have an ex function, which is treated in the same way as any other number.
Worked Example 22
Solve for x, showing working and with a CAS calculator. Express your answers in exact form and correct to 3 decimal places. a ex = 3 b ex − 3e−x = 2 Think
Write/display
Method 1: Technology-free a
b
1
Write the equation.
2
As loge (e) = 1, take loge of both sides of the equation.
3
Rewrite using loga (xp) = p loga (x).
4
Solve for x.
1
Write the equation. 1 − Write e x as . ex
2
152
ex = 3
a
loge (ex) = loge (3) x loge (e) = loge (3) x = loge (3) ≈ 1.099, correct to 3 decimal places. −x
b ex − 3e
ex
=2 3 − x =2 e
3
Multiply every term by ex.
(ex)2 − 3 = 2ex
4
Make the right-hand side equal to zero.
(ex)2 − 2ex − 3 = 0
5
Let ex = a.
a2 − 2a − 3 = 0 where a = ex
6
Factorise and solve for a.
(a − 3)(a + 1) = 0 a − 3 = 0 or a + 1 = 0 a = 3 or a = −1
7
Substitute ex for a.
ex = 3 or ex = −1
8
Solve for x by taking the log of both sides to base e.
loge (ex) = loge (3)
9
ex = 3 is the only solution because ex = −1 has no real solution.
x ≈ 1.099, correct to 3 decimal places.
Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad
Method 2 : Technology-enabled 1
On the Main page, enter the equations in Alg Standard mode as: e x = 3 e x - 3e−x = 2 Highlight each equation and tap: • Interactive • Advanced • solve • OK To get an answer in decimal form either highlight it and tap . or repeat the above process in Alg Decimal mode.
2
Write the answers in exact form.
a Solving ex = 3 for x gives x = loge (3). b Solving ex − 3e−x = 2 for x gives
x = loge (3).
3
Write the answers in approximate form, correct to 3 decimal places.
a Solving ex = 3 for x gives x = 1.099, correct to
3 decimal places.
b Solving ex − 3e−x = 2 for x gives x = 1.099,
correct to 3 decimal places.
REMEMBER
1. Euler’s number e is an irrational number, which is approximated to 2.718 (correct to 3 decimal places). 2. Evaluate e by using the ‘ex’ button on the calculator. 3. The number e is an exact answer. Use the calculator to give an approximation if required. 4. The laws of indices apply in the same way if e is the base. 5. On a calculator, use the LN button to take the log of a number to base e. The LOG button defaults to base 10 if not specified. 6. loge (x) = ln (x). 7. ex > 0, that is, ex = −1 has no real solution. Exercise
3E
Exponential equations (base e) 1 Evaluate the following, giving your answer correct to 3 decimal places. 1
1
4
a e2 7.389
b e4 54.598
c e 2 1.649
d e 3 1.396
f
g ln (4) 1.386
h ln (5) 1.609
i loge (1.5) 0.405 j loge (3.6) 1.281
5
e 1.221
Solve for x in each of the following. 2 WE21 a ex = e 1 b ex = e2 2 c ex − 2 = e4 6 1 − 1 g e3 x + 6 = e 2 e e x + 1 = f e x − 2 = 2 0 e e
e
e
−1
d e2x = e − 11 6
1.284
−1 2
h e 2 x − 1 = e3
1 14
Chapter 3 Exponential and logarithmic equations
153
3 WE22 Solve for x in each of the following, giving your answer correct to 3 decimal places. a ex = 2 0.693
b ex = 5 1.609
c e x =
e ex = 1.3 0.262
f ex = 2.6 0.956
g 2ex = 6 1.099
1 2
d e x =
−0.693
1 4
−1.386
h 3ex = 12 1.386
4 Solve for x in each of the following, giving your answer correct to 3 decimal places. a (ex − 1)(ex + 2) = 0 0 b (e−x − 2)(e2x − 3) = 0 c (3e−x − 2)(2ex − 1) = 0 0.405, −0.693 − g −0.405, 1.099
0.693, 0.549
d (ex)2 − ex = 0 0
e (ex)2 − e × ex = 0 1
f (ex)2 − 7ex + 10 = 0 0.693, 1.609
g 6 − 11ex + 3e2x = 0
h 18 − 23ex + 7e2x = 0 0.251, 0.693
5 Solve for x in each of the following, giving your answer correct to 3 decimal places. d −0.405, 1.099
a ex − 4e−x = 0 0.693
b ex − 15e−x − 2 = 0 1.609
c 5ex − 12e−x − 11 = 0 1.099
d 3ex + 6e−x − 11 = 0
e 4ex + 6e−x − 11 = −0
f ex + 2e−x = 3 0, 0.693
0.288, 0.693
6 Solve for x in each of the following, giving your answer correct to 3 decimal places. a ex > 1 x > 0 d e ≥ 4 x ≥ 0.693 2x
−x
g e > 0.75
b ex < e e
ex + 1 ≤
x 0 3
Worked Example 26
Calculate the inverse of f(x) = 2 log10 (x − 1) + 1: a without the use of technology b using a CAS calculator. Think a
1 2
Write
Interchange x and y to write the inverse equation. In order to make y the subject, begin by subtracting 1 from both sides.
3
Divide both sides by 2.
4
Rewrite using ax = y ⇔ loga (y) = x.
Let y = 2 log10 (x − 1) + 1 Inverse is x = 2 log10 (y − 1) + 1 2 log10 (y − 1) = x − 1 log10 ( y − 1) = 10
x −1 2
x −1 2
= y −1
Chapter 3 Exponential and logarithmic equations
157
b
5
Add 1 to both sides.
6
Write the answer.
f −1 ( x ) = 10
1
Interchange x and y to write the inverse equation.
x = 2 log10 (y - 1) + 1
2
On the Main page, enter the equation as: x = 2 × log10( y - 1) + 1 Highlight it and tap: • Interactive • Advanced • solve Change the Variable to y and tap: • OK
y = 10
x −1 2
x
Write the equation of the inverse.
4
The solution from the CAS can be simplified to obtain the solution found in a . Rewrite the answer with a negative power.
y = 10
5
Add the powers when multiplying indices of the same base.
y = 10
6
1 2
+
10 2 ∴y = 10
3
∴y
Simplify the powers and write the answer.
−
x −1 2
× 10 2
−
1+
+1
+1 x
1
x −1 = 10 2
f −1 ( x )
+1
x 1 + 2 2
+
1 2
+1
+1
+1
x −1 = 10 2
+1
rEmEmBEr
1. The equation y = loge (x) is an inverse function of y = ex. 2. To find the inverse of a function, interchange x and y and then make y the subject. ExErCisE
3G
inverses 1
eBook plus Digital doc
SkillSHEET 3.5 Inverses
x 1 a y = loge 2 1 + log e ( x ) d y= 2 WE25 Find the inverse of the following.
c y = loge (x) + 1
e y = 2 - loge (x)
f y=
b y = ex + 1 c y = ex - 1 d y = e2x - 1 e y = e2 - x f y = e2 - 3x 2 Find the equation of the inverse of the following. a y = 2 + ex y = loge (x - 2) b y = 2 - ex y = loge (2 - x) c y = 1 - 2ex y = loge 1 − x 2 x + 1 d y = 2 + ex + 1 y = loge (x - 2) - 1 e y = 3 - 2ex - 2 y = 2 + log 3 − x f y = 2 - 3e e
3
2 − log e ( x ) 3
a y = 2ex
WE26 Find the inverse of thex following.
a f (x) = 2 loge (x) f ( x) = e 2 b f (x) = loge (x + 1) d f (x) = loge (2x - 1) e f (x) = loge (2 - x) f ( x) =
158
b y = loge (x) - 1
ex + 1 2
maths Quest 12 mathematical methods Cas for the Casio Classpad
(2)
f (x) = 2 - ex
( )
f (x) = e x - 1
( )
y = loge 2 − x - 1 3
c f (x) = loge (x - 1) f (x) = e x + 1 2 − ex f f (x) = loge (2 - 3x) f ( x) =
3
a y = ex - 2
4 Find the equation of the inverse of the following. b y = 2 - loge (x) y = e2 - x a y = 2 + loge (x) d f (x) = 2 - ln (x - 1) 5
x −1 e 5
3
D
7
x−3 2
x −1 e 5
B
+2 −2
E
3
x −1 e 5
3 x+2 e 3
5
x−2 3
f f (x) = 1 - 3 ln (x + 2) f ( x) = e
✔C
−2
y=e
x −1 e 5
1− x 3
−2
+2
3
−1
mC If y = 5e 2x + 1 - 1, the equation of the inverse is:
A
1 x + 1 +1 ln 2 5
B
1 x + 1 −1 ln 2 5
D
1 x + 1 1 + ln 2 5 2
✔E
1 x + 1 1 − ln 2 5 2
(2x)
mC If eloge
A x2
✔D
3h
y = 1+ e
mC If y = 5 loge (3x - 2) + 1, the equation of the inverse is:
A
6
e y = 3 + 2 ln (x - 1)
f (x) = e2 - x + 1
c y = 2 + 3 loge (x)
C 5 ln ( x − 1) + 1 2
= y , then y equals: B loge (2x)
C e2x
E 2ex
2x
literal equations An equation such as ekx = a, where k ∈ R and a ∈ R+, is called a literal equation. It does not have a numerical solution. The solution will be expressed in terms of the other pronumerals, in this case a and k, often called parameters. For this equation, the solution is: kx = ln (a) ∴ x =
1 ln (a), k ≠ 0, a ∈ R + k
WorkEd ExamplE 27 -kx
Solve ekx = 5 + 2e
eBook plus
for x, where k ∈ R\{0}.
Think
Tutorial
WriTE/display
2 e kx
1
Rewrite the equation with positive powers.
e kx = 5 +
2
Multiply both sides by ekx.
(ekx)2 = 5ekx + 2
3
Let y = ekx and make the right-hand side zero to obtain a quadratic equation in terms of y.
y2 - 5y - 2 = 0
4
Solve for y using the quadratic formula.
y= y=
int-0532 Worked example 27
− ( − 5) ±
( − 5)2 − 4 × 1 × − 2 2 ×1
5 ± 33 2
∴y =
Chapter 3
5 + 33 or 5 − 33 y= 2 2
Exponential and logarithmic equations
159
5
Substitute ekx for y.
e kx =
5 + 33 or kx 5 − 33 e = 2 2
6
Only the first solution is valid as ekx > 0.
e kx =
5 + 33 2
7
Re-write the exponential equation in logarithmic form using ax = y ⇔ loga (y) = x or take logs of both sides. Divide both sides by k.
8
Write the solution and state the restriction for k.
9
On the Main page, enter the equation as: e kx = 5 + 2e−kx Highlight it and tap: • Interactive • Advanced • solve • OK
10
Write the answer.
5 + 33 kx = log e 2 ∴x =
5 + 33 1 log e , k ∈ R \ {0}. k 2
∴x =
33 + 5 1 log e k 2
Worked Example 28
1 Solve for x, given that log 2 ( x ) − 5 log 2 ( p) = log 2 ( 6 ) where p > 0, 2 a manually b with a CAS calculator. Think a
160
1
Write/display
Simplify the left-hand side using p loga (m) = loga (mp) and m log a (m) − log a ( n) = log a n
1 log 2 x 2 − log 2 ( p5 ) = log 2 (6) x log 2 5 = log 2 (6) p x =6 p5
2
Equate both sides.
3
Multiply both sides by p5.
x = 6 p5
4
Square both sides to obtain x.
x = (6p5)2
5
Write the solution and state the restriction for p.
∴ x = 36p10 where p > 0
Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad
b
1
On the Main page, enter the equation as: 1 log 2 ( x) − 5 log 2 ( p) = log 2 (6) 2 Highlight it and tap: • Interactive • Advanced • solve • OK
2
Write the answer. Note that the CAS gives the incorrect restriction p ≥ 0. The correct restriction is p > 0 for log2 (p) to be defined.
∴ x = 36p10
Some equations can only be solved using a CAS calculator, that is, no algebraic method exists. The following example demonstrates this. Worked Example 29
Solve the following equations using a CAS calculator. Give your answers correct to 3 decimal places. a ex = x3 b loge (x) = x − 2 Think
Write/display
1
On the Main page, tap: • Action • Equation/Inequality • solve Complete the entry lines as: solve(e x = x3, x) solve(ln(x) = x - 2, x) Press E after each entry.
2
Write the solutions correct to 3 decimal places.
a Solving ex = x3 for x gives x = 1.857 or x = 4.536. b Solving loge (x) = x − 2 for x gives x = 0.159 or
x = 3.146.
REMEMBER
1. An equation such as ekx = a, where k ∈ R and a ∈ R+, is called a literal equation. 2. Literal equations do not have numerical solutions. 3. The solution of a literal equation is expressed in terms of the other pronumerals, in this case a and k, often called parameters.
Chapter 3 Exponential and logarithmic equations
161
ExErCisE
3h
literal equations 1 Solve 5e2x = a for x, where a ∈ R+.
6 a b
4 − 9a 2b 7 x= 3 x=
9 x=
2 If log3 (D) = cy + log3 (Z), solve for D.
a log e , a > 0 5 D = Z × 3cy
1 2
3 Solve emx + n = 3k for x, where m ∈ R\{0} and k ∈ R+. x = 4 Solve for q given that 2 log3 (p + 5q) = 4. q =
9− p 5
log e (3k ) − n m
x2 z2 x5 y4 y log log log − + log − 5 Prove that 10 10 10 10 y 3 z 2 z 4 = 0. z 3 x y − m z3 x4
1 log e R b
16b 10 a = or 5 a=
x=
1 5 × 16b
6 Consider the exponential equation 9xb × 273a = 81. a Find x in terms of a and b, where a ∈ R, b ∈ R\{0}. b Hence find the value of x if a = 2 and b = -3. 7 Solve 42x - b = 20 for x, where b ∈ R. x = 8 Solve 2x - 1 = 3x + a for x, where a ∈ R. 9 If y = m +
Rebx,
8 x=
solve for x.
log e 20 + 2b log e 2 4 log e 2
a log e ( 3) + log e ( 2 ) log e
2
3 10 Solve for a given that (log2 (5a))2 = 16b2. 3 + 9 + 4a a 1 kx x = log e 11 WE27 Solve e = 3 + kx for x, where a ≥ 0, k ∈ R\0. k 2 e 1
12 WE28 Solve for x given that 2 log 4 ( x ) − 3 log 4 ( y ) = log 4 (3).
5 Proof: z2x5 x2y4 y log10 3 5 − log10 3 4 − log10 4 z x z y z x z2x5 x2y4 y = log10 3 5 − log10 3 4 + log10 4 z x z y z x x2y4 z2x5y = log10 3 5 − log10 7 4 y z x z x xy xy = log10 5 − log10 5 z z = log10 (1) =0 x = 9y6
13 Solve for b given that 2 loge (a) - 5 loge (b) - 2 = 0, where a, b ∈ R+. b = 5
a2 e2
14 WE29 Solve the following equations using a CAS calculator. Give your answers correct to 3 decimal places. a ex = 3x x = 1.512 or x = 0.691 b x + 2 = e x -0.443 c x2 - 1 = e2x -1.058 15 WE29 Solve the following equations using a CAS calculator. Give your answers correct to 3 decimal places. a ln (x) = 2 - x x = 1.557 b ln (x - 2) = x - 4 x = 2.159 and x = 5.146 c x2 - 1 = ln (2x) x = 0.191 and x = 1.433 loge ( p) 16 Solve for x given that log7 (x) = log4 (p).
3i eBook plus eLesson
eles-0091 Exponential and logarithmic modelling
162
x=7
loge ( 4 )
Exponential and logarithmic modelling Exponential and logarithmic functions can be used to model many real situations involving natural growth and decay. Continuous growth and decay can be modelled by the equation A = A0ekt, where A0 represents the initial value, t represents the time taken and k represents a constant. For continuous growth, k is positive, but for continuous decay, k is negative. Logarithms to base 10, often called common logarithms, are used in scientific formulas for measuring the intensity of earthquakes, the acidity of solutions and the intensity of sound.
maths Quest 12 mathematical methods Cas for the Casio Classpad
WorkEd ExamplE 30
eBook plus
In the town of Ill Ness, the number of cases of a particular disease, Tutorial D, can be modelled by the equation D = D0ekt, where t is the time in years. int-0533 Using available medication the number of cases is being reduced Worked example 30 by 20% each year. There are 10 000 people with the disease today. a How many people will have the disease after one year? b Find the value of k correct to 3 decimal places. c Write the equation substituting values for k and D0. d Find how long it would take for the number of people with the disease to be halved. Give your answer correct to the nearest year. e How long would it take for the number of people with the disease to be reduced to 100? Give your answer correct to the nearest year. Think a
b
d
a (100 - 20)% = 80%
1
Find the percentage of people with the disease after one year.
2
Find 80% of the original number.
80% of 10 000 = 8000
3
Write a sentence.
Therefore, 8000 people will have the disease after one year.
1
Substitute t = 0 and D = 10 000 into the given equation.
2
c
WriTE
Substitute t = 1 and D = 8000 into [1], and solve for k.
Use the given equation D = D0ekt. 1
Substitute 5000 for D.
2
Simplify by dividing both sides by 10 000.
3
Take loge of both sides.
4
Solve for t.
b
D = D0 ekt When t = 0 and D = 10 000, 10 000 = D0 ek × 0 10 000 = D0 × 1 10 000 = D0 So D = 10 000ekt
[1]
When t = 1 and D = 8000, 8000 = 10 000ek × 1 8000 = ek 10 000 0.8 = ek loge 0.8 = loge (ek) -0.223 = k log (e) e -0.223 = k × 1 k = -0.223 -0.223t
c D = 10 000e
-0.223t
d D = 10 000e
When D = 5000, 5000 = 10 000e 0.223t -0.223t
0.5 = e
loge (0.5) = -0.223t t=
log e (0.5) − 0.223
≈ 3.108 (3 decimal places) 5
Write a sentence.
It would take about 3 years.
Chapter 3
Exponential and logarithmic equations
163
e
−0.223t
D = 10 000e
e
1
Write the equation.
2
Substitute 100 for D.
3
Simplify by dividing by 10 000.
4
Take loge of both sides.
5
Solve for t.
6
Write the answer in a sentence.
When D = 100, − 100 = 10 000e 0.223t −0.223t
0.01 = e
loge (0.01) = −0.223t t ≈ 20.651 (3 decimal places) It would take about 21 years.
REMEMBER
1. Make a note of all the information that has been given. 2. Give your answers to the correct number of decimal places. 3. Continuous growth and decay is modelled by the equation A = A0ekt, A0 represents the initial value (that is, when t = 0) and k represents a constant.
Exercise
3I
Exponential and logarithmic modelling 1 WE30 Changing δ-gluconolactone into gluconic acid can be modelled by the equation − y = y0e 0.6t, where y is the number of grams of δ-gluconolactone present t hours after the process has begun. Suppose 200 grams of δ-gluconolactone is to be changed into gluconic acid. a Find the value of y0. 200 b Write the equation replacing y0 with your answer. y = 200e-0.6t c How many grams of δ-gluconolactone will be present after 1 hour? Give your answer correct to the nearest gram. 110 g d How long will it take to reduce the amount of δ-gluconolactone to 50 grams? Give your answer correct to the nearest quarter of an hour. 2 14 hours −
2 The decay of radon-222 gas is given by the equation y = y0e 0.18t, where y is the amount of radon remaining after t days. When t = 0, y = 10 g. Give all answers to the nearest whole number. a Find the value of y0. 10 g b Write the equation substituting your value of y0. y = 10e−0.18t c What will be the mass after 1 day? 8 g d How many days will it take for the mass to reach 1 g? 13 days 3 The equation y = A + B loge (x) relates two variables x and y. The table below shows values of x and y. x y a b c d
164
1 3
2 4.386
3 m
Find the value of A and B correct to the nearest whole number. 3, 2 Write the equation relating x and y substituting values for A and B. y = 3 + 2 loge (x) Using your new equation, find the value of m correct to 3 decimal places. 5.197 If y = 7.6, find x correct to the nearest whole number. 10
Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad
4 An amount of $1000 is invested in a building society where the 5% p.a. interest paid is compounded continuously. The amount in the account after t years can be modelled by the equation A = A0ert, where r is the continuous interest rate. a Find the value of A0 and r. 1000, 0.05 b Write the equation substituting values of A0 and r. A = 1000e0.05t c Find the amount in the bank after i 1 year ii 10 years. Give your answer correct to the nearest dollar. 1051, 1649 d How long will it take for the investment to double in value? Give your answer to the nearest year. 14 5 The number of people living in Boomerville at any time, t years, after the first settlers arrived can be modelled by the equation P = P0ekt. Suppose 500 people arrived on 1 January 1850, and by 1 January 1860 there were 675 people. a What is the value of P0? 500 b Find the value of k correct to 2 decimal places. 0.03 c Write the equation substituting values for P0 and k. P = 500e0.03t d What will be the population on 1 January 1900? Give your answer to the nearest 10 people. 2240 e When will the population be 2000? 1896 6 A cup of soup cools to the temperature of the surrounding air. Newton’s Law of Cooling can be − written as T − TS = (T0 − TS)e kt, where T is the temperature of the object after t minutes, and TS is the temperature of the surrounding air. The soup cooled from 90 °C to 70 °C after 6 minutes in a room with an air temperature of 15 °C. a Find the values of TS, T0 and k correct to 2 decimal places. 15 °C, 90 °C, 0.05 b Write the equation substituting the values for TS, T0 and k. T = 15 + 75e−0.05t c Find the temperature of the soup after 10 minutes. Give your answer to the nearest degree. 60 °C d How long would it take for the soup to be 40 °C? Give your answer to the nearest minute. 22 e If the soup is placed in a refrigerator in which the temperature is 2 °C, how long will it take for the soup to reach 40 °C? Use the same value of k and give your answer to the nearest minute. 17 7 The diameter of a tree for a period of its growth can be modelled by the equation D = D0ekt, where t is the number of years after the beginning of the period. The diameter of the tree grew from 50 cm to 60 cm in the first 2 years that measurements were taken. a Find the values of D0 and k. 50, 0.09 b Write the equation using these values. D = 50e0.09t c How much will it have grown in the first 5 years? Round to the nearest centimetre. 78 cm d How long will it take for the tree’s diameter to double? Round to the nearest year. 8 years −
8 The decay of a radioactive substance can be modelled by the equation M = M0e kt, where M grams is the mass of the substance after t years. After 10 years the mass of the substance is 98 grams and after 20 years the mass is 96 grams. a What was the mass of the substance initially? Give your answer to the nearest gram. 100 b Find the value of k. Give your answer to 3 decimal places. 0.002 c Write the equation using these values. M = 100e−0.002t d Find the mass of the substance after 50 years. 90 g e How long would it take for the mass to be halved? 347 years 9 The number of bacteria present in a culture at any time, t hours, can be modelled by the equation N = N0ekt. a If the original number is doubled in 3 hours, find k correct to 2 decimal places. 0.23 b Write the equation substituting the value of k. N = N0e0.23t c Find the original number of bacteria if there were 2500 bacteria after 4 hours. Give the answer correct to the nearest thousand. 1000
Chapter 3 Exponential and logarithmic equations
165
d Write the equation substituting your value for the original population. N = 1000e0.23t e Find the number of bacteria present after 8 hours. Give your answer correct to the nearest thousand. 6000 10 The intensity of light d metres below the surface of the sea can be modelled by the equation − I = I0e kd. Divers in the Sea of Loga have found that the intensity of light is halved when a diver is 5 metres below the surface of the water. a Find the value of k correct to 4 decimal places. 0.1386 b Write the equation substituting the value of k. I = I0e−0.1386d c Find the percentage of light available at a depth of 10 metres. 25% d If artificial light is necessary when the intensity of light is less than 0.1 of the intensity at the surface (I 1 then loga (x) > 0. • It is not possible to take the logarithm of a negative number. Exponential equations (base e) n
1 • Euler’s number e = lim 1 + = 2.718 281 828459 ... h →∞ n • The laws of indices and logarithms apply in the same way when using e. Equations with natural (base e) logarithms
• To solve logarithmic equations use the laws of logarithms and indices. Inverses
• • • •
The equation y = loge (x) is an inverse function of y = ex. To find an inverse, interchange x and y. a log ( x ) = x, x ∈ R + log a (a x ) = x, x ∈ R a
Literal equations
• An equation such as ekx = a, where k ∈ R and a ∈ R+, is called a literal equation. • Literal equations do not have numerical solutions. • The solution of a literal equation is expressed in terms of the other pronumerals, in this case a and k, often called parameters.
Chapter 3 Exponential and logarithmic equations
167
chapter review Short answer −2
33 x5
3 2 − 4 1 Simplify 4 x 5 y 3 × 2 x 3 y 5 , leaving your answers with positive indices.
14 y 15
✔ A
6x B log3 x 6
0
8x D log3 x 6
C log3 (6x - x6)
2 If log2 (5) = 2.321 and log2 (9) = 3.17, find log 2 95 .
2 2 log3 (x) + 4 log3 (x) − log3 (x6) is equal to:
E 6log3 (x - x6)
−
0.849
3 The solution of the equation 3e2x = 4 is closest to: B −0.405 A −0.406 C 0.143 ✔ D 0.144 E 0.575 [© VCAA 2005]
3 Solve 3 × 2x − 7 = 17 for x. 3
log 2 (32) 1 2 . 3 log 2 (8) 1 5 Solve log x (2) = for x. 8 3 4 Evaluated to 3 decimal places, log3 (24) is: 6 If 4e(2 − x) = 128, find x, giving your answer in exact A 2.892 ✔ B 2.893 form. 2 − 5 ln (2) x + a 1 C 0.345 D 0.346 8 y = 2 loge 3 E 1.380 7 Solve for x in loge (5) + loge (x) − loge (2) = loge (10). 4 5 The solution set of the equation (2x − 1)(22x − 4) = 0 4 Evaluate
( )
9 Find the rule of the inverse function of y = loge (1 − x) + 3. y = 1 − e(x − 3) 10 Solve the equation loge (3x + 5) − loge (2) = 2 for x.
b + ln (2k ) , a ε R \{0} a bεR k ε R+
10 x =
2e 2 − 5 3
8 Find the rule of the inverse function to y = 3e2x − a.
Exam tip Common mistakes include cancelling out the logarithms, adding to obtain 3x + 7, or multiplying to get 6x + 5. [Assessment report 1 2007]
[© VCAA 2007]
12
−
11 Solve 6e3x = k for x, where k ∈ R+.
1 ln 3
k 6
−
6 The solution set of the equation ex − 12e x = −4 over R is: A loge (2), loge (6) B ex − 2, ex + 6 − ✔ D loge 2 C 2, 6 E loge 6 7 If loge (x) = a, then e2a + 3ea − 2e−a is equal to: A a 2 + 3a −
a
14 Solve for x given that log2 (x) = y + log2 (z). State the restrictions for the parameters. z × 2y, x, z ε R+ Multiple choice
1 If a > 1, the solution of x for the equation x = a2 is: A 1 B a negative number less than 1 C a positive number less than 1 D a negative number greater than 1 ✔ E a positive number greater than 1
168
B {0, 2} D {1, 4}
2 a 2 x 2 + 3x − 8 − b , a ≠ 0 ✔ B x
12 Solve 3eax + b − 6k = 0 for x and state the necessary restrictions for the parameters a, b and k. 13 Solve for x given that 4 log2 (ax + b) = 12.
over R is: {0, 1} C {1, 2} E {2, 4}
✔ A
C 2 log e (a) + 3 log e (a) −
2 log e (a)
d log e ( x 2 ) + 3 log e ( x ) −
2 log e ( x )
E (ea + 2)(e + 1) 8 If loge (2x) = a, then x is equal to: A 2ea ae D 2
Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad
B 2ae E e2a
✔ C
ea 2
9 If ex + 4 = e2x − 1, then x is equal to: − 5 ✔ B 5 A 3 D -5
C e5 −
14 If y = loga (7x − b) + 3, then x is equal to: 1 A 1 a y − 3 + b B (a y − 3) + b 7 7 b 1 y−3 (a + b) D a y − 3 − ✔ C 7 7 y−3 E log a (7 − b) [© VCAA 2007]
5
E e 3 10 The solution(s) to the equation 2 ln (x) = ln (x + 4) + ln 2 is/are: B 2, −4 A -2, 4 C 1 D 2 ✔ E 4 11 If 2a =
x2 , then y is equal to: y
x 2a x2 C a 2 A
x 2a x D 22 a
2
✔ B
x2 4a 12 The equation which is the inverse of y = ex − 1 is: A y = loge (x) − 1 B y = loge (x − 1) C y = loge (x) + 1 ✔ D y = loge (x + 1) − E y = e x − 1 E
13 The air pressure P cm of mercury at h km above sea level can be modelled by the equation − P = 76e 0.13h. One kilometre above sea level the pressure has: A increased by approximately 9 cm ✔ B decreased by approximately 9 cm C increased by approximately 41 cm D decreased by approximately 41 cm E neither increased nor decreased significantly
15 If 2 loge (x) − loge (x + 2) = 1 + loge (y), then y is equal to: x2 A 10( x + 2) 2x −1 B x+2 x2 C x+2 2x D x+2 x2 ✔ E e( x + 2) [© VCAA 2003] 16 If 3log ( x + 4 ) = y , then y equals: A log (x + 4) ✔ B x + 4 C (x + 4)3 D 3 log (x + 4) E −4 3
17 The solution set of the equation e4x − 5e2x + 4 = 0 over R is: A {1, 4} B {−4, −1} C {−2, −1, 1, 2} D {−loge (2), 0, loge (2)} [© VCAA 2007] ✔ E {0, loge (2)}
Extended response
a 1 log e = x . If loge (a) = 0.6932, find the value of x, giving your answer correct to 2 decimal places. 3 x x 2 x3 + + ... 2 e x = 1 + + 1! 2! 3! x 4 x 5 x6 a Write the next 3 terms. 4! + 5! + 6! 1 1 1 1 1 b Substitute x = 1 in the equation using the first 7 terms. e = 1 + 1 + + + + + 2 6 24 120 720 c Show that e ≈ 2.7182.
−0.41
3 The apparent brightness of a star can be found using the formula B = 6 - 2.5 log10 A, where A is the actual brightness of that star. Find the apparent brightness of a star with actual brightness of 3.16. 4.75 a 4 Earthquake magnitude is often reported on the Richter scale. The magnitude, M, is given by M = log10 + B, T where a is the amplitude of the ground motion in microns at the receiving station, T is the period of the seismic wave in seconds, and B is an empirical factor that allows for the weakening of the seismic wave with the increasing distance from the epicentre of the earthquake.
Chapter 3 Exponential and logarithmic equations
169
Find the magnitude of the earthquake if the amplitude of the ground motion is 10 microns, the period is 1 second and the empirical factor is 6.8. 7.8 5 Five grams of a radioactive substance is decaying so that the amount, A grams, that is left after t days, is given by the formula A = 5e kt. a Find the value of A when the number of grams of the radioactive substance has been halved. 2.5 b Rewrite the equation with the new value of A. e-kt = 0.5 ln (2) c Rearrange the equation so that t is the subject of the equation. t = k d If k = 0.005, find how long it will take for the number of grams of the radioactive substance to be halved. Give your answer correct to the nearest day. 139 days 6 A school in the suburb of Bienvenue opened with 30 students in February 1995. It has been found for the first years after opening that the number of students enrolled in the school t years after opening can be modelled by the equation N = N0ekt. There were 45 students enrolled in February 1996. a Find the values of N0 and k. 30, 0.4055 b Write the equation substituting the values for N0 and k. N = 30e0.4055t c How many students will there be 5 years after the opening? 228 d How many years will it take for the school to have 1000 pupils? 9 years Another school in the suburb of Enbaisse has a declining student population. The number of students enrolled at any one time can be modelled by the equation E = E0e rt. There are 1000 students enrolled in February 1995 and 900 in February 1996. e Find the values of E0 and r. 1000, 0.1054 f Write the equation substituting the values for E0 and r. E = 1000e-0.1054t g How many students will be enrolled after 5 years? 590 h How many years will it take for the two schools to have approximately the same number of pupils? 7 years i What will the population be then? Use the calculator value in the working and do not round off until the final answer. 485 -kx
7 Solve ekx = 4 + ke
1 k
(
for x, where k ∈ R+. x = ln 2 + 4 + k
)
eBook plus Digital doc
Test Yourself Chapter 3
170
maths Quest 12 mathematical methods Cas for the Casio Classpad
eBook plus
aCTiviTiEs
Chapter opener Digital doc
• 10 Quick Questions: Warm up with ten quick questions on exponential and logarithmic equations. (page 132) 3A
The index laws
Tutorial
• WE 5 int-0528: Watch a worked example on writing expressions with positive indices. (page 135) 3B
Logarithm laws
Tutorial
• WE 10 int-0529: Watch a worked example on simplifying logarithmic expressions. (page 139) 3C
Exponential equations
Tutorial
• WE 14 int-0530: Watch a worked example on solving exponential equations. (page 143) Digital docs
• SkillSHEET 3.1: Practise writing expressions using index form. (page 146) • SkillSHEET 3.2: Practise solving equations. (page 146) • SkillSHEET 3.3: Practise solving indicial equations by equating the bases. (page 146) • SkillSHEET 3.4: Practise solving linear inequations. (page 146) 3D
Logarithmic equations using any base
Tutorial
• WE 18 int-0793: Watch a worked example on solving logarithmic equations using a CAS calculator. (page 147) Digital doc
• WorkSHEET 3.1: Simplify exponential and logarithmic expressions, and solve logarithmic and exponential equations. (page 149) 3F
Equations with natural (base e) logarithms
3G
Inverses
Interactivity
• Inverses int-0248: Consolidate your understanding of inverses using the interactivity. (page 156) Digital docs
• SkillSHEET 3.5: Practise finding inverses. (page 158) 3H
Literal equations
Tutorial
• WE 27 int-0532: Watch a worked example on solving literal equations. (page 159) 3I
Exponential and logarithmic modelling
eLesson
• Exponential and logarithmic modelling eles-0091: Learn about how exponential and logarithmic modelling is used. (page 162) Tutorial
• WE 30 int-0533: Watch a worked example on exponential modelling. (page 163) Chapter review Digital doc
• Test Yourself Chapter 3: Take the end-of-chapter test to test your progress. (page 170) To access eBookPLUS activities, log on to www.jacplus.com.au
Digital doc
• WorkSHEET 3.2: Solve logarithmic and exponential equations and application questions. (page 156)
Chapter 3
Exponential and logarithmic equations
171
