7A Index laws 7B Negative and rational powers 7C Indicial equations 7D Graphs of exponential functions 7E Logarithms 7F Solving logarithmic equations 7G Applications of exponential and logarithmic functions

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7

syllabus

rref

efer

erence

ence

Topic:

• Exponential and

logarithmic functions and

applications

In this

cha

chapter

pter

7A Index laws

7B Negative and rational

powers

7C Indicial equations

7D Graphs of exponential

functions

7E Logarithms

7F Solving logarithmic

equations

7G Applications of exponential

and logarithmic functions

Exponential

and

logarithmic

functions

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258

M a t h s Q u e s t M a t h s B Y e a r 1 1 f o r Q u e e n s l a n d

Introduction

There are one million millimetres in a kilometre. It would take a person almost two weeks of non-stop counting, day and night, to count to one million.

Although it is difficult to comprehend numbers of this size, at times we must do so. Consider, for example, measuring the intensity of sound ranging from a quiet room to a jet engine.

How can we express the loudness of these sounds without using numbers so large or so small that the mind cannot comprehend them?

The answer involves the use of a scale based on exponents or powers of a number. Exponents or, more specifically, exponential functions provide a powerful means of describing phenomena such as radioactive decay, population growth or the value of investments.

Index laws

Recall that a number, a, which is multiplied by itself n times can be represented in index notation.

a × a × a ×…× a = an n lots of a

where a is the base number and n is the index (or power or exponent). an is read as ‘a to the power of n’ or ‘a to the n’.

Multiplication

When multiplying two numbers in index form

with the same base, add the indices. am×an=am+n

For example, 23× 24= 2 × 2 × 2 × 2 × 2 × 2 × 2 = 27

Division

When dividing two numbers in index form

with the same base, subtract the indices. am÷an=am−n

For example,

Raising to a power

To raise an indicial expression to a power,

multiply the indices. (am)n=am×n=amn

For example, (24)3= 24× 24× 24= 24 + 4 + 4= 212

      

26÷₂2 2×2×2×2×2×2 2×2

--- 24

= =

Index (or power or exponent) Base

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C h a p t e r 7 E x p o n e n t i a l a n d l o g a r i t h m i c f u n c t i o n s

259

Raising to the power of zero

Any number raised to the power of zero is equal to one. a0= 1, a≠ 0 For example 23÷ 23= 23 − 3= 20 [1]

or 23÷ 23= (2 × 2 × 2) ÷ (2 × 2 × 2) = 8 ÷ 8

= 1

So 23÷ 23= 1 [2]

Using [1] and [2] we have 20= 1.

Products and quotients

Note the following. (ab)n=anbn

For example, (2 × 3)4= (2 × 3) × (2 × 3) × (2 × 3) × (2 × 3) = 2 × 3 × 2 × 3 × 2 × 3 × 2 × 3 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 = 24× 34

a b

---   n _an

bn ---=

Simplify each of the following.

a 2x3_y2_×₄_x2_y _b ₍₂_x2_y3₎2_×_xy4 _c ₍₃_a₎5_b6_÷₉_a4_b3 _d

Continued over page

THINK WRITE

a Collect ‘plain’ numbers (2 and 4) and terms with the same base.

a 2x3y2× 4x2y

= 2 × 4 ×x3×x2×y2×y

Simplify by multiplying plain numbers and adding powers with the same base.

= 8x5y3

b Remove the bracket by multiplying the powers. (The power of the 2 inside the bracket is 1.)

b (2x2y3)2×xy4

= 22×x4×y6×xy4

Convert 22 to a plain number (4) first and collect terms with the same base.

= 4 ×x4×x×y6×y4

Simplify by adding powers with the same base.

= 4x5y10

c Write the quotient as a fraction. c (3a)5_b6_÷₉_a4_b3₌

Remove the bracket by multiplying the powers.

=

8p6_m2_¥₍₃_p₎3_m5 6p4_m

---1

2

1

2

3

1 ( )3a

5_b6

9a4_b3

---2 243a

5_b6

9a4_b3

---1

WORKED

E

xample

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260

Expressions involving just numbers and numerical indices can be simplified using index laws and then evaluated.

THINK WRITE

Simplify by first cancelling plain numbers.

=

Complete simplification by subtracting powers with the same base. (Note: a1= a.)

= 27ab3

d Expand the brackets by raising each term to the power of 3.

d =

Convert 33 to 27 and collect ‘like’ pronumerals.

=

Simplify by first reducing the plain numbers, and the

pronumerals by adding the indices for multiplication and subtracting the indices for division.

= 36p6 + 3 − 4m2 + 5 − 1

Simplify the indices of each base. = 36p5m6

3 27a

5_b6

a4_b3

---4

1 8p

6_m2×( )₃_p 3_m5

6p4_m

--- 8p6m2×33p3m5 6p4_m

---2 8 27 p

6_×_p3_×_m2_×_m5

× ×

6p4_m

---3

4

Simplify .

THINK WRITE

Write the expression.

Change the division sign to multiplication and replace the second term with its reciprocal (turn the second term upside down).

=

Remove the brackets by multiplying the powers.

=

Collect plain numbers and terms with the same base.

=

Cancel plain numbers and apply index laws. ₌

Simplify. =

6a4_b3 16a7_b6

--- 3a2b 2a3_b2

--- 

 

∏

3

1 6a

4_b3

16a7_b6

--- 3a2b 2a3_b2

--- 

 3

÷

2 6a

4_b3

16a7_b6

--- 2a3b2 3a2_b

--- 

 3

×

3 6a

4_b3

16a7_b6

--- 23a9b6 33_a6_b3

---×

4 6 8a

4+9–7–6_b3+6–6–3 ×

16×27

---5 a0b0

9

---6 1₉

---2

WORKED

E

xample

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261

Complex expressions involving terms with different bases have to be simplified by replacing each base with its prime factors.

Write in simplest index notation and evaluate.

a 23× 162 b

THINK WRITE

a Rewrite the bases in terms of their prime factors.

a 23× 162= 23× (2 × 2 × 2 × 2)2

Simplify the brackets using index notation.

= 23× (24)2

= 23× 28

Simplify by adding the powers. = 211

Evaluate as a basic number. = 2048

b Rewrite the bases in terms of their prime factors.

b =

=

Write in simplest index form. = 35

95_¥₃4 273

---1

2

3

4 5

1 9

5_×₃4

273

--- (3×3)5×34 3×3×3

( )3

---2 3

2

( )5×₃4

33

( )3

---3 3

10_×₃4

39

---4 5

3

WORKED

E

xample

Simplify .

THINK WRITE

Rewrite the bases in terms of their prime factors.

=

Remove the brackets by multiplying powers.

=

Collect terms with the same base by adding the powers in the products and subtracting the powers in the quotients.

=

Simplify. =

= =

34n_¥₁₈n+1 63n–2

---1 3

4n×₁₈n+1

63n–2

--- 34n×(3×3×2)n+1 2×3

( )3n–2

---2 3

4n_×₍₃2_×₂1₎n+1

2×3

( )3n–2

---3 3

4n×₃2n+2×₂n+1

23n–2_×₃3n–2

---4 34n+2n+2–(3n–2)_×₂n+1–(3n–2)

5 36n₊2–3n₊2_×₂n₊1–3n₊2

33n+4_×₂–2n+3

33n+4×₂3–2n

4

WORKED

E

xample

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262

Index laws

1 Simplify each of the following.

a x2_×_x5_×_x3 _b _m3_×_m2_p_×_p4

c 52× 57× (53)3 d 4y3× 2y × y7

e (xy)3× x4y5 f (2x4)2× (4x2)5

g 3m2p5× (mp2)3× 2m4p6 h 5x2y3× (5xy2)4× (5x2y)2

a a7b8÷ a2b5 b 2a12b9÷ (2a)3b4

c (3x5)y11÷ 6x2y2 d p13q10÷ (pq4)2

e (4mn4)2÷ 14n3 f

g 25r15s10t4÷ r5(s5)2(5t)3 h

i

a b

c d

e f

remember

1. Index laws:

(a) am×an=am+n

(b) am÷an=am −n

(c) a0= 1 (d) (am)n=amn

(e) (ab)n=anbn

(f)

2. To simplify indicial expressions:

(a) when dealing with questions in the form (expression 1) ÷ (expression 2), replace expression 2 with its reciprocal and change ÷ to ×

(b) remove brackets using laws (d), (e) and (f) (c) collect plain numbers and terms of the same base (d) simplify using laws (a), (b) and (c).

a b

---   n _an

bn

---=

remember

7A

Mathc ad

Indices

W WORKEDORKED

E Example

1a, b

W WORKEDORKED

E Example

1c

a3_b4 ab2 ---15a6_b7

3a3_b4 ---24x4_y3

20x2_y3

---W WORKEDORKED

E Example

1d 6p8_m4_×₂_p7_m6 9p5_m2

--- ( )3x 2y2×5x6y3 10x7_y ---14u11_v9_×₍_3u2₎3_v

21u6_v5

--- 5e 3

( )2_f4_×_8e4_f3 20e f5 ---6w2_t7_×_9w4_t12

3w ( )5_t13

--- ( )2x 4y×(3x7y)2 18x5( )_2y 3

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263

5

a The fraction can be simplified to:

b The product can be simplified to:

c The quotient is equal to:

7 Write in simplest index notation.

8 Write in simplest index notation and evaluate.

g h

i j

a b

c d

e f

A 3p2m2 B 3p4m6 C 3p3m8 D 3p3m2 E 3

A 2x5y4 B 3x5y4 C x30y16 D E

A −a18 B −3a6 C −3a6b _D E 3a6

a b

a 24× 42× 8 b 37× 92× 273× 81 c 53× 152× 32

d 205× 84× 125 e f

a b c d

e f g h

3x3_y2 –

( )3

2x3_y6

--- 6x7y5 x2_y ( )2

---× (–3mp)2×4m4p

12(mp)2

---m3_p4_×₍_{m p}3₎2 m p2 –

( )4

--- 4 u 7_v6

( )3

2u3_v2 –

( )2_×_u4₍_3v5₎2

---W WORKEDORKED

E Example

2 15a8_b3 9a4_b5 --- 2a 3_b 3ab2 ---   

÷ 2 5k12d

2k3 ( )2

--- 6kd4 25(k2_d3)3 ---÷

4g4(₂_p11)2 g3_p7

--- 8g4p 2gp

( )3

---÷ 3jn2

n5

--- 

 3 (4j2_n)2 n13_{( )}₂_j 4 ---÷

x4_y7 x3_y2 --- x3y2

x5_y

---÷ 6x3y8

x2_y3

( )3

--- (2xy3)2 8x5_y7 ---÷

m

multiple choiceultiple choice

3p3_m4 p1_m2

---6x6_y5 x5_y3 --- x4

2y ( )2 ---×

3x9_y10 2 --- 3x 5 2 ---3ab3 ab –

--- a2b a5

--- 

 2

÷

3a15 –

b6

---xn+1_×_y5_×_z4–n

xn–2×_y4–n×_z3–n

--- x

n

ym+3

( )2

xn+2y3–m

--- x 2

y

xn–5×y5–3m ---×

W WORKEDORKED

E Examplexample

3

34_×₂₇2 64×₃5

--- 8×52 23×₁₀

---45 27

--- 94×₃5÷₂₇ (162)

3 25 ( )4

--- 272 32 ( )3

---625

( )4

53 ( )5

--- ( )25 4 125

( )3

--- 4 11_÷₈2

163

--- 27 2_×₈₁ 93×₃5

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264

*Hint: Factorise the numerator and denominator first.

10

In simplest index notation, is equal to:

Negative and rational powers

Negative powers

Wherever possible, negative index numbers should be expressed with positive index numbers using the simple rule:

When an index number is moved from the numerator to denominator or vice versa, the sign of the power changes.

, a≠ 0

This is easily verified as follows:

= since a0= 1

= a0 −n using division rule for indices = a−n simplifying the index.

In other words, = and =

Note: Change the level, change the sign.

a b

c d

e f

g h*

i*

A 216n+ 5 B 65n+ 1 C 62n+ 5 D 69 E 62n+ 9

W WORKEDORKED

E Example

4

2n_×₉2n+1 6n–2

--- 253n×5n–3 54n+3

---12x–2_×₄x

6x–2

--- 12n–3×271–n 92n×₈n–1×₁₆n

---4n_×₇n–3_×₄₉3n+1 14n+2

--- 352×55×76 254×₄₉3

---35n–4_×₁₆n_×₉3 4n+1×₁₈1–n×₆3–2n

--- 3n+3n+1 3n+₃n–1

---5n_–₅n+1 5n+1+₅n

---m

362n×₆n+3 216n–2

---a–n 1

an ---=

1 an

--- a0 an

---a–n

1 --- 1

an

--- 1

a–n

--- a

n

1

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265

Rational powers

Until now, the indices have all been integers. In theory, an index can be any number. We will confine ourselves to the case of indices which are rational numbers (fractions).

, where n is a positive integer, is defined as the nth root of a.

For example, we know that × =a

but × =

=a1 =a

Therefore, = .

Similarly, , . . . etc.

is defined for all a≥ 0 if n is a positive integer. In general, for any rational number,

Express each of the following with positive index numbers.

a b

THINK WRITE

a Remove the brackets by raising the denominator and numerator to the power of −4.

a =

Interchange the numerator and denominator, changing the signs of the powers.

=

Simplify by expressing as a fraction to the power of 4.

=

b Remove the brackets by multiplying powers.

b =

Collect terms with the same base by adding the powers on the numerator and

subtracting the powers on the denominator.

=

= =

Rewrite the answer with positive powers. = 5 8 --- 

 –4 x4y–2¥(x2y)–5

x–3_y3

---1 _{ } 5₈--- –4 5

4 –

8–4

---2 8 4 54 ---3 8 5 --- 

 4

1 x

4_y–2×(_x2_y)–5

x–3_y3

--- x4y–2×x–10y–5

x–3_y3

---2 x

6 – _y–7

x–3_y3

---x–6–_{( )}–3_y–7–3

x–3_y–10

3 1

x3_y10

---5

WORKED

E

xample

a 1 n ---a 1 n ---a n = a a a 1 2 ---a 1 2 ---a 1 2 --- 1 2 ---+ a a 1 2 ---a

3 = _a13--- 4 _a _a 1 4 ---= a 1 n ---a m n

----(

a 1 n

---)

m

(

n a

)

m n _am

= = =

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266

Evaluate each of the following without a calculator.

a b

THINK WRITE

a Rewrite the base number in terms of its prime factors.

a =

= 26

b Rewrite the base numbers of the fraction in terms of their prime factors.

b =

=

Rewrite with positive powers by interchanging the numerator and denominator.

=

Evaluate the numerator and denominator as basic numbers.

= 16 3 2 --- 9 25 --- 

  3

2 ---– 1 16 3 2 ---24 ( )32

---2 3 1 9 25 --- 

  3

2

---– ₃2

52

--- 

  3

2 ---– 2 3 3 –

5–3

---3 5 3 33 ---4 125 27

---6

WORKED

E

xample

Simplify the following, expressing your answer with positive indices.

a b

THINK WRITE

a Write the expression. a

Write using fractional indices. =

Write 64 and 512 in index form. =

Multiply the powers. Simplify the powers.

= =

b Write the expression. b

Express the roots in index notation. =

Remove the brackets by multiplying the powers. = Collect terms with the same base by subtracting the powers. =

Simplify the powers. =

Rewrite with positive powers. =

64

3 _¥4 ₅₁₂ 3 _x2_y6_∏ _x3_y5

1 3 64_×4 ₅₁₂

2 64

1 3

---512

× 14

---3 ( )26

1 3

---29

( ) × 14 ---4

5

22×2

9 4 ---2 17 4

---1 3 x2_y6÷ _x3_y5

2 (x2_y6)

1 3

---x3_y5

( )

÷ 12

---3 x

2 3

---y2 _x 3 2 ---y 5 2 ---÷ 4 x 2 3 --- 3 2 ---– y2 5 2 ---– 5 x 5 6 ---– y 1 2 ---– 6 1 x 5 6 ---y 1 2 ---

---7

WORKED

E

xample

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267

Negative and rational powers

1 Express each of the following with positive index numbers.

2 Simplify each of the following, expressing your answer with positive index numbers.

3 Evaluate the following without a calculator.

a 6−3 b 5−4 c ( )−2 d ( )−5

e ( )−2 _f _g (−3)−1 h

a b c

d e f

a b c d

e f g h

i j k l

m n o p

Math_ca

d

Negative and rational powers The Maths Quest CD-Rom

con-tains a Mathcad file which can be used to evaluate numbers raised to negative or rational powers. A sample screen is shown at right.

remember

a−n= , a≠ 0

=

=( )m=( )m= 1 an ---a 1 n ---a n a m n ----a 1 n ---a

n n _am

remember

7B

SkillS

HEET

7.1

W WORKEDORKED E

Example

5a 35

---7 4

---1 9

--- (₆₄–2)3 34

23 ---   –4

Math_ca

d Negative and rational powers W WORKEDORKED E Example

5b ( )–2 3×₂–4 2–3

--- x 2 –

( )3×( )_y4 –2

x–5_×₍_y–2₎3

--- (–m)2×m–3 p–2

( )–1_×_p–4

---x5

x–3

--- ( )x4 –2 x2

( )–3

---÷ (3–2)2×(2–5)–1 24

( )–2×( )₃4 –3

--- x3y–2×(xy2)–3 2x3

( )2×(_y–3)2 ---W WORKEDORKED E Example 6 9 1 2 ---27 1 3 ---625 1 4 ---256 1 8 ---8 2 3 ---81 3 4 ---125 4 3 --- ₈ 125 ---   13

---16 81 --- 

 14--- 25

16 --- 

 32--- 27

64 --- 

 23--- ₃₂ 2 5 ---– 81 3 4 ---– 8 27 --- 

 –23--- 16

121 --- 

 –12--- 125

216 ---   –13

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268

4

a The exact value of 6−2 is:

b The exact value of is:

c simplifies to:

5 Simplify each of the following, expressing your answer with positive indices.

Indicial equations

We can solve equations of the form: =2 as follows:

Take the cube of both sides: =23

The left-hand side becomes x, so x=8.

However, when the unknown (or variable) is not a base number but is an index number, a different approach is required.

Method 1: Exact solutions without a calculator

To attempt to solve index equations exactly, express both sides of the equation to the same base and equate the powers.

If am=an, then m=n (unless a= -1, 0 or 1).

A −12 B −36 C − D E −

A − B 6 C D − E

A B C D E

a b c

d e f

g h i

j k l

m n o

p q r

m

1 6 --- 1 36 --- 1 36 ---27 8 ---   –13

---2 3 --- 2 3 --- 3 2 --- 3 2 ---25

3 _× ₁₂₅

25 5 6 ---5 7 6 ---5 3 2 ---5 11 6 ---5 13 6 ---W WORKEDORKED E Examplexample 7

9_×3 81

x 2 3 ---x 1 6 ---× x 3 4 ---– x 9 8 ---× x 5 2 ---x 1 3 ---( )

÷ 4 3 ₍xy3₎_÷ ₍_x2_y₎ 5 ₃₂_×4 ₈

2 5 4 ---4 1 2 ---– 8 2 3 ---– × × 27 1 4 ---– 9 2 3 ---× 3 5 4 ---– × 18 1 2 ---9 4 3 ---4 3 4 ---× ---x3 4

( )23--- ₍3 _x4₎

3 8

---× (64m6)

4 3

---4m–2

--- x

3

x

---1

x–4

--- (x+1)2

x+1

--- x 1

x

---–

x+2 x

x+2

---+ (y–4) y–4 (p+3)(p+3)

2 5 ---– x 1 3 ---x 1 3

---( )3

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269

More complicated equations can be solved using the same technique. Find the value of x in each of the following equations.

a 3x= 81 b 4x− 1= 256 c 63x− 1= 362x− 3

THINK WRITE

a Write the equation. a 3x=81

Express both sides to the same base. 3x=34

Equate the powers. Therefore, x=4.

b Write the equation. b 4x − 1=256

Express both sides to the same base. 4x − 1=44

Equate the powers. Therefore, x− 1=4.

Solve the linear equation for x by adding one to both sides.

x=5

c Write the equation. c 63x − 1= 362x − 3

Express both sides to the same base. 63x − 1= (62)2x − 3

63x − 1= 64x − 6

Equate the powers. Therefore, 3x− 1=4x− 6

Subtract 3x from both sides to make x the subject.

−1=x− 6

Add 6 to both sides to solve the equation.

x=5

1

2

3

1

2

3

4

1

2

3

4

5

6

8

WORKED

E

xample

Solve for n in the following equation: 23n× 16n+ 1= 32

THINK WRITE

Write the equation. 23n × 16n + 1=32

Express both sides using the same base, 2.

23n × (24)n + 1=25

23n× 24n + 4=25

Multiply the terms on the left-hand side by adding the powers.

27n + 4=25

Equate the powers. Therefore, 7n+ 4 = 5

Solve the linear equation for n. 7n= 1

n=

1 2

3

4

5

6

1 7

---9

WORKED

E

xample

(14)

270

In some cases indicial equations can be expressed in a quadratic form and solved using the Null Factor Law. Look for numbers in index form similar to a2x and ax appearing in different terms.

Note that in step 8, the possible solution 5x= −1 was rejected because there is no value of x for which it will be satisfied. Recall that exponential functions such as 5x are always positive.

Method 2: Using a calculator and trial and error

Indicial equations which cannot have both sides expressed to the same base number do not generally have exact, rational solutions. A trial and error method using a calculator can find solutions to a desired degree of accuracy.

Solve for x if 52x− 4(5x) − 5 = 0.

THINK WRITE

Write the equation. ()52x− 4(5x) − 5 = 0

Rewrite the equation in quadratic form. Note that 52x= (5x)2.

(5x)2− 4(5x) − 5 = 0

Substitute y for 5x. Let y = 5x.

Rewrite the equation in terms of y. Therefore, y2− 4y− 5 = 0.

Factorise the left-hand side. (y− 5)(y+ 1) = 0

Solve for y using the Null Factor Law. Therefore, y= 5 or y= −1.

Substitute 5x for y. 5x= 5 or 5x = −1

Equate the powers. ⇒ 5x= 51 and 5x= −1 has no solution.

State the solution(s). ⇒x= 1

1

2

3

4

5

6

7 8

9

10

WORKED

E

xample

Solve 2x= 5 to 2 decimal places.

THINK WRITE

Write the equation. 2x= 5

Get a rough estimate of the solution. Since 22= 4 and 23= 8 then x is between 2 and 3.

Try x = 2.5 and evaluate 22.5. 22.5 = 5.657 — too big, so try 2.3

Repeat step 3 until an estimate of desired accuracy is found.

22.3= 4.925 — too small, so try x= 2.4 22.4= 5.278 — too big, so try x= 2.35 22.35= 5.098 — too big, so try x= 2.32 22.32= 4.993 — too small, so try x= 2.33 22.33= 5.028 — too big, so try x= 2.325

22.325= 5.011 — too big

Select the value of x closest to 5. Since x= 2.320 is too small, and x= 2.325 is too big, x= 2.32, to 2 decimal places.

1 2 3 4

5

11

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xample

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Solving indicial equations using the solve( command

The solve( command is found in the CATALOG. The correct syntax is:

solve(expression, variable, guess). The following steps show how to find the solution to the equation 2x= 5.

1. Rearrange the equation so it is in the form f(x) = 0. In this example, we have 2x− 5 = 0.

2. Press [CATALOG] (above the zero key). 3. Press the S key (no need to first press ). 4. Scroll down until you find solve(.

5. Press to paste solve( to the home screen.

6. Fill in the arguments of the solve function. In this example, type the following:

2^X–5,X,1).

Choose 1 as our guess at the solution (not a bad one, as 22− 5 = −1, not a long way from 0!) If there are several solutions, the solve function will return the one closest to the guess value.

7. Press .

Indicial equations

1 Solve for x in each of the following equations.

2 Solve for n in each of the following equations.

3 Find x in each of the following.

a 2x= 32 b 5x= 625 c 3x= 243

d 10−x= e 4−x= 16 f 6x=

g 3−x= h 2−x= 1 i 8x= 26

j 25x= 5 k 81x= 33 l 125x= 54

a 23n+ 1= 64 b 52n+ 3= 25 c 32 −n= 27

d 16n+ 3= 23 e 495 − 3n= f 364n− 3= 216

a 42x= 8x− 1 b 274 −x= 92x+ 1 c 163x+ 1= 128x− 2

d 252x− 3= e 325 −x= 43x+ 2 f 642 − 3x= 16x+ 1

g 93x+ 5= h 164 − 3x=

CASIO

Solving indicial equations 2nd

ALPHA ENTER

ENTER

remember

1. If am=an, then m=n (unless a= −1, 0 or 1). 2. Inexact solutions require the use of a calculator.

remember

7C

Math_ca

d

Equation solver

W WORKEDORKED

E Examplexample

8a

1 100

--- 1

216

---1 81

---Math_ca

d

Indicial equations

W WORKEDORKED

E Examplexample

8b 1

7

---W WORKEDORKED

E Examplexample

8c ₁

125

---1 243

--- 1

8x+3

(16)

272

4 Solve for x in each of the following equations:

5 Solve for x in each of the following.

6

Consider the indicial equation 32x− 12(3x) + 27 = 0. The equation can be solved by making the substitution:

7

The quadratic equation formed by the appropriate substitution in question 6 is:

8

The solutions to the equation in question 7 are x equals:

9 Solve each of the following to 2 decimal places.

10

The nearest solution to the equation 3x= 10 is:

a 2x× 83x− 1= 64 b 52x× 1253 −x= 25

c 34x× 27x+ 3= 81 d 16x+ 4× 23 + 2x= 45x

e 3125 × 252x+ 1= 53x+ 4 f = 92x

g 493 − 4x× 72x+ 3= 3432 −x h 57x+ 5÷ 6252x+ 1= 253 −x

i 2562 −x× 43x+ 1= 64x− 1 _j

a 32x− 4(3x) + 3 = 0 b 22x− 6(2x) + 8 = 0

c 3(22x) − 36(2x) + 96 = 0 d 2(52x) − 12(5x) + 10 = 0

e 3(42x) = 15(4x) − 12 f 25x− 30(5x) + 125 = 0

g 4x − 16(2x) = −64 h 2(25x) + 10 = 12(5x)

A y = 3x B y = 2x C y = 32x D y = 2x E y = 3x

A y2− 3y + 27 = 0 B y2− 11y + 27 = 0 C y2+ 12y + 27 = 0 D y2− 12y + 27 = 0 E y2− 9y + 3 = 0

A 2 or 3 B 1 or 2 C 1 or 3 D 0 or 1 E 0 or 2

a 2x= 3 b 2x= 12 c 10x = 45

d 10x= 45 e 10x= 19 f 4x= 10

A x = 2.5 B x = 2.3 C x = 1.9 D x = 2 E x = 2.1

Simulating radioactivity

Consider the imaginary element Braggium. It is an unstable element and every hour there is a 1 in 6 chance that an atom will decay. We will use random numbers to simulate the decay of Braggium.

Suppose that initially there are 100 atoms of Braggium.

Generate a random number from 1 to 6 using a die, a calculator or a spreadsheet. If the number is 6 this means that the atom decays.

Repeat this process 100 times. W

WORKEDORKED

E Example

9

812–x

27x+3

---362x+1 216x–2 --- 1

6 ---=

W WORKEDORKED

E Example

10

m

W WORKEDORKED

E Example

11

m

(17)

273

1 After one hour, how many atoms have decayed?

2 After one hour, how many atoms of Braggium remain?

What happens to this sample of 100 atoms by the end of the second hour? To simulate the decay of atoms in the second hour, count the number of atoms not yet decayed at the end of the first hour, n say. Then generate n random numbers between 1 and 6. The number of 6s generated will give the number of atoms that have decayed during the second hour.

3 Copy the following table and complete it by generating random numbers between 1 and 6 as described above.

4 Use the data obtained in the table for t = 0 and t = 4 to devise a model for the simulation. That is, find values of a and b in N = a bt.

5 Devise a theoretical model for the simulation. That is, find values of a and b in N = a bt given that, in theory, 1 in 6 of the Braggium atoms decays each hour.

6 Compare the theoretical model devised in question 5 with the data. Which value of N in the table differs most from the theoretical prediction obtained in question 5?

Time (t) in hours

No. of atoms remaining (N)

No. of atoms decayed this hour

0 100

1 2 3 4 5 6 7 8 9 10

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274

Graphs of exponential functions

Functions of the form f(x) = ax, where a is a positive real number other than 1 and x is a real number, are called exponential functions.

In general, there are two basic shapes for exponential graphs: y = ax, a >1 or y = ax, 0 < a < 1

Increasing exponential Decreasing exponential

However, in both cases:

• the y-intercept is (0, 1) • the asymptote is y = 0 (x-axis)

• the domain is R • the range is R+.

Verify the shapes of these graphs by graphing, say y = 2x, y = 3x, y = x and y = x on a graphics calculator.

The following sections on graphing exponential functions show the range of variation of exponential graphs.

Reflections of exponential functions

The graph of y = a−x is obtained by The graph of y = −ax is obtained by reflecting y = ax through the y-axis. reflecting y = ax through the x-axis.

Horizontal translations of exponential functions

The graph of y = ax+b is obtained by translating y = ax:

1. b units to the right if b < 0 2. b units to the left if b > 0.

For example, the graph of y = 2x− 3 is obtained by translating y = 2x to the right 3 units.

Check this graph using a graphics calculator. Note also that 2x− 3= (2x)(2−3) = ( )2x so that the effect is identical to that of multiplying by a constant.

y

x 1

0

y = ax_,_a_{> 1} y

x 1

0

y = ax_{, 0 <}_a_{< 1}

1 2

---   

1 3

---   

y

x 0

1 y = a–x, a > 1 y = ax_,_a_{> 1}

y

x 0

1

–1

y = –ax_,_a_{> 1}

y = ax_,_a_{> 1}

y

x 0

1 2

1 –1 2 3 4

y = 2x _{y =}₂x – 3

3 units

1 8

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275

Vertical translations of exponential

functions

The graph of y = ax+ c is obtained by translating y = ax: 1. up by c units if c > 0

2. down by c units if c < 0.

Furthermore the equation of the asymptote becomes y = c. For example, the graph of y = 10x− 5 is obtained by translating y = 10x down by 5 units.

The equation of the asymptote is y = −5. Check this graph using a graphics calculator.

y

x 1

5

–5 –4 10

1 –1

y = 10x

y = 10x_{– 5}

(Asymptote) –5 units

Find the equation of the asymptote and the y-intercept. Hence, sketch the graph of

y= 2x+ 3− 5 and state its domain and range.

THINK WRITE

Write the rule. y= 2x + 3− 5

The graph is the same as y = 2x

translated 3 units left and 5 units down.

State the asymptote. Asymptote is y= −5.

Evaluate y when x = 0 to find the y-intercept.

When x= 0, y= 23− 5

= 3

Therefore, the y-intercept is (0, 3).

Locate the y-intercept and asymptote on a set of axes.

Sketch the graph of the exponential function using the y-intercept and asymptote as a guide.

Use the graph to state the domain and range.

Domain is R.

Range is (−5, ∞).

1 2

3 4

5 y

x 0

3

–5

y = 2x+ 3 _{– 5}

6

7

12

WORKED

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xample

CASIO

WE 7–13

Use a graphics calculator to solve 2x= 15 using the intersection of two graphs. Give the answer rounded to 2 decimal places.

THINK DISPLAY/WRITE

In the Y= menu select Y1= and enter 2^X. Select Y2= and enter 15.

Set suitable WINDOW values. Press .

Press [CALC] and select 5:intersect. Press 3 times (or follow the prompts).

Write the solution to 2 decimal places. Solution: x= 3.91

1 2 3

4 GRAPH

5 2nd

6 ENTER

7

13

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xample

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276

Graphs of exponential

functions

1 Sketch the graph of each of the following on separate axes. (Use a table of values or

copy a graphics calculator screen).

2 Sketch the following graphs, using a table of values or by copying a graphics calculator screen. State the equation of the asymptote and the y-intercept for each.

3 Find the equation of the asymptote and the y-intercept for each of the following. Hence, sketch the graph of each and state its domain and range.

4

a The rule for the graph at right is:

A y= 3x − 2

B y= 3x

C y= 2x − 3

D y= 3x + 2

E y= 3x − 1

a y= 3x b y= 5x c y= 6x

d y= 10x e y= 2−x f y= 4−x

g y= −3x h y= −2x i y= −3−x

j y= 0.5x k y= 2.7x l y= ( )x

a y= 2(3x) b y= 3(2x) c y= 0.5(4x)

d y= 4(5x) e y= (2x) f y= 4( )x

a y= 2x − 1 b y= 3x + 2 c y= 51 − x

d y= 2x + 3 e y= 3x − 3 f y= 2x + 3 − 1

g y= 6−x + 3 h y= 102 − x + 5 i y= 3x − 4 − 2

j y= −2x + 2 + 1

remember

General shapes of graphs of exponential functions: If f(x) = ax, a > 1 If f(x) = ax, 0 < a < 1

y

x

0

f(x) = ax_,_a_{> 1}

1

y

x

0

f(x) = ax_{, 0 <}_a_{< 1}

1

In both cases,

the y-intercept is (0, 1) the asymptote is y = 0 the domain = R the range = R+.

remember

7D

SkillS

HEET

7.2

2 3

---EXCEL

Sprea_dshe et

Exponential functions

1 4

--- 1

3

---W WORKEDORKED

E Example

12

Mathc

ad

Exponential functions

G Cpro

gram Exponential functions

m

y

x

2 1 3

0 1 2 3

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277

b The rule for the graph at right is:

A y = 2x− 3

B y = 3x− 2

C y = 2x+ 1 − 3

D y = 2x− 1 + 3

E y = 2x− 1 − 3

5 Use a graphics calculator to solve the following indicial equations using the intersection

of two graphs. Give answers rounded to 2 decimal places.

a 2x= 10 b 2x= 21 c 2x= 0.7

d 10x= 20 e 10x= 8 f 10x= 45

g 5x= 9 h 3x= 12 i 2x= x + 3

j 3x= x + 4

A world population model

The statistics below describe P, the estimated world population (in billions) at

various times, t.

y

x

–3 –4 –2 0

(1, –2)

W WORKEDORKED

E

Example

13 Work

SHEET

7.1

EXCEL Spreadshe_et

World popu-lation

t 0 1000 1250 1500 1750 1800 1850 1900 1910

P 0.30 0.31 0.40 0.50 0.79 0.98 1.26 1.65 1.75

t 1920 1930 1940 1950 1960 1970 1980 1990 2000

P 1.86 2.07 2.30 2.52 3.02 3.70 4.45 5.30 6.23

(22)

278

1 Use a graphics calculator or the Maths Quest Excel file ‘Exponential model’ to plot the data and fit an exponential curve. A ‘not yet well fitted’ model is shown at right. If using a graphics calculator: Press then select

1:Edit, and enter the years in

L1, and the populations in L2. Press , select [CALC]

and 0:ExpReg, then L1,

L2, Y1 and . Y1

is found under VARS/Y-VARS/1:Function.

2 Use the equation for the curve to predict the world population in 2050.

3 What limitations are there on the use of the equation to predict future populations?

4 If using the spreadsheet, comment on the effect of each part of the equation on the shape of the graph.

Bode’s Law

In 1772, Johann Bode discovered a curious

relationship between pure numbers and the distance of planets from the sun.

His law consisted of a simple formula relating the number of the planet to its distance from the sun.

The actual distances of the planets from the sun are given in the table below. By graphing the distance against 2 raised to the power of the planet number, discover the relationship that Bode found. (Hint: Use either a spreadsheet or a graphics calculator to graph the data and then find the regression line.)

STAT

2nd 2nd ENTER

Planet

number Planet

Distance in AU (1 AU = distance from the Earth to the sun)

0 Mercury 0.39

1 Venus 0.72

2 Earth 1

3 Mars 1.52

4 Ceres (asteroid) 2.77

5 Jupiter 5.2

6 Saturn 9.54

7 Uranus 19.18

8 Pluto 39.4

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279

Logarithms

The index, power or exponent (x) in the indicial equation y = ax is also known as a logarithm.

This means that y = ax can be written in an alternative form: log_ay = x which is read as ‘the logarithm of y to the base a is equal to x’.

For example, 32= 9 can be written as log₃9 = 2. 105= 100 000 can be written as log₁₀100 000 = 5.

In general, for a> 0 and a≠ 1: ax=y is equivalent to x= loga y.

Using the indicial equivalent, it is possible to find the exact value of some logarithms.

Logarithm laws

The index laws can be used to establish corresponding rules for calculations involving logarithms. These rules are summarised in the following table.

Name Rule Restrictions

Logarithm of a product loga(mn) = logam + logan m, n > 0

a > 0, a ≠ 1 Logarithm of a quotient

loga = logam – logan

m, n > 0 a > 0 and a ≠ 1 Logarithm of a power log_amn= nlog_am m > 0

a > 0 and a ≠ 1 Logarithm of the base logaa = 1 a > 0 and a ≠ 1

Logarithm of one loga1 = 0 a > 0 and a ≠ 1

y

=

a

x

Base numeral

Base Logarithm

Evaluate the following without a calculator.

a log6216 b log2( )

THINK WRITE

a Let x equal the quantity we wish to find. a Let x= log₆216

Express the logarithmic equation as an indicial equation.

6x= 216

Express both sides of the equation to the same base.

6x= 63

Equate the powers. x= 3

b Write the logarithm as a logarithmic equation. b Let x= log₂

( )

Express the logarithmic equation as an indicial equation.

2x=

= =

Express both sides of the equation to the same base.

2x= 2−3

Equate the powers. x= −3

1 8

---1 2

3

4

1 1₈

---2 1₈

---1 2 ---   3

(

2–1

)

3

4

14

WORKED

E

xample

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d

Logarithm laws

m n ----   

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280

It is important to remember that each rule works only if the base, a, is the same for each term. Note that it is the ‘logarithm of a product’ and ‘logarithm of a quotient’ rules that formed the basis for the pre-1970s calculation device for multiplication and division — the slide rule.

Simplify, and evaluate where possible, each of the following without a calculator.

a log₁₀5 + log₁₀4 b log₂12 + log₂8 − log₂3

THINK WRITE

a Apply the ‘logarithm of a product’ rule. a log₁₀ 5 + log₁₀4

= log₁₀(5 × 4)

Simplify. = log₁₀20

b Multiply the base numerals of the logs being added since their bases are the same.

b log₂12 + log₂8 − log₂3

= log₂(12 × 8) − log₂3

Apply the ‘logarithm of a quotient’ law. = log₂(96 ÷ 3)

Simplify, noting that 32 is a power of 2. = log₂32

= log₂25

Evaluate using the ‘logarithm of a power’ and ‘logarithm of the base’ laws.

= 5 log₂2

= 5

1

2 1

2 3

4

15

WORKED

E

xample

Simplify 3 log25 − 2 log210.

THINK WRITE

Express both terms as logarithms of index numbers.

3 log₂5 − 2 log₂10 = log₂53− log₂102

Simplify each logarithm. = log₂125 − log₂100

Apply the ‘logarithm of a quotient’ law. = log₂(125 ÷ 100)

Simplify. = log ₂ or log₂1.25

1

2 3

4

(

5₄---

)

16

WORKED

E

xample

Simplify each of the following.

a b 2 log₁₀x+ 1

THINK WRITE

a Express each base numeral as powers to the same base, 7.

a =

Apply the ‘logarithm of a power’ law. = Simplify by cancelling out the common

factor of log₈7.

=

log849 log8 343

---1

log₈ 49 log₈ 343

--- log8 7

2

log₈ 73

---2 2 log8 7

3 log₈7

---3 2₃

---17

WORKED

E

xample

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281

Logarithms

1 Express the following indicial equations in logarithmic form.

2 Express the following logarithmic equations in indicial form.

3

The value of log₅25 is:

4

When expressed in logarithmic form, 83= 512 is:

5

When expressed in indicial form, log₁₀10 000 = 4 is:

a 23= 8 b 35= 243 c 50= 1

d 0.01 = 10−2 e bn= a f 2−4=

a log₄16 = 2 b log₁₀1 000 000 = 6 c log₂ = −1

d log₃27 = 3 e log₅625 = 4 f log₂128 = 7

g log₃ = −2 h log_ba = x

A −2 B 5 C 1 D 2 E 4

A log₃8 = 512 B log₃512 = 8 C log₈512 = 3

D log₅₁₂3 = 8 E log₈3 = 512

A 104= 10 000 B 10 0004= 10 C 10 00010= 4 D 1010 000= 4 E 410= 10 000

THINK WRITE

b Express 2 log₁₀x as log₁₀x2 and 1 as a logarithm to base 10 also.

b 2 log₁₀x+ 1 = log₁₀x2+ log₁₀ 10

Simplify using the ‘logarithm of a

product’ law. =

log₁₀ 10x2

1

2

remember

1. If y=ax then log_ay=x where a= the base, x= the power, index or logarithm and y= the base numeral. Note that a> 0, a≠ 1, and therefore y> 0.

2. Log laws:

(a) log_am+ log_an= log_a(mn) (b) log_am− log_an= log_a m

n

----   

remember

(c) log_amn=nlog_am

(d) log_aa= 1 (e) log_a1 = 0

7E

1 16

---Math_ca

d

Logarithms to any base

1 2

---1 9

---m

multiple choiceultiple choice

m

(26)

282

6 Evaluate each of the following without a calculator.

7 Simplify, and evaluate where possible, each of the following without a calculator.

9 Simplify the following.

10

The expression log₁₀ xy is equal to:

11

The expression log₅ xy is equal to:

a log₂16 b log₃81 c log₅125 _d _log

2

e log₁₀1000 f log₁₀(0.000 01) g log₂0.25 h log₃

i log₂32 _j log₂ k log₃(−3) _l log_nn5

a log₂8 + log₂10 b log₃7 + log₃15 c log₁₀20 + log₁₀5

d log₆8 + log₆7 e log₂20 − log₂5 f log₃36 − log₃12

g log₅100 − log₅8 _h log₂ + log₂9 i log₄25 + log₄

j log₁₀5 − log₁₀20 k log₃ − log₃ l log₂9 + log₂4 − log₂12

m log₃8 − log₃2 + log₂5 n log₄24 − log₄2 − log₄6

a 3 log₁₀5 + log₁₀2 b 2 log₂8 + 3 log₂3

c 2 log₃2 + 3 log₃1 d log₅12 − 2 log₅2

e 4 log₁₀2 + 2 log₁₀8 f log₃42+ 3 log₃2

g log₂27 − log₂36 h log₂(x − 4) + 3 log₂x

i log₃16 + 2 log₃4 j 2 log₁₀(x + 3) − log₁₀(x − 2)

a b c

d e f

g h _i

j

A log10 x × log10 y B log10 x − log10 y C

D y log₁₀ x E log₁₀ x + log₁₀ y

A xlog₅ y B ylog₅ x C 5 log_x y

D log₅ x + log₅ y E 5y

W WORKEDORKED

E Example

14 14

---1 243 ---1

64

---W WORKEDORKED

E Example

15

1 3

--- 1

5 ---4

5

--- 1

5

---W WORKEDORKED

E Example

16

1 3

--- 1

2 ---1

2

---W WORKEDORKED

E Example

17a log₃ 25 log₃ 125

--- log2 81 log₂9

--- log4 36 log₄ 6 ---2 log₁₀ 8

log₁₀ 16

--- 3 log5 27 2 log₅ 9

--- 4 log3 32 5 log₃ 4

---log₃ x6 log₃ x2

--- log10 x 3

log₁₀ x

--- log5 x 3 2

---log₅ x

---2 log₂(x+1)3 log₂(x+1)

---m

log₁₀ x log₁₀ y

---m

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283

12

The expression log₂64 + log₂5 can be simplified to:

13

The expression can be simplified to:

14 Express each of the following in simplest form:

Solving logarithmic equations

Logarithms to the base 10

Logarithms to the base 10 are called common logarithms and can be evaluated using the LOG function on a calculator.

For example, to evaluate log₁₀8, correct to 3 decimal places:

1. Press on the calculator and press . (On some calculators, press .)

2. The display shows 0.903 089 887.

This means log₁₀ 8 = 0.903 to 3 decimal places, or 100.903≈ 8.

When solving logarithmic equations involving bases other than 10, the following steps should be followed:

Step 1 Simplify the equation using logarithm laws.

Step 2 Express the equation in index form if required.

Step 3 Solve by either:

(a) evaluating if the base numeral is unknown (b) equating the powers if possible

(c) equating the bases if possible.

Note: The logarithm of a negative number or zero is not defined. Therefore:

log_ax is defined for x> 0, if a> 0 This can be seen more clearly using index notation as follows: Let n = log_a x.

Therefore, an= x (indicial equivalent of logarithmic expression).

However, an> 0 for all values of n if a > 0 (positive based exponentials are always positive). Therefore, x > 0.

A log₂40 B 1 C log₂ D log₂20 E log₂

A log₄ x3 B log₄ C D log₄ (x5− x2) E log₄ x7

a log₃ 27 + 1 b log₄ 16 + 3 c 3 log₅ 2 − 2

d 2 + 3 log₁₀ x e 2 log₂ 5 − 3 f 4 log₃ 2 − 2 log₃ 6 + 2

g 2 log₆ 6 − log₆ 4 h + 3 log₁₀ x2 m

multiple choiceultiple choice

1 3

---64 15

--- 320

3

---m

log₄ x5 log₄ x2

---x 5 2

--- ₅

2

---W WORKEDORKED

E Example

17b

1 2

---8 LOG ENTER

LOG 8

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284

Solving exponential equations using log

₁₀

on the

calculator

We have already seen three methods for solving exponential equations: 1. equating the bases, which is not always an option, for example, 2x= 7 2. using a calculator and trial and error, which can be time consuming 3. using a graphical technique and a graphics calculator.

An efficient method for solving equations involves the use of logarithms and the log₁₀ function on the calculator. This is outlined in the following examples.

Find x if log39 =x− 2.

THINK WRITE

Write the equation. log₃9 =x− 2

Simplify the logarithm using the ‘logarithm of a power’ law and the fact that log₃3 = 1.

log₃32= x – 2 2 log₃3=x – 2 2 =x – 2

Solve for x by adding 2 to both sides. Therefore, x= 4

1 2

3

18

WORKED

E

xample

Solve for x if log6 x= −2.

THINK WRITE

Write the equation. log₆x = −2

Express in index form. Therefore, x = 6−2.

Evaluate the index number. x =

= 1

2

3 1

62

---1 36

---19

WORKED

E

xample

Find x if 2 logx 25 = 4, x> 0.

THINK WRITE

Write the equation. 2 log_x25 = 4

Divide both sides by 2. log_x25 = 2

Write as an index equation. Therefore, x2= 25.

Express both sides of the equation to the same base, 5.

x2= 52

Equate the bases.

Note that x = −5 is rejected as a solution, because x > 0.

x= 5

1 2 3 4

5

20

WORKED

E

xample

(29)

285

Therefore, we can state the following rule:

If ax=b, then x=

This rule applies to any base, but since your calculator has base 10, this is the most commonly used for this solution technique.

Solving logarithmic equations

1 Find x in each of the following.

2 Solve for x.

a log₂4 = x b log₉1 = x c log₃27 = x d log₄256 = x

e log₁₀ = x f log₃ = x g 2 log₂8 = x h log₃81 = 2x

i log₁₀1000 = 2x − 1 j 2 log₂32 = 3x + 1

a log₂x = 3 b log₃x = 2 c log₅x = 4 d log₁₀x = 1

e log₈x = −1 f log₃x = −3 g log₂x = 6 h log₁₀x = 4

i log₃(x − 3) = 3 j log₂(3x + 1) = 4 k log₁₀(2x) = 1 l 2 log₆(3x) = 1

m log₅x = log₅4 + log₅6 n log₃5 − log₃4 = log₃x – log₃8 Solve for x, correct to 3 decimal places, if 2x= 7.

THINK WRITE

Write the equation. 2x= 7

Take log₁₀ of both sides. log₁₀2x= log₁₀7

Use the ‘logarithm of a power’ law to bring the power, x, to the front of the logarithmic equation.

xlog₁₀2 = log₁₀7

Divide both sides by log₁₀2 to get x by itself. Therefore, x=

Evaluate the logarithms correct to 4 decimal places, at least one more than the answer requires.

=

Solve for x. x= 2.808

1

2 3

4 log107

log₁₀2

---5 0.8451

0.3010

---6

21

WORKED

E

xample

log₁₀ b

log₁₀ a

---remember

1. Logarithmic equations are solved more easily by: (a) simplifying using log laws

(b) expressing in index form (c) solving as required.

2. If ax=b, then x= log10 b. log₁₀ a

---remember

7F

W WORKEDORKED

E Example

18

1 10

--- 1

9