Exponential and logarithmic functionsmathsbooks.net/Maths Quest 11 Methods/By Chapter/Ch... · exponential functions. For example: f (x) = a x where a > 0 and a ≠ 1 is an exponential

4VCEVCEcocovverageerageAreas of studyUnit 2 • Functions and graphs

• Algebra

In thisIn this chachapterpter4A Index laws4B Negative and rational

powers4C Indicial equations4D Graphs of exponential

functions4E Logarithms4F Solving logarithmic

equations4G Applications of exponential

and logarithmic functions

Exponential and logarithmicfunctions

168

M a t h e m a t i c a l M e t h o d s U n i t s 1 a n d 2

Introduction

Functions in which the independent variable is an index number are called

indicial

or

exponential functions

. For example:

f

(

x

)

=

a

x

where

a

>

0 and

a

≠

1is an exponential function.

It can be shown that quantities whichincrease or decrease by a constant per-centage in a particular time can bemodelled by an exponential function.

Exponential functions have applicationsin science and medicine (exponentialdecay of radioactive material, exponentialgrowth of bacteria like those shown in thephoto), and finance (compound interestand reducing balance loans).

Index laws

We recall that a number,

a

, which is multiplied by itself

n

times can be represented inindex notation.

where

a

is the base number and

n

is the index (or power or exponent).

a

n

is read as ‘

a

to the power of

n

’ or ‘

a

to the

n

’.

Multiplication

When multiplying two numbers in index formwith the same

base

,

add

the indices.

a

m

×

a

n

=

a

m

+

n

For example, 2

3

×

2

4

=

2

×

2

×

2

×

2

×

2

×

2

×

2

=

2

7

Division

When dividing two numbers in index form with the same

base

,

subtract

the indices.

a

m

÷

a

n

=

a

m

−

n

For example,

Raising to a power

To raise an indicial expression to a power,

multiply

the indices.

(

a

m

)

n

=

a

m

×

n

=

a

mn

For example, (2

4

)

3

=

2

4

×

2

4 × 24 = 24 + 4 + 4 = 212

Raising to the power of zeroAny number raised to the power of zero is equal to one. a0 = 1, a ≠ 0For 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.

Index (or power or exponent)

Basea × a × a ... × a = an

n lots of a

26 22÷ 2 2 2 2 2 2×××××2 2×

------------------------------------------------- 24= =

C h a p t e r 4 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 169

Products and quotientsNote 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

ab---

n an

bn-----=

Simplify.a 2x3y2 × 4x2y b (2x2y3)2 × xy4 c (3a)5b6 ÷ 9a4b3 d

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)5b6 ÷ 9a4b3 =

Remove the bracket by multiplying the powers.

=

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

8 p6m2 3 p( )3× m5

6 p4m---------------------------------------------

1

2

1

2

3

13a( )5b6

9a4b3-------------------

2243a5b6

9a4b3--------------------

327a5b6

a4b3-----------------

4

18 p6m2 3 p( )3m5×

6 p4m------------------------------------------- 8 p6m2 33 p3m5×

6 p4m-----------------------------------------

28 27 p6 p3 m2 m5×××××

6 p4m-----------------------------------------------------------------

3

4

1WORKEDExample

170 M a t h e m a t i c a l M e t h o d s U n i t s 1 a n d 2

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

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. =

6a4b3

16a7b6-----------------

3a2b2a3b2--------------

÷3

16a4b3

16a7b6----------------- 3a2b

2a3b2--------------

3÷

26a4b3

16a7b6----------------- 2a3b2

3a2b--------------

3×

36a4b3

16a7b6----------------- 23a9b6

33a6b3-----------------×

46 8a4 9 7– 6–+ b3 6 6– 3–+×

16 27×--------------------------------------------------------------

5a0b0

9-----------

619---

2WORKEDExample

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

45

195 34×

273----------------- 3 3×( )5 34×

3 3 3××( )3-------------------------------

232( )5 34×

33( )3------------------------

3310 34×

39-------------------

45

3WORKEDExample

C h a p t e r 4 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 171Complex expressions involving terms with different bases have to be simplified byreplacing each base with its prime factors.

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–

-----------------------------

134n 18n 1+×

63n 2–----------------------------- 34n 3 3 2××( )n 1+×

2 3×( )3n 2–-------------------------------------------------

234n 32 21×( )n 1+×

2 3×( )3n 2–---------------------------------------------

334n 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–×

4WORKEDExample

remember1. 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)

ab---

n an

bn-----=

remember


Index laws

1 Simplify.

2 Simplify.

3 Simplify.

4 Simplify.

a x2 × x5 × x3 b m3 × m2p × 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 × (4xy2)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

g h

i j

a b

c d

e f

4AWORKEDEExample

1a, b

Mathca

d

Indices

WORKEDExample

1c

a3b4

ab2-----------

15a6b7

3a3b4-----------------

24x4y3

20x2y3-----------------

WORKEDExample

1d 6 p8m4 2 p7m6×9 p5m2

--------------------------------------- 3x( )2y2 5x6y3×10x7y

---------------------------------------

14u11v9 3u2( )3v×21u6v5

-------------------------------------------- 5e3( )2 f 4 8e4 f 3×20e f 5

-------------------------------------------

6w2t7 9w4t12×3w( )5t13

------------------------------------- 2x( )4y 3x7y( )2×18x5 2y( )3

------------------------------------------

3x3y2–( )3

2x3y6------------------------ 6x7y5

x2y( )2---------------× 3mp–( )2 4m4 p×

12 mp( )2-----------------------------------------

m3 p4 m p3( )2×m p2–( )4

------------------------------------- 4 u7v6( )3

2u3v2–( )2 u4 3v5( )2×----------------------------------------------------

WORKEDExample

2 15a8b3

9a4b5----------------- 2a3b

3ab2------------

÷2 5k12d

2k3( )2--------------- 6kd4

25 k2d3( )3------------------------÷

4g4 2 p11( )2

g3 p7--------------------------- 8g4 p

2gp( )3-----------------÷ 3 jn2

n5-----------

3 4 j2n( )2

n13 2 j( )4---------------------÷

x4y7

x3y2---------- x3y2

x5y----------÷ 6x3y8

x2y3( )3------------------ 2xy3( )2

8x5y7-------------------÷


a can be simplified to:

b can be simplified to:

c is equal to:

6 Simplify each of the following.

7 Write in simplest index notation and evaluate.

8 Write in simplest index notation and evaluate.

9 Simplify.

*Hint: Factorise the numerator and denominator first.

10

In simplest index notation, is equal to:

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

a b

c d

e f

g h*

i*

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

mmultiple choiceultiple choice3 p3m4

p1m2----------------

6x6y5

x5y3-------------- x4

2y( )2-------------×

3x9y10

2---------------- 3x5

2--------

3ab3

ab–------------ a2b

a5--------

2÷

3a15–b6

--------------

xn 1+ y5 z4 n–××xn 2– y4 n– z3 n–××----------------------------------------------- xnym 3+( )2

xn 2+ y3 m–------------------------- x2y

xn 5– × y5 3m–----------------------------------×

WORKEDExample

334 272×64 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---------------- 411 82÷

163------------------- 272 81×

93 35×--------------------

WORKEDExample

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+-----------------------

mmultiple choiceultiple choice362n 6n 3+×

216n 2–-----------------------------


Negative and rational powersNegative powersWherever possible negative index numbers should be expressed with positive indexnumbers 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 n– 1an-----=

1an----- a0

an-----

a–n

1------- 1

an----- 1

a–n------- a

n

1-----

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.

=

58---

4– x4 y 2– x2 y( ) 5–×x 3– y3

---------------------------------------

158---

4– 5 4–

8 4–-------

284

54-----

385---

4

1x4y 2– x2y( )× 5–

x 3– y3------------------------------------- x4y 2– x 10– y 5–×

x 3– y3-------------------------------------

2x 6– y 7–

x 3– y3---------------

x–6– 3–( )y 7– 3–

x 3– y 10–

31

x3y10-------------

5WORKEDExample


Rational powersUntil 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,

a1n---

a1n---

an=

a a

a12---

a12---

a12--- 1

2---+

a a12---

a3 a13---= a4 a

14---=

a1n---

amn---- (a

1n---)

m( an )m

amn= = =

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.

=

1632--- 9

25------

32---–

1 1632---

24( )32---

2

3

19

25------

32---– 32

52-----

32---–

23 3–

5 3–-------

353

33-----

412527

---------

6WORKEDExample


Simplify the following expressing your answer with positive indices.

a bTHINK 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. =

647 5124× x2 y63 x3 y5÷

1 647 5124×

2 6417---

512×14---

3 27( )17---

26( )×14---

4

5

21 2×64---

21 2×32---

252---

1 x2y63 x3y5÷

2 x2y6( )13---

x3y5( )÷12---

3 x23---y2 x

32---y

52---÷

4 x23--- 3

2---–y

2 52---–

5 x56---–y

12---–

61

x56---y

12---

----------

7WORKEDExample

The Maths Quest CD Rom con-tains a Mathcad file which can beused to evaluate numbers raisedto negative or rational powers. Asample screen is shown at right.

Mathca

d

Negativeandrationalpowers

remembera−n = , a ≠ 0

=

= ( )m = ( )m =

1

an

-----

a1n---

an

amn----

a1n---

an amn

remember


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.

4a 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.

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

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

4BWORKEDExample

5a SkillSH

EET 4.135--- 7

4---

19--- 64 2–( )3 34

23-----

4–

WORKEDExample

5bMathcad

Negativeand

rationalpowers

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×-------------------------------------

WORKEDExample

6 912---

2713---

62514---

25618---

823---

8134---

12543--- 8

125---------

13---

1681------

14--- 25

16------

32--- 27

64------

23---

3225---–

8134---– 8

27------

23---– 16

121---------

12---– 125

216---------

13---–

mmultiple choiceultiple choice

16--- 1

36------ 1

36------

278------

13---–

23--- 2

3--- 3

2--- 3

2---

253 125×

2556---

576---

532---

5116

------5

136

------

WORKEDExample

7 9 813× x23---

x16---× x

34---–

x98---×

x52---

x13---( )÷

4xy3( )3 x2y( )÷ 325 84×

254---

412---–

823---–×× 27

14---–

923---× 3

54---–× 18

12---

943---

434---×

----------------

x34( )23---

x43( )38---

×64m6( )

43---

4m 2–--------------------

x3

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+( )

25---–


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

If am = an, then m = n.

More complicated equations can be solved using the same technique.

x13---

x13---( )

3

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

8WORKEDExample


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

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

Method 2: Using a calculator and trial and errorIndicial equations which cannot have both sides expressed to the same base number donot generally have exact, rational solutions. A trial and error method using a calculatorcan find solutions to a desired degree of accuracy.

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

THINK WRITE

Write the equation. 23n × 16n + 1 = 32Express 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 = 5Solve the linear equation for n. 7n = 1

n =

1

2

3

4

5

617---

9WORKEDExample

Solve for x if 52x − 4(5x) − 5 = 0.

THINK WRITE

Write the equation. ()52x − 4(5x) − 5 = 0Rewrite 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) = 0Solve for y using the Null Factor Law. Therefore, y = 5 or y = −1Substitute 5x for y. 5x = 5 or 5x = −1Equate 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

10WORKEDExample


Solving indicial equations using the solve( commandThe solve( command is found in the CATALOG. The correct syntax is:solve(expression, variable, guess). The following steps show how to find the solution tothe 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 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)1 is 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 theguess value.

7. Press .

Solve 2x = 5 to 2 decimal places.

THINK WRITE

Write the equation. 2x = 5Get 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 2.3Repeat step 3 until an estimate of desired accuracy is found.

22.3 = 4.925 — too small, so try x = 2.422.4 = 5.278 — too big, so try x = 2.3522.35 = 5.098 — too big, so try x = 2.3222.32 = 4.993 — too small, so try x = 2.3322.33 = 5.028 — too big, so try x = 2.32522.325 = 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.

1234

5

11WORKEDExample

2ndS ALPHA

ENTER

ENTER

remember1. If am = an, then m = n.2. Inexact solutions require the use of a calculator.

remember


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.

4 Solve for x in each of the following equations:

5 Solve for x in each of the following.

6Consider the indicial equation 32x − 12(3x) + 27 = 0. The equation can be solved bymaking the substitution:

7The quadratic equation formed by the appropriate substitution in question 6 is:

8The solutions to the equation in question 7 are x equals:

9 Solve each of the following to 2 decimal places.

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

a 2x = 32 b 5x = 625 c 3x = 243d 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 = 27d 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 =

a 2x × 83x − 1 = 64 b 52x × 1253 − x = 25c 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 = 0c 3(22x) − 36(2x) + 96 = 0 d 2(52x) − 12(5x) + 10 = 0e 3(42x) = 15(4x) − 12 f 25x − 30(5x) + 125 = 0g 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 = 0D 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 = 45d 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

4CWORKEDExample

8a

Mathcad

Equationsolver

1100--------- 1

216---------

181------

WORKEDExample

8bMathcad

Indicialequations

17---

WORKEDExample

8c 1125---------

1243---------

1

8x 3+-------------

WORKEDExample

9

812 x–

27x 3+--------------

362x 1+

216x 2–----------------- 1

6---=

WORKEDExample

10




WORKEDExample

11



Graphs of exponential functionsFunctions of the form f (x) = ax, where a is a positive real number other than 1 and x isa 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 , , and on a graphics calculator.

The following sections on graphing go beyond the requirements of the study design,but are included to show the range of variation of exponential graphs.

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

Horizontal translations of exponential functionsThe graph of y = ax + b is obtained by translating y = ax:1. b units to the right if b < 02. b units to the left if b > 0.

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

Check this graph using a graphics calculator. Note also that2x − 3 = (2x)(2−3) = ( )2x so that the effect is identical to thatof multiplying by a constant.

Vertical translations of exponential functionsThe graph of y = ax + c is obtained by translating y = ax:1. up by c units if c > 02. 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

0

y = ax, a > 1y

x

1

0

y = ax, 0 < a < 1

y 2x= y 3x= y (12---)

x=

y (13---)

x=

y

x0

1 y = a–x, a > 1

y = ax, a > 1

y

x0

1

–1y = –ax, a > 1

y = ax, a > 1

y

x0

12

1–1 2 3 4

y = 2x y = 2x – 3

3 units

18---

y

x1

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 − 5The 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 RRange is (−5, ∞)

12

34

5 y

x0

3

–5

y = 2x+3 – 5

6

7

12WORKEDExample

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

1234 GRAPH5 2nd6 ENTER

7

13WORKEDExample

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

y

x0

f (x) = ax, a > 1

1

y

x0

f (x) = ax, 0 < a < 11

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

remember


Graphs of exponential functions

1 Sketch the graph of each of the following on separate axes. (Use a table of values orcopy a graphics calculator screen).

2 Sketch the following graphs, using a table of values or by copying a graphics calculatorscreen. 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

b The rule for the graph at right is:A y = 2x − 3B y = 3x − 2C y = 2x + 1 − 3D y = 2x − 1 + 3E y = 2x − 1 − 3

5 Use a graphics calculator to solve the following indicial equations using the intersectionof two graphs. Give answers rounded to 2 decimal places.

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 − 1g y = 6−x + 3 h y = 102 − x + 5 i y = 3x − 4 − 2j y = −2x + 2 + 1

a 2x = 10 b 2x = 21 c 2x = 0.7d 10x = 20 e 10x = 8 f 10x = 45g 5x = 9 h 3x = 12 i 2x = x + 3j 3x = x + 4

4D

SkillSH

EET 4.2

23---

EXCEL

Spreadsheet

Exponentialfunctions

14--- 1

3---

WORKEDExample

12Mathca

d


GCpro

gram



y

x

21

3

0 1 2 3

y

x

–3–4

–2

0

(1, –2)

WORKEDEExample

13WorkS

HEET 4.1


A world population model

The statistics below describe P, the estimated world population (in billions) at various times t.

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.

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

STAT

STAT2nd

2nd ENTER

EXCEL Spreadsheet

Worldpopulation


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

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

For example, 32 = 9 can be written as log3 9 = 2.105 = 100 000 can be written as log10 100 000 = 5.

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

y = ax

Base numeralBase

Logarithm

Career profileA L I S O N H E N N E S S Y — Au d i o l o g i s t

Qualifications:Bachelor of ScienceGraduate Diploma in AudiologyMaster of ScienceEmployer:Bionic Ear Institute, East MelbourneCompany website:http://www.medoto.unimelb.edu.au/oto/http://www.medoto.unimelb.edu.au/crc/Audiology appealed to me as it combined my love of science and my desire to work with people. Part of my day involves clinical work: people attending our clinic for hearing tests, specialised diagnostic tests, counselling about hearing loss, and for the fitting and evaluation of hearing aids. I also give lectures to our university students and am involved in attracting research students to our Cooperative Research Centre. I read to keep up-to-date with advances in hearing aids.

I use a number of different formulas for prescribing the amplification required from hearing aids for

hearing-impaired clients. The amount of amplification is dependent on a number of factors: the hearing loss at a particular frequency, the hearing loss at other frequencies, the growth of loudness and the listening situation (for example, quiet or noisy, soft or loud sound). Formulas are becoming more complicated as hearing aid technology improves.

Another area where a formula is used is in the calculation of sound intensity levels, which are measured in decibels. One formula is:

where L is the sound intensity

level in decibels (dB), I is the sound intensity in watts per square metre (W/m2) and I0 is the reference sound intensity which is usually the threshold of hearing and so has a value of 1.0 × 10–12 W/m2.

To interpret journal articles and marketing materials sent by hearing aid manufacturers, I need to understand the mathematical processes involved in the statistical tests reported. This knowledge helps me to assess whether the claims are supported by the results.

Maths is involved in a subtle way in my job. The emphasis is on correctly assessing hearing loss and providing the best possible assistance in improving communication. In order to do this I spend a lot of time talking to people and conducting tests. At the same time it is critical to the success of my job that I understand scientific principles and mathematical processes.

Questions1. List two reasons why Alison finds

mathematics useful in her profession.2. Use the formula given to calculate the

sound intensity level for a jackhammer, at a distance of 10 m, that has an intensity of 1.5 × 10–3 W/m2.

3. Find out what subjects are recommended to be undertaken in Year 12 to continue on to a tertiary course in this field.

L 10 10II--0

log=

C h a p t e r 4 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 187Using the indicial equivalent, it is possible to find the exact value of some logarithms.

The following section goes beyond the requirements of the study design but isimportant particularly for Mathematical Methods Units 3 & 4.

Logarithm lawsThe index laws can be used to establish corresponding rules for calculations involvinglogarithms. These rules are summarised in the following table.

It is important to remember that each rule works only if the base, a, is the same foreach 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 anddivision — the slide rule.

Name Rule Restrictions

Logarithm of a product loga (mn) = loga m + loga n m, n > 0a > 0, a ≠ 1

Logarithm of a quotientloga = loga m – loga n

m, n > 0a > 0 and a ≠ 1

Logarithm of a power loga mn = n loga m m > 0a > 0 and a ≠ 1

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

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

Evaluate the following without a calculator.a log6 216 b log2 ( )

THINK WRITE

a Let x equal the quantity we wish to find. a Let x = log6 216Express 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

18---

12

3

4

118---

218---

12---

3

(2 1– )3

3

4

14WORKEDExample

Mathcad

Logarithmlaws

mn----


Simplify, and evaluate where possible, each of the following without a calculator.a log10 5 + log10 4 b log2 12 + log2 8 − log2 3THINK WRITE

a Apply the ‘logarithm of a product’ rule.

a log10 5 + log10 4 = log10 (5 × 4)

Simplify. = log10 20b Multiply the base numerals of the

logs being added since their bases are the same.

b log2 12 + log2 8 − log2 3 = log2 (12 × 8) − log2 3

= log2 96 − log2 3

Apply the ‘logarithm of a quotient’ law.

= log2 (96 ÷ 3)

Simplify, noting that 32 is a power of 2.

= log2 32= log2 25

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

= 5 log2 2= 5

1

2

1

2

3

4

15WORKEDExample

Simplify 3 log2 5 − 2 log2 10.THINK WRITE

Express both terms as logarithms of index numbers.

3 log2 5 − 2 log210 = log2 53 − log2 102

Simplify each logarithm. = log2 125 − log2 100Apply the ‘logarithm of a quotient’ law. = log2 (125 ÷ 100)Simplify. = log 2 or log2 1.25

1

234 (5

4---)

16WORKEDExample

Simplify each of the following.

a b 2 log10 x + 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 log8 7.

=

b Express 2 log10 x as log10 x2 and 1 as a logarithm to base 10 also.

b 2 log10 x + 1 = log10 x2 + log10 10

Simplify using the ‘logarithm of a product’ law.

= log10 10x2

log8 49log8 343---------------------

1log8 49

log8 343---------------------

log8 72

log8 73-----------------

22 log8 7

3 log8 7-------------------

3 23---

1

2

17WORKEDExample


Logarithms

1 Express the following indicial equations in logarithmic form.

2 Express the following logarithmic equations in indicial form.

3The value of log525 is:

4When expressed in logarithmic form, 83 = 512 is:

5When expressed in indicial form, log10 10 000 = 4 is:

6 Evaluate each of the following without a calculator.

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

a 23 = 8 b 35 = 243 c 50 = 1d 0.01 = 10−2 e bn = a f 2−4 =

a log4 16 = 2 b log10 1 000 000 = 6 c log2 = −1

d log3 27 = 3 e log5 625 = 4 f log2 128 = 7g log3 = −2 h logb a = x

A −2 B 5 C 1 D 2 E 4

A log3 8 = 512 B log3 512 = 8 C log8 512 = 3D log512 3 = 8 E log8 3 = 512

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

a log2 16 b log3 81 c log5 125 d log2

e log10 1000 f log10 (0.000 01) g log2 0.25 h log3

i log2 32 j log2 k log3 (−3) l logn n5

a log2 8 + log2 10 b log3 7 + log3 15 c log10 20 + log10 5d log6 8 + log6 7 e log2 20 − log2 5 f log3 36 − log3 12

g log5 100 − log5 8 h log2 + log2 9 i log4 25 + log4

j log10 5 − log10 20 k log3 − log3 l log2 9 + log2 4 − log2 12

m log3 8 − log3 2 + log2 5 n log4 24 − log4 2 − log4 6

remember1. If y = ax then loga y = 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) loga m + loga n = loga (mn)

(b) loga m − loga n = logamn----

remember

(c) loga mn = n loga m(d) loga a = 1(e) loga 1 = 0

4E

116------

Mathcad

Logarithmsto anybase

12---

19---




WORKEDExample

1414---

1243---------

164------

WORKEDExample

15

13--- 1

5---

45--- 1

5---


8 Simplify each of the following.

9 Simplify the following.

10The expression log10 xy is equal to:

11The expression log5 x

y is equal to:

12

The expression log2 64 + log2 5 can be simplified to:

13

The expression can be simplified to:

14 Express each of the following in simplest form:

a 3 log10 5 + log10 2 b 2 log2 8 + 3 log2 3

c 2 log3 2 + 3 log3 1 d log5 12 − 2 log5 2

e 4 log10 2 + 2 log10 8 f log3 42 + 3 log3 2

g log2 27 − log2 36 h log2 (x − 4) + 3 log2 x

i log3 16 + 2 log3 4 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 CD y log10 x E log10 x + log10 y

A x log5 y B y log5 x C 5 logx yD log5 x + log5 y E 5y

A log2 40 B 1 C log2 D log2 20 E log2

A log4 x3 B log4 C D log4 (x

5 − x2) E log4 x7

a log3 27 + 1 b log4 16 + 3 c 3 log5 2 − 2d 2 + 3 log10 x e 2 log2 5 − 3 f 4 log3 2 − 2 log3 6 + 2g 2 log6 6 − log6 4 h + 3 log10 x

2

WORKEDExample

16

13--- 1

2---

12---

WORKEDExample

17a log3 25

log3 125---------------------

log2 81

log2 9------------------

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 x3

log10 x--------------------- log5 x

32---

log5 x-------------------

2 log2 x 1+( )3

log2 x 1+( )-----------------------------------


log10 x

log10 y-----------------


mmultiple choiceultiple choice13---

6415------ 320

3---------


log4 x5

log4 x2-----------------

x52--- 5

2---

WORKEDExample

17b

12---

C h a p t e r 4 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 191Solving logarithmic equations

This section goes beyond the study design requirements but could be considered aspreparation for Mathematical Methods Unit 3.

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

For example, to evaluate log10 8, 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 followingsteps should be followed:1. Simplify the equation using log laws.2. Express the equation in index form if required.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:

loga x is defined for x > 0, if a > 0

This can be seen more clearly using index notation as follows:Let n = loga x.Therefore, an = x (indicial equivalent of logarithmic expression).However, an > 0 for all values of n if a > 0 (positive based exponentials are alwayspositive). Therefore, x > 0.

8 LOG ENTERLOG 8

Find x if log3 9 = x − 2.THINK WRITE

Write the equation. log3 9 = x − 2Simplify the logarithm using the ‘logarithm of a power’ law and the fact that log3 3 = 1.

log3 32 = x – 22 log3 3 = x – 2

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

1

2

3

18WORKEDExample

Solve for x if log6 x = −2.THINK WRITE

Write the equation. log6 x = −2Express in index form. Therefore, x = 6−2.

Evaluate the index number. x =

=

12

31

62-----

136------

19WORKEDExample


Solving exponential equations using log10 on the calculatorWe have already seen three methods for solving exponential equations:1. equating the bases, which is not always an option, for example, 2x = 72. using a calculator and trial and error, which can be time consuming3. using a graphical technique and a graphics calculator.

An efficient method for solving equations involves the use of logarithms and thelog10 function on the calculator. This is outlined in the following examples.

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 mostcommonly used for this solution technique.

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

THINK WRITE

Write the equation. 2 logx 25 = 4Divide both sides by 2. logx 25 = 2Write 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

1234

5

20WORKEDExample

Solve for x, correct to 3 decimal places, if 2x = 7.

THINK WRITE

Write the equation. 2x = 7Take log10 of both sides. log10 2x = log10 7Use the ‘logarithm of a power’ law to bring the power, x, to the front of the logarithmic equation.

x log10 2 = log10 7

Divide both sides by log10 2 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

4log107

log102---------------

50.84510.3010----------------

6

21WORKEDExample

log10 b

log10 a-----------------


Solving logarithmic equations

1 Find x in each of the following.

2 Solve for x.

3 Solve for x given that:

4a The solution to the equation log7 343 = x is:

b If log8 x = 4, then x is equal to:

c Given that logx 3 = , x must be equal to:

d The solution to the equation log3 x − 2 = log3 (x − 8) is:

5 Solve the following equations correct to 3 decimal places.

6The nearest solution to the equation 4x = 5 is:

7The nearest solution to the equation 0.62x − 1 = 2 is:

a log2 4 = x b log9 1 = x c log3 27 = x d log4 256 = x

e log10 = x f log3 = x g 2 log2 8 = x h log3 81 = 2x

i log10 1000 = 2x − 1 j 2 log2 32 = 3x + 1

a log2 x = 3 b log3 x = 2 c log5 x = 4 d log10 x = 1e log8 x = −1 f log3 x = −3 g log2 x = 6 h log10 x = 4i log3 (x − 3) = 3 j log2 (3x + 1) = 4 k log10 (2x) = 1 l 2 log6 (3x) = 1m log5 x = log5 4 + log5 6 n log3 5 − log3 4 = log3 x – log3 8

a logx 36 = 2 b logx 125 = 3 c 3 logx 16 = 6

d −2 logx = 4 e logx 64 = 3 f 5 logx 625 = 10

g logx + 127 = 3 h −log3x − 1 = 5

A x = 2 B x = 3 C x = 1 D x = 0 E x = −2

A 4096 B 512 C 64 D 2 E

A 3 B 6 C 81 D 1 E 9

A x = 8 B x = 6 C x = 9 D x = −4 E x = 2

a 2x = 11 b 2x = 0.6 c 3x = 20d 3x = 1.7 e 5x = 8 f 0.7x = 3g 10x − 1 = 18 h 3x + 2 = 12 i 22x + 1 = 5j 43x + 1 = 24 k 10−2x = 7 l 82 − x = 0.75

A x = 0.86 B x = 1.2 C x = 1.25 D x = 1 E x = 0.5

A x = 0.18 B x = 0.13 C x = −0.18 D x = −0.71 E x = −0.13

remember1. 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

4FWORKEDExample

18110------ 1

9---

WORKEDExample

19

WORKEDExample

201

100--------- 1

2---

132------


12---

12---

WORKEDExample

21



WorkS

HEET 4.2


Further work on logarithmic graphs is available on the Maths Quest CD Rom. Click on the ‘Extension — Logarithmic graphs’ panel.

Logarithmic graphs1 Using a graphics calculator or graphing software produce graphs of the

following on the same set of axes. Ensure equal axis scales if possible (if using a graphics calculator, use ZOOM and 5:ZSquare). Copy the screen view into your work book.a y = log10x b y = 10x c y = x

2 Copy and complete:a The graph of y = log10 x is the r of the graph of y = 10x in the line

y = x. Such functions are called inverses of each other.b An asymptote is a line that a graph never quite intersects. The line x =

is an asymptote for the graph of y = log10x.3 The following is beyond the scope of the study design for units 1 and 2

Mathematical Methods, but is still of interest here.Use technology to investigate the shape of the graph of

y = Aloga (x + b) + B

for various values of the pronumerals A, a, b and B.Sketch several examples into your workbook, showing asymptotes. The

Maths Quest spreadsheet ‘Logarithmic graphs’ is ideal for this.a What is the effect of A on the graph?b What is the effect of a on the graph?c What is the effect of b on the graph?d What is the effect of B on the graph?

4* Try sketching graphs of the following without using technology.(Hint: Find x and y intercepts by putting y = 0 and x = 0 respectively.)a 2 log10 (x – 3) + 1 b 3log10 (x + 2) – 5 c y = –log10 x – 2

EXCEL

Spreadsheet

Logarithmicgraphs

extensioneextensionxtension — Logarithmic graphs


Applications of exponential and logarithmic functions

Exponential and logarithmic functions can be used to model many practical situationsin science, medicine, engineering and economics.

A square sheet of paper which is 0.1 mm thick is repeatedly folded in half.a Find a rule which gives the thickness, T mm, as a function of the number of folds, n.b What is the thickness after 10 folds?c How many folds are required for the thickness to reach 6 cm?

THINK WRITE

a T = 0.1 when n = 0 and doubles with each fold. This doubling implies that the base should be 2.

a When n = 0, T = 0.1 and as n increases by 1, T doubles.

Complete a table of values showing the thickness, T, for values of n from 0 to 5.

Determine the rule for T(n). There is a doubling term (2n) and a multiplying constant for the starting thickness (0.1).

T(n) = 0.1(2n)

Compare the rule for T(n) against the table of values in step 2.

b Substitute n = 10 into the formula for T.

b When n = 10, T(10) = 0.1(210)

Calculate T. T = 102.4 mm

c Change 6 cm to millimetres. c 6 cm = 60 mmSubstitute T = 60 into the formula. When T = 60,

60 = 0.1 (2n)Divide both sides by 0.1. 600 = 2n

Take log10 of both sides. log10 600 = log10 2n

Use the ‘logarithm of a power’ law to bring the power n to the front of the logarithm.

log10 600 = n log10 2

Divide both sides by log10 2. n =

Evaluate. n = 9.23Round the answer up to the nearest whole number since the number of folds are positive integers and if you round down the thickness will not have reached 60 mm.

Therefore, n = 10 folds.

1

2n 0 1 2 3 4 5

T 0.1 0.2 0.4 0.8 1.6 3.2

3

4

1

2

1

2

3

4

5

6log10 600

log10 2-----------------------

7

8

22WORKEDExample


The price of gold P (dollars per ounce) since 1980 can be modelled by the function: P = 400 + 50 log10 (5t + 1), where t is the number of years since 1980.

a Find the price of gold per ounce in 1980.b Find the price of gold in 1999.c In what year will the price pass $550

per ounce?

THINK WRITE

a State the modelling function. a P = 400 + 50 log10 (5t + 1)Determine the value of t represented by the year 1980.

In 1980, when t = 0,

Substitute t into the modelling function. P = 400 + 50 log10 [5(0) + 1] = 400 + 50 log10 1

Evaluate P. P = 400

b Repeat part a by determining the value of t represented by the year 1999.

b t = 1999 − 1980 t = 19

Substitute the value of t into the modelling function and evaluate P.

When t = 19,P = 400 + 50 log10 [5(19) + 1]

= 400 + 50 log10 96= 400 + 99.114= $499.11

c Since P = 550, substitute into the modelling function and solve for t.

c .550 = 400 + 50 log10 (5t + 1)

Simplify by isolating the logarithm part of the equation.

. 150 = 50 log10 (5t + 1). 3 = log10 (5t + 1)

Express this equation in its equivalent indicial form.

103 = 5t + 1

Solve this equation for t. .1000 = 5t + 1.999 = 5t

199.8 = tConvert the result into years. The price will reach $550 in 199.8 years after 1980.

The price of gold will reach $550 in 1980 + 199.8 = 2180 (approximately).

1

2

3

4

1

2

1

2

3

4

5

23WORKEDExample

remember1. Read the question carefully.2. Use the skills developed in the previous sections to answer the question being

asked.

remember


Applications of exponential and logarithmic functions

1 Prior to a mice plague which lasts 6 months, the popu-lation of mice in a country region is estimated to be10 000. The mice population doubles every monthduring the plague. If P represents the mice populationand t is the number of months after the plague starts:a express P as a function of tb find the population after:

i 3 monthsii 6 months

c calculate how long it takes the population to reach100 000 during the plague.

2 The population of a town, N, is modelled by the function N = 15 000(20.01t) where t isthe number of years since 1980.a Find the population in 1980.b Find the population in: i 1985 ii 1990.c What is the predicted population in 2005?d In what year will the population reach 20 000?

3 The weight of a baby, W kg, t weeks after birth can be modelled by W = 3 log10 (8t + 10).a Find the initial weight.b Find the weight after: i 1 week ii 5 weeks iii 10 weeks.c Sketch the graph.d When will the baby reach a weight of 7 kg?

4 If $A is the amount an investment of $P grows to after n years at 5% p.a.:a write A as a function of Pb use the function from a to find the value of $10 000 after 10 yearsc calculate how many years it will be until an investment of $10 000 reaches $26 500.

5 The value of a car, $V, decreases according to the function V = 25 000 0.1t.a Find the value of the car when new.b Find the value of the car after 6 years.c In how many years will the car be worth $10 000?

6 The temperature, T (°C), of a cooling cup of coffee in a room of temperature 20°C canbe modelled by T = 90(3−0.05t), where t is the number of minutes after it is poured.a Find the initial temperature.b Find the temperature: i 3 minutes after pouring ii 6 minutes after pouring.c How long is it until the temperature reaches half its initial value?

7 A number of deer, N, are introduced to a reserve and its population can be predictedby the model N = 120(1.1t), where t is the number of years since introduction.a Find the initial number of deer in the reserve.b Find the number of deer after: i 2 years ii 4 years iii 6 years.c How long does it take the population to treble?d Sketch the graph of N versus t.e Explain why the model is not reliable for an indefinite time period.

4GWORKEDExample

22

WORKEDExample

23

25---


8 After a recycling program is introduced theweight of rubbish disposed of by a house-hold each week is given by W = 80(2−0.015t),where W is the weight in kg and t is thenumber of weeks since recycling wasintroduced.a Find the weight of rubbish disposed of

before recycling starts.b Find the weight of rubbish disposed of

after recycling has been introduced for:i 10 weeks ii 40 weeks.

c How long is it after recycling starts until the weight of rubbish disposed of is halfits initial value?

d i Will the model be realistic in 10 years time?ii Explain.

9 The number of hectares (N) of forest land destroyedby fire t hours after it started, is given by N = 40 log10 (500t + 1).a Find the amount of land destroyed after:

i 1 hour ii 2 hoursiii 10 hours.

b How long does the fire take to burn out 155hectares?

10 A discus thrower competes at several competitionsduring the year. The best distance, d metres, that heachieves at each consecutive competition is modelled byd = 50 + log10 (15n), where n is the competition number.a Find the distance thrown at the:

i 1st ii 3rd iii 6thiv 10th competition.

b Sketch the graph of d versus n.c How many competitions does it take for the thrower to reach a distance of 53 metres?

11 The population, P, of a certain fish t months after being introduced to a reservoir isP = 400(100.08t), 0 ≤ t ≤ 20. After 20 months, fishing is allowed and the population isthen modelled by P = 15 000 + 924 log10 [10(t − 19)], t ≥ 20.a Find the initial population.b Find the population after:

i 5 months ii 15 months iii 25 months iv 40 months.c How long does it take the population to pass 10 000?

12 A ball is dropped from a height of 5 metres and rebounds to of its previous height.a Find the rule that describes the height of the ball (h metres) after n bounces.b Find the height after: i 4 bounces ii 8 bounces.c Sketch the graph of the height of the ball after n bounces.

13 A computer appreciates in value by 10% per year. If the computer costs $5000 whennew, find:a the rule describing the value, V, of the computer at any time, t years, after purchase.b the value of the computer after 6 years.c the number of years it takes to reach double its original value.

710------


The Richter scaleThe Richter scale is used to describe the ‘strength’ of earthquakes. A formula for the Richter scale is:

K – 0.9, where R is the Richter scale value for an earthquake that releases K kilojoules (kJ) of energy.

1 Find the Richter scale value for an earthquake that releases the following amounts of energy:a 1000 kJ b 2000 kJ c 3000 kJd 10 000 kJ e 100 000 kJ f 1 000 000 kJ

2 Does doubling the energy released double the Richter scale value?

3 Find the energy released by an earthquake of:a magnitude 4 on the Richter scaleb magnitude 5 on the Richter scalec magnitude 6 on the Richter scale

4 What is the effect (on the amount of energy released) of increasing the Richter scale value by 1?

5 Why is an earthquake measuring 8 on the Richter scale so much more devastating than one that measures 5?

Mathcad

TheRichterscale

R 23--- log =


Index laws• am × an = am + n

• am ÷ an = am − n

• (am)n = amn

• a0 = 1• (ab)n = anbn

•

Negative and rational powers

• , a ≠ 0

•

•

Indicial equations• If am = an, then m = n.• A graphics calculator may be used to solve indicial equations, using the solve(

function.

Graphs of exponential functions• If f (x) = ax, a > 1

• If f (x) = ax, 0 < a < 1

• y-intercept is (0, 1)• Asymptote is y = 0 (x-axis)• Domain = R• Range = R+

summary

ab---

n an

bn-----=

a n– 1an-----=

a1n---

an=

amn---- (a

1n---)m ( an )m

amn= = =

y

x0

1f (x) = ax, a > 1

y

x0

1f (x) = ax, 0 < a < 1

C h a p t e r 4 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 201Logarithms• If y = ax then loga y = x where a = the base, x = the power, index or logarithm and

y = the base numeral.Log laws:• loga m + loga n = loga (mn) m, n > 0

• loga m − loga n = loga m, n > 0

• loga mn = n loga m m > 0• loga a = 1• loga 1 = 0

Solving logarithmic equations• Logarithmic equations are solved more easily by:

1. simplifying using log laws2. expressing in index form3. solving as required.

• If ax = b, then x = .

Logarithmic graphs• The logarithmic function f (x) = loga x is the inverse function of the exponential

function g(x) = ax.

• x-intercept is (1, 0)• Asymptote is y = 0• Domain = R+

• Range = R

mn----

log10 b

log10 a-----------------

f (x) = log ax, a > 1

g(x) = ax, a > 1y

x0 1

1

y = x


Multiple choice

1 When simplified, is equal to:

2 may be simplified to:

3 The value of 5−2 is:

4 If 252 − x = 125, then x is equal to:

5 If 42x − 17(4x) + 16 = 0, then x is equal to:

6 The rule for the graph at right could be:A y = 3x − 2

B y = 3x + 2

C y = 3x − 2D y = −2x

E y = 2x + 2

Questions 7 to 9 refer to the function defined by the rule y = 2x + 3 − 1.

7 The graph which best represents this function is:A B C

D E

A B C D E

A B C D E

A B 5 C D E −5

A 1 B C −1 D 2 E 5

A 1 or 16 B 0 or 1 C 2 or 8 D 1 or 4 E 0 or 2

CHAPTERreview

4A2xy3( )2

7x3------------------- 3x5y2

4y--------------×

x4y7

7---------- 3x4y7

7-------------- 3y7

x2-------- 3x4

y6--------

xy--

4A5m4 p2

2m3 p---------------- ÷ 5m2 p6( )3

3m7 p-----------------------

m2

47 p16-------------- 3m10

2 p32------------ 3m2

50 p16-------------- m15

p29-------- m20 p24

25-----------------

4B64125---------

13---–

120------ 4

5--- 5

4---

4C 12---

4C

4D y

x0–1

–2

4Dy

x0–3

7

y

x0

y

x0–3–1

y

x0 3–1

y

x0–3

1

C h a p t e r 4 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 2038 The domain is:

9 The range is:

10 When expressed in log form, 5x = 250 becomes:

11 The value of log7 49 + 3 log2 8 − 4 is:

12 The value of is nearest to:

13 The expression simplifies to:

14 The solution to log5 x = 4 is:

15 The value of x if 2 logx 343 = 6 is:

16 If log3 (2x − 1) + log3 2 = 2, then x is equal to:

17 The solution to the equation 43 − 2x = 12 is nearest to:

Short answer

1 Simplify the following expression with positive indices.

2 Solve the following equations.a 2x5 = 100b 8x + 1 × 22x = 43x − 1

3 Find the solution to 9x − 5(3x) + 6 = 0 correct to 2 decimal places.

A (3, ∞) B [−1, ∞) C R+ D R E R\{−3}

A [−1, ∞) B R C R+ D (1, ∞) E (−1, ∞)

A logx 5 = 250 B log5 x = 250 C log5 250 = x D logx 250 = 5 E log250 x = 5

A 3 B 7 C 0 D 69 E 1

A 3 B 5 C 2 D 9 E 20

A B log7 C log7 D E

A 25 B 125 C 1 D 625 E 20

A 3 B 7 C 5 D 14 E

A 2 B 1 C 3 D E

A x = −1 B x = 0.35 C x = 0.604 D x = 0.2 E x = 0

4D

4D

4E

4E

4Flog3 25

log3 5------------------

4Flog7 x

54---

log7 x-------------------

52--- x

52---

x34--- 3

4--- 5

8---

4F

4F7

4F52--- 11

4------

4F

4A,B16x 6– y10( )

12---

27x3y9( )3÷

4C

4C


4 For the function with the rule f (x) = 3x − 2 + 1:a find the y-interceptb state the equation of the asymptotec sketch the graph of f (x)d state the domain and range.

5 a Evaluate log3b Express y in terms of x if log10 x + log10 y = 2 log10 (x + 1).

6 Simplify the following.a 3 log4 5 − 2 log4 6

b

7 Solve each of the following.a log6 x = 3b 2 logx 125 = 6c log2 (3x + 6) − log2 5 = 2

8 If 5x = 4, find x correct to 3 decimal places.

9 Solve for x where 102 − 3x = 8.

10 The number of bacteria in a culture, N, is given by the exponential functionN = 1500(20.18t), where t is the number of days.

a Find the initial number of bacteria in the culture.b Find the number of bacteria (to the nearest 100) after: i 5 days ii 10 days.c How many days does it take for the number of bacteria to reach 9000?

AnalysisThe number of lions, L, in a wildlife park is given by L = 20(100.1t), where t is the number of years since counting started. At the same time the number of cheetahs, C, is given by C = 25(100.05t).a Find the number of:

i lions ii cheetahswhen counting began.

b Find the numbers of each after i 1 year ii 18 months.

c Which of the animals is the first to reach a population of 40 and by how long?

d After how many months are the populations equal and what is this population?

4D

4E127------

4E2 log5 x2

13--- log5 x---------------------

4F

4F4F4G

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CHAPTERyyourselfourself

testyyourselfourself

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Documents

Exponential and logarithmic functionsmathsbooks.net/Maths Quest 11 Methods/By Chapter/Ch... · exponential functions. For example: f (x) = a x where a > 0 and a ≠ 1 is an exponential