Chapter 14 Indices Exponentials and Logarithms Part 2

14
CH APTER
Number and Algebra
Indices, exponentials and

logarithms – part 2
In Chapter 9, starting with integer powers of numbers, we developed the ideas
of the exponential function and the logarithmic function. We learned basic
properties, such as:
2x 2y = 2x + y and log 2 ( xy) = log 2 x + log 2 y
In this chapter, we will investigate the change of base formula and meet a range
of new applications, especially applications to science.
ICE-EM Mathematics 10 3ed ISBN 978-1-108-40434-1 © The University of Melbourne / AMSI 2017 Cambridge University Press
417
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14A Logarithm rules
In Section 9G, we introduced logarithms. Logarithms are closely related to indices. Recall that the
logarithm of a number to base a is the index to which a is raised to give this number. For example:
34 = 81 is equivalent to log3 81 = 4
106 = 1 000 000 is equivalent to log10 1 000 000 = 6
1 1
5−3 = is equivalent to log5 = −3
125 125
3
3
16 4 = 8 is equivalent to log16 8 =
4
In general, the logarithmic function is defined as follows:
If a > 0, a ≠ 1 and y = a x , then loga y = x
Logarithms obey a number of important laws. Each one comes from a property of indices.
Index laws
If a and b are positive numbers and x and y are rational numbers, then:
ax
Index law 1 a x a y = a x + y Index law 2 = ax− y
ay
Index law 3 ( a x ) y = a xy Index law 4 ( ab ) x = a x b x
x
 a ax
Index law 5   =
 b bx
The first three index laws have a direct correspondence to the first three logarithmic laws, which are
developed below.
Suppose a > 0 and a ≠ 1 for the rest of this section.
Logarithmic Law 1 If x and y are positive numbers, then loga xy = loga x + loga y.
That is, the logarithm of a product is the sum of the logarithms.
Suppose that loga x = c and loga y = d

That is, x = ac and y = a d
Then xy = ac × a d
= ac + d (by Index law 1)
So loga xy = loga ac + d
=c+d
= loga x + loga y
41 8 I C E - E M M at h em at ic s y e a r 1 0
1 4 A L o g a r i t h m r u le s
x
Logarithmic Law 2 If x and y are positive numbers, then loga y = loga x − loga y.
That is, the logarithm of a quotient is the difference of their logarithms.
Suppose that loga x = c and loga y = d

That is, x = ac and y = a d
x ac
Then = d
y a
= ac − d (by Index law 2)
x
So loga = loga a c − d
y
=c−d
= loga x − loga y
Logarithmic Law 3 If x is a positive number and n is any rational number, then

loga ( x n ) = n loga x .
This follows from index law 3. Suppose that loga x = c. That is, x = ac.
Then x n = ( ac ) = a cn (by Index law 3)
n
So loga ( x n ) = loga (acn )

Hence, loga ( x n ) = cn
= n loga x , as required
1
Logarithmic Law 4 If x is a positive number, then loga = – loga x .
x
This follows from logarithm law 3.

1
loga = loga x −1 (definition)
x
= − loga x (logarithm law 3)
Logarithmic Law 5 loga 1 = 0 and loga a = 1
Let the base a be a positive number, with a ≠ 1.

Since a 0 = 1, we have loga 1 = 0 .
Similarly, since a1 = a, we have loga a = 1.
Example 1
Write each statement in logarithmic form.

1
a 24 = 16 b 53 = 125 c 10 –3 = 0.001 d 2 −4 =
16
C h a p t e r 1 4 I n d ic e s , e x p o n e n t ia l s a n d l o g a r i t h m s – pa r t 2
419
Solution
a 24 = 16 so log2 16 = 4 b 53 = 125 so log5 125 = 3

1 1
c 10 –3 = 0.001 so log10 0.001 = –3 d 2 −4 = so log2 = −4
16 16
Example 2
Evaluate each logarithm.
a log2 256 b log2 3 2 c log3 81

1 1
d log9 81 e log5 f log7
5 49
Solution
Method 1
1
1
a 256 = 28, so log2 256 = 8 b 3
2 = 2 3, so log2 3 2 =
3
c 81 = 34, so log3 81 = 4 d 81 = 9 , so log9 81 = 2
2
1 1
e log5 = log5 5−1 f log7 = log7 7 −2
5 49
= −1 = −2
Method 2
The following method introduces a pronumeral x.
a Let x = log2 256 b Let x = log2 3 2
so 2 x = 256 = 28 3
1
so 2x = 2 = 23
x =8
1
x =
3
c Let x = log3 81 d Let x = log9 81
so 3x = 81 = 34 so 9 x = 81 = 92
x =4 x = 2
Example 3
Solve each logarithmic equation.

1
a log2 x = 5 b log7 ( x − 1) = 2 c log x 64 = 6 d log x = –2
25
Solution
a log2 x = 5 b log7 ( x − 1) = 2
so x = 25 so x − 1 = 72
= 32 = 49
x = 50
1
c log x 64 = 6 d log x = −2
25
so x 6 = 64 1
so x −2 =
x 6 = 26 25
x = 2, since x > 0 x = 25
2
x = 5, since x > 0
Example 4
Write each statement in logarithmic form.

3
a y = bx b ax = N c 70 =1 d 3 3 = 32
Solution
a y = b x becomes x = logb y b a x = N becomes x = loga N

3
3
c 70 = 1 becomes log7 1 = 0 d 3 3 = 3 2 becomes log3 3 3 =
2
Example 5
Given log7 2 = α , log7 3 = β and log7 5 = γ , express each in terms of α , β and γ .

15
a log7 6 b log7 75 c log7
2
Solution
a log7 6 = log7 (2 × 3) b log7 75 = log7 (3 × 25)

= log7 2 + log7 3 = log7 3 + log7 52
= α+β = log7 3 + 2 log7 5
= β + 2γ
15
c log7 = log7 15 − log7 2
2
= log7 (3 × 5) − log7 2
= log7 3 + log7 5 − log7 2
=β+γ −α
421
Exercise 14A
Example 2 1 Calculate each logarithm.

a log2 8 b log3 27 c log2 2048 d log7 1
e log5 625 f log 7 343 g log10 10 000 h log10 1 000 000
2 Calculate:
1 1 1
a log2 b log3 c log10 d log10 0.01
16 27 10
1 1 1
e log5 f log6 g log2 h log10 0.0001
125 36 1024
3 Evaluate:
a log2 2 2 b log3 9 3 c log6 36 6 d log2 4 2
 1 
e log3 (27 3) f log10  g log5 (52 × 3 5) h log8 2
 100 10 
Example
3a, b
4 Solve each equation for x.
a log2 x = 5 b log3 x = 6 c log10 x = 3
d log10 x = –3 e log10 x = – 4 f log5 x = 4
g log2 ( x – 3) = 1 h log2 ( x + 4) = 6 i log2 ( x – 5) = 3
Example
3c, d
5 Solve each equation.
a log x 81 = 2 b log x 8 = 6 c log x 1024 = 5
d log x 1024 = 10 e log x 9 = 2 f log x 1000 = 3
Example
1, 4
6 Write each statement in logarithmic form.
1 −1
( 2) c   = 2
2
a 2 = b 0.001 = 10 –3 d 1024 = 322
 2
3
e 10 x = N f 5 2 = 5 2 g 50 = 1 h 131 = 13
7 Write each statement in exponential form.
a log2 32 = 5 b log3 81 = 4 c log10 0.001 = –3
7
d log3 27 3 = e logb y = x f loga N = x
2
8 Simplify:
a log3 7 + log3 5 b log2 3 + log2 5 c log2 9 + log2 7
1
d log10 5 + log10 20 e log6 4 + log6 9 f log3 7 + log3
7
9 Simplify:
a log3 100 – log3 10 b log7 20 – log7 10 c log7 21 – log7 3
d log3 17 – log3 51 e log5 100 – log5 10 f log5 10 – log5 2
10 Simplify:
a log2 3 + log2 5 + log2 7 b log3 100 – log3 10 – log3 2
c log5 7 + log5 343 – 2 log5 49 d log7 25 + log7 3 – log7 75
Example 5 11 Given that log10 2 = α , log10 3 = β, log10 5 = γ and log10 7 = δ , express in terms
of α , β, γ and δ:
a log10 12 b log10 75 c log10 210
d log10 6 000 000 e log10 1875 f log10 1050
g log10 (2a3b5c 7d ) h What does α + γ equal?
12 Find a relation between x and y that does not involve logarithms.

a log3 x + log3 y = log3 ( x + y) b 2 log10 x – 3log10 y = –1
c log5 y = 3 + 2 log5 x d log7 (1 + y) – log7 (1 – y) = x
4 3
13 V = πr is the volume of a sphere of radius r. Express log2 V in terms of log2 r .
3
14 If y = a × 10 bx, express x in terms of the other pronumerals.
15 Solve log10 A = bt + log10 P for A.
14B Change of base

In Section 14A we studied logarithms to one base (which was a positive number other than 1) and
their relationships, such as:
loga x + loga y = loga xy
Often we need to work with different bases and, in particular, calculate quantities such as log5 8,
which is clearly between 1 and 2. It is of immediate concern that some calculators do not have the
capacity to calculate log5 8 directly, but they can calculate log10 8 and log10 5.
log10 8
We will show that log5 8 = ≈ 1.2920.
log10 5
This is a special case of the change of base formula:
loga c
logb c =
loga b
where a, b and c are positive numbers, a ≠ 1 and b ≠ 1.
The change of base formula is very important in later mathematics.
423
1 4 B C h an g e o f ba s e
Proof 1 Proof 2
Let x = logb c so, b x = c If loga b = e , then ae = b
Taking logarithms to base a of both sides: Similarly, if logb c = f , then b f = c
loga b x = loga c Hence, c = b f = (a e ) f = a ef
x loga b = loga c (Logarithm law 3) So loga c = ef = loga b × logb c
loga c loga c
x = and logb c =
loga b loga b
loga c
That is, logb c =
loga b
Change of base formula

• If a, b and c are positive numbers, a ≠ 1 and b ≠ 1 then:
log a c
log b c =
log a b
• This formula can also be written as:

log a c = log a b × log b c
These formulas are called ‘change of base’ formulas, since they allow the calculation of logarithms
to the base b from knowledge of logarithms to the base a.
Example 6
By changing to base 2, calculate log16 8.
Solution
log2 8 = 3 and log2 16 = 4,

log2 8
hence, log16 8 =
log2 16
3
=
4
3
So log16 8 =
4
3 3
As a check, 16 4 = (24 ) 4 = 23 = 8
1 4 B C h an g e o f ba s e
Example 7
Calculate log7 8 , correct to four decimal places, using base 10 logarithms
Solution
Changing from base 7 to base 10:

log10 8
log7 8 =
log10 7
≈ 1.0686
As a check, 71.0686 ≈ 7.9997 with a calculator.
Example 8
If 3x = 7, calculate x, correct to four decimal places.
Solution
log10 7
x = log3 7 =
log10 3
≈ 1.7712
Example 9
Suppose that a > 0. Find the exact value of loga a3. 2
Solution
loga a 3
loga a3 =
2
loga a 2
3 loga a
=
2 loga a
3
=
2
3
As a check, 2
(a ) 2 = a3
425
Exercise 14B
In this exercise, a, b and c are positive and not equal to 1.

Example 6 1 a By changing to base 3, calculate log9 243.
b By changing to base 2, calculate log8 32.
Example 7 2 Use the change of base formula to convert to base 10 and calculate these logorithms, correct
to four decimal places.
a log7 9 b log5 3 c log3 5
1
d log3 13 e log19 17 f log7
4
Example 8 3 Solve for x, correct to four decimal places.
a 2 x = 5 b 3x = 18 c 5x = 2
d 5x = 17 e 2 – x = 7 f 3– x = 5
4 Solve for x, correct to four decimal places.
a ( 0.01)x = 7 b 51– 2 x = 3 c 4 2 x –1 = 7 x –3 d 33 x – 3 = 55 x – 5
5 Simplify:
a (loga b)(logb a) b (loga b)(logb c)(logc a)
Example 9 6 Change to base a and simplify.
11
a loga a3
2 b loga a 7
2 c loga a53 d log 3
a a
3 5
e loga a8 – loga a 7 + loga a11 f log a a + log 3
a
4
a + log 4
a a
14 C Graphs of exponential and

logarithm functions
We saw the basic shape of the graph of an exponential function in Chapter 9.
For example, y = 2 x is graphed to the right. y
The graph has the following features:

y = 2x
• The y-intercept is 1.
(1, 2)
• There is no x -intercept.
1
• The y- values are always positive.
0 x
• As x takes large positive values, 2 x becomes very large.
• As x takes large negative values, 2 x becomes very small.
• The x -axis is an asymptote to the graph.
1 4 C G r a p h s o f e x p o nen t ial an d l o g a r i t h m f u nc t i o n s
y
Here are the graphs of y = 3x and y = 3– x drawn on the same axes. y = 3x
Notice that y = 3x is the reflection of y = 3– x in the y- axis.
y = 3−x
(−1, 3) (1, 3)
(0, 1)
0 x
Simple logarithm graphs

We can also draw the graph of y = log2 x . As usual, we begin with y y = log2 x
(8, 3)
a table of values. (4, 2)
1 1 1 1 (2, 1)
x 1 2 4 8 16
16 8 4 2
0 (1, 0) x
y = log 2 x – 4 –3 –2 –1 0 1 2 3 4 1,
2
−1
How are the graphs of y = log2 x and y = 2 x related? y

y = 2x
Here is a table of values of y = 2 x. The graphs of y = 2 x (2, 4) x

=
y
and y = log2 x are shown on the one set of axes. y = log2x
(1, 2)
(4, 2)
x –4 –3 –2 –1 0 1 2 3 4 (0, 1)
(2, 1)
1 1 1 1
y= 2x 1 2 4 8 16 0 (1, 0) x
16 8 4 2
If the point (a, b) lies on y = 2 x, then b = 2a.

Hence, we can write a = log2 b , so (b, a) lies on the graph of y = log2 x .
Thus, each point on y = log2 x can be obtained by taking a point on y = 2 x and interchanging the x
and y values.
a + b b + a
The midpoint of (a, b) and ( b, a ) is  ,  , and thus always lies on the line y = x .
 2 2 
Graphically this means (a, b) is the reflection of ( b, a ) in the line y = x and vice versa. This is
evident in the above pair of graphs.
From this we can list some of the features of the graph of y = log2 x .
• The graph is to the right of the y- axis. (This is because the function is only defined for x > 0 .)
• The y- axis is a vertical asymptote to the graph.
• The x-intercept is (1, 0 ), corresponding to log2 1 = 0.
• The graph does not have a y- intercept.
• As x takes very large positive values, log2 x becomes large positive.
• As x takes very small positive values, log2 x becomes large negative.
• The graph is a reflection of y = 2 x in the line y = x .
427
Example 10
Use the graph of y = 3x to assist in sketching y = log3 x.
Solution
First draw the graph of y = 3x .

y y=x
(2, 9)
y = 3x
(1, 3)
y = log3 x (9, 2)
–1, 13 (0, 1)
(3, 1)
0 (1, 0) x
1
3, –1
The two graphs are reflections of each other in the line y = x .
Example 11
Sketch the graph of y = log2 ( x – 3) .
Solution
Translate the graph of y = log2 x three units to the right.

y
y = log2 (x − 3)
x=3
0 3 4 x
Note that the line x = 3 is an asymptote to the graph.
Example 12
Sketch the graphs of y = log3 x and y = log5 x on the same set of axes.
Solution
1 1
x 1 5 25
25 5
y = log 3 x −2.93 −1.46 0 1.46 2.93
y = log 5 x −2 −1 0 1 2
y y = log3 x
log10 5
log3 5 = ≈ 1.46
log10 3
log3 25 = log3 52 = 2 log3 5 ≈ 2.93 y = log5 x
1
log3 = − log3 5 ≈ −1.46 0 x
5 (1, 0)
1
log3 = log3 5−2 = −2 log3 5 ≈ −2.93
25
The table of values shows that:

log3 x > log5 x if x > 1 and log5 x > log3 x if 0 < x < 1
Exercise 14C
Example 10 1 a Use the graph of y = 4 x to draw the graph of y = log4 x .

b Use the graph of y = 5x to draw the graph of y = log5 x .
2 For each of these logarithm functions, produce a table of values for ( x , y), using the
following y-values: –2, –1, 0, 1, 2. Use the table to draw the graph of the function.
a y = log10 x b y = log6 x
3 Draw each set of graphs on the same axes.
1
a y = 3x , y = 3x + 1, y = 3x – 2 b y = 5x , y = 2 × 5x , y = × 5x
2
x −x
 1  1
c y = 2 x , y = 2 – x d y =   , y =  
2 2
Example 11 4 a Sketch the graphs of y = log2 x and y = log3 x on the same set of axes, for y values
between –3 and 3.
b In what ways are the graphs similar?
c How do the graphs differ?
d Without using a table of values, sketch the graph of y = log4 x on the same set of axes
used in part a.
429
Example 12 5 Sketch the following graphs.
a y = log3 x , x > 0 b y = log3 ( x – 1), x > 1 c y = log3 ( x + 5), x > –5
d y = 2 log3 x , x > 0 e y = log3 ( x ) + 2, x > 0
6 Sketch y = 2 x , y = 3x , y = log2 x and y = log3 x on the one set of axes.
14D Applications to science,

population growth and finance
In Section 9F of you saw that in a given experiment, the growth in bacteria could be described using
an exponential function, such as N = 1000 × 2t .
Here, N is the number of bacteria at time t, measured in hours.
Equations of this type arise in many practical situations in which we know the value of N, but want
to solve for t.
Logarithms are needed for such calculations.
Example 13
Initially there are 1000 bacteria in a given culture. The number of bacteria, N, is doubling
every hour, so N = 1000 × 2t , where t is measured in hours.
a How many bacteria are present after 24 hours? Give your answer correct to three
significant figures.
b How long is it until there are one million bacteria? Give your answer correct to
three significant figures.
Solution
a After 24 hours, N = 1000 × 224

≈ 1.68 × 1010
b If N = 106 , then 106 = 1000 × 2t
2t = 1000
log10 2t = log10 1000
t log10 2 = 3
3
t =
log10 2
≈ 9.97 hours
There are one million bacteria after approximately 9.97 hours.
1 4 D A p p licat i o n s t o s cience , p o p u lat i o n g r o w t h an d f inance
The following example illustrates the use of logarithms in estimating the age of fossils.
Example 14
The carbon isotope carbon-14, C14, occurs naturally but decays with time. Measurements of
carbon-14 in fossils are used to estimate the age of samples.
If M is the mass of carbon-14 at time t years and M 0 is the mass at time t = 0 ,
then M = M 0 10 – kt where k = 5.404 488 252 × 10 –5 .
All 10 digits are needed to achieve reasonable accuracy in these calculations.
a Calculate the fraction left after 100 years as a percentage.
b Calculate the fraction left after 10 000 years as a percentage.
1
c Calculate the half-life of C14. That is, after how long does M = M 0?
2
Solution
a When t = 100, M = M 0 10 –100 k

M
= 10 −100 k
M0
≈ 0.987 63
≈ 98.76%
That is, the fraction left after 100 years is 98.76%.
b When t = 10 000 , M
= 10 −10 000k
M0
≈ 0.288 11
≈ 28.81%
That is, the fraction left after 10 000 years is 28.81%.
1 1
c M = M 0 when = 10 − kt
2 2
1
log10 = − kt
2
kt = log10 2
log10 2
t =
k
≈ 5570.000 001
≈ 5570 years
That is, the half-life of C14 is about 5570 years.
431
Compound interest
In Section 1D, we introduced the compound interest formula:
An = P (1 + R)n
where An is the amount that the investment is worth after n units of time, P is the principal
and R is the interest rate.
Logarithms can be used to find the value of n in this formula given R, P and An.
Example 15
$50 000 is invested on 1 Jan at 8% per annum. Interest is only paid on 1 Jan of each year.
At the end of how many years will the investment be worth
a $75 000? b $100 000?
Solution
a An = P(1 + R)n
An = 75 000, P = 50 000 and R = 0.08, so
75 000 = 50 000(1.08)n
3
= (1.08)n
2
3
log10 = n log10 (1.08) (Take logarithms of both sides.)
2
3
log10  
 2
n=
log10 (1.08)
= 5.268 44 …
At the end of the sixth year, the investment will be worth $50 000 (1.08 )6 = $79 343.72 .
At the end of the fifth year, the investment will be worth $50 000 (1.08 )5 = $73 466.40 .
The investment will be worth more than $75 000 at the end of the sixth year.
b An = P(1 + R)n
An = 100 000, P = 50 000 and R = 0.08, so
100 000 = 50 000(1.08)n

2 = (1.08)n
log10 (2) = n log10 (1.08) (Take logarithms of both sides.)
log10 (2)
n=
log10 (1.08)
= 9.006 46 …
At the end of the tenth year, the investment will be worth $50 000 (1.08 )10 = $107 946.25.
At the end of the ninth year, the investment will be worth $50 000 (1.08 )9 = $99 950.23 .
The investment will be worth more than $100 000 at the end of the tenth year.
Exercise 14D
Example 13 1 A culture of bacteria initially has a mass of 3 grams and its mass doubles in size every hour.
How long will it take to reach a mass of 60 grams?
2 A culture of bacteria initially weighs 0.72 grams and is multiplying in size by a factor of
five every day.
a Write down a formula for M, the weight of bacteria in grams after t days.
b What is the weight after two days?
c How long will the culture take to double its weight?
d The mass of the Earth is about 5.972 × 10 24 kg. After how many days will the culture
weigh the same as the Earth?
e Discuss your answer to part d.
3 The population of the Earth at the beginning of 1976 was four billion. Assume that the rate
of growth is 2% per year.
a Write a formula for P, the population of the Earth in year t , t ≥ 1976.
b What will be the population in 2076?
c When will the population reach 10 billion?
4 The population of the People’s Republic of China in 1970 was 750 million. Assume that its
rate of growth is 4% per annum.
a Write down a formula for C , the population of China in year t , t ≥ 1970.
b When would the population of China reach two billion?
c With the assumptions of question 3, when would the population of China be equal to half
the population of the Earth?
d When would everyone in the world be Chinese? (Discuss your answer.)
Example 14 5 The mass M of a radioactive substance is initially 10 g and 20 years later its mass is 9.6 g.
If the relationship between M grams and t years is of the from M = M o 10 − kt, find:
a M o and k
b the half-life of the radioactive substance
Example 15 6 An amount of $80 000 is invested on 1 Jan at a compound interest rate of 7% per annum.
Interest is only paid on 1 Jan of each year. At the end of how many years will the investment
be worth:
a $110 000 ? b $200 000?
7 A man now owes the bank $47 000, after taking out a loan n years ago with an interest rate
of 10% per annum. He borrowed $26 530. Find n.
433
8 The formula for the calculation of compound interest is An = P(1 + R)n. Find, correct to
one decimal place:
a An if P = $50 000, R = 8% and n = 3
b P if An = $80 000, R = 5% and n = 4
c n if An = $60 000, R = 2% and P = $20 000
d n if An = $90 000, R = 4% and P = $20 000
Review exercise
1 Calculate each logarithm.
a log2 16 b log5 125 c log 2 512 d log7 1
1 1
e log3 f log2 g log10 10 000 h log10 ( 0. 001)
27 64
2 Solve each logarithmic equation.
a log x 16 = 2 b log x 64 = 6
c log x 2048 = 11 d log x 512 = 3
e log x 25 = 2 f log x 125 = 3
3 Write each statement in logarithmic form.
a 1024 = 210 b 10 x = a
c 60 = 1 d 111 = 11
e 3x = b f 54 = 625
4 Write each statement in exponential form.
a log3 81 = 4 b log2 64 = 6
c log10 0. 01 = –2 d logb c = a
e loga b = c
5 Simplify:
a log2 11 + log2 5 b log2 7 + log2 5
c log6 11 + log6 7 d log3 8 – log3 32
e log5 200 – log5 40 f log5 30 – log5 6
6 Simplify:
a log2 5 + log2 4 + log2 7 b log5 1000 – log5 100 – log5 10
c log7 7 + log 7 343 – 3log7 49 d log3 25 + 2 log3 5 – 2 log3 75
Re v ie w E x e r ci s e
7 Use the change of base formula to convert to base 10 and calculate each to
four decimal places.
a log7 11 b log5 7 c log3 24
1
d log3 35 e log16 8 f log3
4
8 Solve for x, correct to four decimal places.
a 2 x = 7 b 3x = 78
c 5x = 28 d 5x = 132
e 2 – x = 5 f 3– x = 15
9 Solve for x.
a log2 (2 x – 3) = 4 b log3 3 x = 4
c log2 (3 – x ) = 2 d log10 x = 4
e log4 (5 – 2 x ) = 3 f log2 ( x – 6) = 2
10 Sketch each graph.

a y = log5 x , x > 0
b y = log3 ( x – 2), x > 2
c y = log2 ( x + 4), x > – 4
d y = log2 ( x ) + 5, x > 0
11 Express y in terms of x when:

a log10 y = 1 + log10 x
b log10 ( y + 1) = 2 + log10 x
8 3 3
12 Simplify log2   − 2 log2   − 4 log2   .
 75   5  2
 x2 
13 If log10 x = 0.6 and log10 y = 0.2, evaluate log10  .
 y 
14 a Express 3 + log2 5 as a single logarithm.

b Express 5 − log2 5 as a single logarithm.
15 An amount of $120 000 is invested on 1 Jan at a compound interest rate of 8% per annum.
Interest only paid on 1 Jan of each year. At the end of how many years will the investment
be worth:
a $160 000? b $200 000?
435
Challenge exercise
Throughout this exercise, the bases a and b are positive and not equal to 1.
1 Consider a right-angled triangle with side lengths a, b and c, with c the hypotenuse.
1 1
Prove that log10 a = log10 (c + b) + log10 (c − b).
2 2
2 Simplify loga (a 2 + a) – loga (a + 1).
1  x 3 y2 
3 Show that 3log10 x + 2 log10 y – log10 z = log10  .
2  z 
4 Solve for x:
a log2 ( x + 1) – log2 ( x – 1) = 3
b (log10 x )(log10 x 2 ) + log10 x 3 – 5 = 0
c (log2 x 2 )2 – log2 x 3 – 10 = 0
d (log3 x )2 = log3 x 5 – 6
5 Solve each set of simultaneous equations.

a 9 x = 27 y –3 , 16 x +1 = 8 y × 2
b 8 x = 32 y +1 , 5x –1 = 25 y
c 49 x + 3 = 343 y –1 , 2 x + y = 8 x − 2 y
d 8 x = 4 y , 73 x + 3 = 343 y
6 Solve the equation (loga x )(logb x ) = loga b for x where a and b are positive numbers
different from 1.
7 If a = log8 225 and b = log2 15, find a in terms of b.
8 a Show that log10 3 cannot be a rational number.

b Show that log10 n cannot be a rational number if n is any positive integer that is not a
whole number power of 10.
 xy  yz  zx 
9 Prove that loga   + loga   + loga   = loga x + loga y + loga z.
 z  x  y
10 If x and y are distinct positive numbers, a > 0 and

log x loga y loga z
a = = , show xyz = 1 and x x y y z z = 1.
y−z z−x x−y
11 If 2 loga x = 1 + loga (7 x – 10 a), find x in terms of a, where a is a positive constant and
x is positive.

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14

Number and Algebra

Indices, exponentials and

Suppose a > 0 and a ≠ 1 for the rest of this section.

Suppose that loga x = c and loga y = d

Suppose that loga x = c and loga y = d

Logarithmic Law 3 If x is a positive number and n is any rational number, then

So loga ( x n ) = loga (acn )

This follows from logarithm law 3.

Logarithmic Law 5 loga 1 = 0 and loga a = 1

Let the base a be a positive number, with a ≠ 1.

Write each statement in logarithmic form.

a 24 = 16 so log2 16 = 4 b 53 = 125 so log5 125 = 3

Evaluate each logarithm.

a log2 256 b log2 3 2 c log3 81

Solve each logarithmic equation.

Write each statement in logarithmic form.

a y = b x becomes x = logb y b a x = N becomes x = loga N

Given log7 2 = α , log7 3 = β and log7 5 = γ , express each in terms of α , β and γ .

a log7 6 = log7 (2 × 3) b log7 75 = log7 (3 × 25)

Example 2 1 Calculate each logarithm.

12 Find a relation between x and y that does not involve logarithms.

15 Solve log10 A = bt + log10 P for A.

14B Change of base

Change of base formula

• This formula can also be written as:

By changing to base 2, calculate log16 8.

log2 8 = 3 and log2 16 = 4,

Calculate log7 8 , correct to four decimal places, using base 10 logarithms

Changing from base 7 to base 10:

If 3x = 7, calculate x, correct to four decimal places.

Suppose that a > 0. Find the exact value of loga a3. 2

In this exercise, a, b and c are positive and not equal to 1.

14 C Graphs of exponential and

The graph has the following features:

Simple logarithm graphs

How are the graphs of y = log2 x and y = 2 x related? y

Here is a table of values of y = 2 x. The graphs of y = 2 x (2, 4) x

If the point (a, b) lies on y = 2 x, then b = 2a.

Use the graph of y = 3x to assist in sketching y = log3 x.

First draw the graph of y = 3x .

The two graphs are reflections of each other in the line y = x .

Sketch the graph of y = log2 ( x – 3) .

Translate the graph of y = log2 x three units to the right.

Note that the line x = 3 is an asymptote to the graph.

y = log 3 x −2.93 −1.46 0 1.46 2.93

The table of values shows that:

Example 10 1 a Use the graph of y = 4 x to draw the graph of y = log4 x .

6 Sketch y = 2 x , y = 3x , y = log2 x and y = log3 x on the one set of axes.

14D Applications to science,

a After 24 hours, N = 1000 × 224

a When t = 100, M = M 0 10 –100 k

That is, the half-life of C14 is about 5570 years.

10 Sketch each graph.

11 Express y in terms of x when:

14 a Express 3 + log2 5 as a single logarithm.

5 Solve each set of simultaneous equations.

7 If a = log8 225 and b = log2 15, find a in terms of b.

8 a Show that log10 3 cannot be a rational number.