YEAR 12 MATHS
WEEK 1
LAWS OF LOGARITHMS
Lesson objectives
At the end of this lesson, students should be able to


State the laws of logarithms
Apply the laws of logarithms to simplify calculations
DEFINITION: The logarithm to a given base of a number is the power to which the base must
be raised to make the number. If N= ax, then logaN = x
Example 1
1
Evaluate: a) log216 (b) log432 (c) log24√2 (d) log22
SOLUTION
(a) Let log216 = x
∴ 2x = 16
2x = 24
x=4
(b) Let log432 = x
4x = 32
2x = 25
x=5
(c) Let log24√2 = x
2x = 4√2
2x = √32
= 321/2
x
5/2
2 =2
x = 5/2
PRACTICE EXERCICE 1
Find the values of the following:
(1) Log832 (2) log1/216 (3) log80.25 (4) log7√7 (5) log100.001 (6) log80.0625
Since logarithms are exponents, the Laws of Exponents give rise to the Laws of Logarithms.
Let a be a positive number, with a ≠ 1.
Let A, B, and C be any real numbers with A > 0 and B > 0.
•
Law 1: Addition Law: logax + logay=logaxy
The logarithm of a product of numbers is the sum of the logarithms of the
numbers.
Examples
1) log10 4 + log10 25
= log10 ( 4 × 25)
= log10 100
= log10 10 ²
= 2 log10 10
=2
•
Law 2: Subtraction Law
•
logax - logay=logax/y
The logarithm of a quotient of numbers is the difference of the logarithms
of
the numbers.
Example 2
Log428-log414
28
=log4 ( 14 )
=log42
1
=2
•
Law 3
• Power law: logaxn = nlogax
•
The logarithm of a power of a number is the exponent times the logarithm of
the number.
Example 3
a) Log283 = 3log28
= 3log223
= 3×3log22
=9 (since log22= 1)
1
1
b) Log8192 = 2log819
1
1
1
= ( 2 × 2) since log819=2
1
=4
Law 4: Logarithm to its base
Logaa = 1
Example 4:
log22= 1
Expanding and Combining Logarithmic Expressions
The laws of logarithms allow us to write the logarithm of a product or a quotient as the sum or
difference of logarithms
This process—called expanding a logarithmic expression—is illustrated in the next example.
EXAMPLE 2: Use the Laws of Logarithms to expand each expression:
log16
log32
Solution
Log16
log32
log24
log25
= 4log2
5log 2
=
= 4/5
PRACTICE EXERCICE:2
Evaluate the following:
(1)
(2) log 2 48 - log 2 3
log 32
(3) log49 (4) log3 +log4
log8
log√7
(5) Log 8÷ log 4 (6) log√5÷ log 5
Expanding Log Expressions:
 ab 
ln  3   ln(ab )  ln 3 c
 c
Law 2
 ln a  ln b  ln c1/ 3
Law 1
 ln a  ln b  31 ln c
Law 3
Simplification of Logarithm
Example 1)
Combine: 3 log x + ½ log(x + 1) into a single logarithm.
Solution
3log x  21 log( x  1)
 log x 3  log( x  1)1/ 2
(Law 3)
 log x 3  x  1
(Law 1)

1/ 2

Example 2) simplify: 3 ln s + ½ ln t – 4 ln (t2 + 1)
Solution
3ln s  21 ln t  4ln(t 2  1)
 ln s 3  ln t 1/ 2  ln(t 2  1)4
(Law 3)
 ln(s 3t 1/ 2 )  ln(t 2  1)4
(Law 1)
 3

s t 

 ln
 t2 1 4 


(Law 2)


Example (3)
Evaluate without using tables:
2log5+log36-log9
Solution
2log5+log36-log9
=log52+log36-log9
=log25+log36-log9
=log(
25×36
9
)
=log100
=2
Example 4: Simplify
Log10√35- log10√7+log10√2
Solution
Log10√35- log10√7+log10√2
= log10√35+log10√2-log10√7
= log0√35 × 2 ÷ 7
= log10√70/√7
= log10√10
= log10101/2
=1/2log1010
=½
example : evaluate
l𝐥𝐨𝐠 𝟏𝟎
𝟑𝟎
𝟏𝟔
𝟓
𝟒𝟎𝟎
− 𝟐 𝐥𝐨𝐠 𝟏𝟎 𝟗 + 𝐥𝐨𝐠 𝟏𝟎 𝟐𝟒𝟑
Solution
𝟑𝟎
𝟓
𝟒𝟎𝟎
− 𝟐 𝐥𝐨𝐠 𝟏𝟎 + 𝐥𝐨𝐠 𝟏𝟎
𝟏𝟔
𝟗
𝟐𝟒𝟑
𝟑𝟎
𝟓
𝟒𝟎𝟎
=𝐥𝐨𝐠 𝟏𝟎 𝟏𝟔 - 𝐥𝐨𝐠 𝟏𝟎 (𝟗) 𝟐 + 𝐥𝐨𝐠 𝟏𝟎 𝟐𝟒𝟑
𝟑𝟎
𝟐𝟓
𝟒𝟎𝟎
𝟑𝟎
𝟖𝟏
𝟒𝟎𝟎
=𝐥𝐨𝐠 𝟏𝟎 (𝟏𝟔 ÷ 𝟖𝟏 × 𝟐𝟒𝟑)
=𝐥𝐨𝐠 𝟏𝟎 ( 𝟏𝟔 × 𝟐𝟓 × 𝟐𝟒𝟑)
=𝐥𝐨𝐠 𝟏𝟎 𝟏𝟎
=1
PRACTICE EXERCISE:3
EVALUATE the following
(1)
Log91-log13
Log243-log5
(2) 2log216+4log28-3log264
(3) 3log2+log20-log1.6
(4) log68+2log63+log612-log624
(5) log24+log42-log255
(6) 2log52/5+log59-log572/125+log510
SUBSTITUTION AND EQUATION IN LOGARITHM
EXAMPLES:
1) Given that log 2=0.3010, log 3= 0.4771 and log 7=0.8451, evaluate the following
using tables:
(a) Log 0.72
Solution
(b) log 21
(a) Log 0.72= log
72
100
=log 72- log 100
= log (9×8) – log 100
= log (32×23) – log100
= log 32 + log 23 – log 100
= 2 log 3 + 3log 2 – log 100
= 2(0.4771) + 3(0.3010) – 2
=1.8572 – 2,
= - 0.1428
(b) Log 21 = log(3×7)
= log 3 + log 7
= 0.4771 + 0.8451
= 1.3222
(2)
Solve for x given that
2 log10x + log102 = 2
Solution
2 log10x + log102 = 2
Log10x2 + log102 = log10100
Log102x2 = log10100
2x2 = 100
x2 = 100/2
x = √50
x = 7.1
(3)
If log10(2x+1) – log10(3x-2) = 1, find x
Solution
log10 (2x+1) – log10 (3x-2) = 1
2𝑥+1
log10 (
3𝑥−2
2𝑥+1
3𝑥−2
) = log1010
= 10
10(3x - 2) = 2x + 1
30x – 20 = 2x + 1
28x = 21
21
x = 28
x=
3
4
PRACTICE EXERCISE 4
(1)
(2)
(3)
(4)
(5)
Given that log 2 = 0.3010, log 3= 0,4771 and log 7= 0,8451, evaluate
the following:
(a) Log 5 (b) log8 (c) log 35 (d) log 16
Find the value of x if log10(10+9x) – log10(11- x) = 2
Evaluate log2(y-3) + log2y = 2
Solve for x if log24 + log27 – log2x = 1
Solve 4log5n – log5m= 1, 5log5n + 3log5m =14
ASSIGNMENT
If log 3  a and log 5  b , determine an expression for the following in terms of a
and b :
1.
log 15
2.
log 75
If log 2 5  x and log 2 3  y , determine an expression for the following in terms of
x and y :
3.
4.
log 2 20
log 2 24
5.
log 2 7.5
6.
If log 3 5  x and log 2 36  y , determine an expression for
log 9 5  log 2 6 .
7.
If log 4 25  a , determine an expression for log 2 25  log 16 5 .
If log 3 8  k , determine an expression for the following in terms of k :
8.
log 3 2
Write as a single logarithm:
log 5 c  2 log 5 b  2
10
11..
2 log 3 k  3  log 3 p
9..
log 9 8