IL1.4: LOGARITHMS
Logarithms
Logarithms are closely related to indices.
The logarithm of a number is the power to which the base must be raised to give the number.
This means that a logarithm is an index.
Index or logarithm
Now, let’s revise what it means.
8 = 23
Number
base
This familiar statement may be written as
log 2 8 = 3
Index or logarithm
base
Generally:
log a n = x
ax = n
(follow the arrows and read
ax = n
)
( a ⟩ 0)
Examples
1.
Re-write these equations in logarithmic notation:
(a)
53 = 125
log 5 125 = 3
(b)
3− 2 =
log 3
Using the above general expression:
1
9
1
= −2
9
Using the above general expression
1
(c)
25 2 = 5
log 25 5 =
IL 1.4 - Logarithms
1
2
Using the above general expression
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Examples
To evaluate a logarithm:
1. write the number in index form, with same base as the logarithm.
2. use the definition of a logarithm to evaluate.
Evaluate
log5 125
125 = 53
∴
125 in index form with base 5
equivalent logarithm statement
log 5 125 =
3
Evaluate
log10 0.0001
0.0001 = 10– 4
∴
0.0001 in index form with base 10
equivalent logarithm statement
log10 0.0001 =
−4
Two important logs to remember
a0 = 1
∴ a 0 = 1 ⇔ log a 1 = 0
In words: For any base, the log of 1 is zero.
a = a1 .
∴ a1 = a ⇔ log a a = 1
In words:
The log of any number with itself as base is 1.
Logarithm Laws
Corresponding to the three index laws, there are three laws of logarithms to help in manipulating logarithms.
First Logarithm Law
log a (mn ) = log a m + log a n
Second Logarithm Law
m
log a   = log a m − log a n
n
Third Logarithm Law
( )
log a m p = p log a m
IL 1.4 - Logarithms
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Examples
log a 15 = log a (5 × 3)
= log a 5 + log a 3
Using the first logarithm law
 12 
log a   = log a 12 − log a 5 Using the second logarithm law
5
( )
log a 9 2 = 2 log a 9
( )
= 2 log a 32
Using the third logarithm law
= 4 log a 3
Note:
To use the logarithm laws, all logarithms must have the same base.
These laws together with the definition of a logarithm can be used to simplify and evaluate logarithmic
expressions and to solve equations involving logarithms.
Examples
1.Solve
log 2 x = 5
log 2 x = 5 ⇒ 25 = x using the definition of a logarithm
Therefore x = 32
2.Simplify
log10 6 + log10 2
log10 6 + log10 2 = log10 ( 6 × 2 ) using first log. law
= log10 12
3. Simplify
log 3 6a + log 3 b − log 3 2a
Use the laws for adding and subtracting
logarithms.
= log 3 ( 6a × b ) − log 3 2a
 6a × b 
= log 3 

 2a 
= log 3 3b
IL 1.4 - Logarithms
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4. Simplify
1
log10 36 − log10 15 + 2 log10 5
2
1
2
1
log 10 36 and 2 log10 5 must be written as log10 36 2
and log10 5 before using addition and subtraction laws.
2
1
= log10 36 2
− log10 15 + log10 52
use laws for adding and subtracting logs
=log10 6 − log10 15 + log10 25
6 × 25
= log10
15
= log10 10
=1
5.Simplify
3log10 a − 2 log10 b + 2 log10 5
3log10 a − 2 log10 b + 2 log10 5 = log10 a 3 − log10 b 2 + log10 52
 25a 3 
2 
 b 
= log10 
6. Solve for x
log10 (2 x + 1) = log10 3
Therefore
2x +1 =
3
both sides of the equation are log10 so can equate (2x+1) and 3
x =1
Exercises
Exercise 1
Write in logarithm form.
(a) 32
9=
(b) 104 10 000
(c) 10−2
0.01
=
(d) e a b
Exercise 2
Evaluate without using a calculator.
(a) log 7 49
IL 1.4 - Logarithms
(b) log10 10
(c) log 5 1
(d) log10 100 000 (e) log 5 5
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Exercise 3
1. Simplify.
1
log10 25 − log10 4 + 2 log10 3
2
(a) log 4 8 + log 4 3 − log 4 2
(b)
(c) log e 2e3 + 2 log e 3 − log e 18
(d) log a 4 + 2 log a 3 − 2 log a 6
(e)
1
log10 a 2 + 3log10 b − log10 ab 2
2
2.Solve for x
(b) log 2 2 x − 2 log 2 3 =
log 2 6
(a) log10 x = log10 4 − log10 2
Answers
Exercise 1
(a) log 3 9 = 2 (b) log10 10, 000 = 4
Exercise 2
(a) 2
(b)
1/2
(c) 0
(c) log10 0.01 = −2
(d) 5
(d) log e b = a
(e) 1
Exercise 3
1.(a) log 4 12
2. (a) x = 2
IL 1.4 - Logarithms
45
4
(b) x = 27
(b) log10
(c) 3
(d) 0
(e) log10 b
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