Download logs defined

LOGARITHMS
We have learnt indices or exponents in the algebra material. If you haven’t check then, we recommend
you to do so. We will need those concepts for progressing with logs. The idea of logarithms (or simply
logs) is based on indices. In fact, as you will find out very soon, the rules for logarithms are very similar
to the rules for indices. Therefore, a recap of the concept of indices will be useful for us to understand
how logarithms works.
EXPONENTS REVISITED
First, a number can always be expressed as another number (called base) raised to a certain number
of power (called index or exponent). For example:
32 = 25
1331 = 113
10,000 = 104
1000 = 103
2 = 10 0.301030
3 = 10 0.477121
You can use your calculator to check the last two examples are correct.
In general, a number (N) can be represented in index form:
N = Bx
where B is the base and x is the index (or power).
In the section on equations, we have introduced the method of solving simple power equations. We
know that we can solve for N if B and x are given; similarly we can solve B if N and x are known. But
what if we want to solve for x, given N and B? Unless N and B are very ‘friendly’ numbers (as in 16 =
4x), there is no way for us to find out the value of x. Try this for yourself.
Solve the following equation for x:
2 = 3x
To solve the above equation, we need logarithms!
LOGS DEFINED
Given:
N = Bx
The log of a number (N) with base (B), written as “log B N”, would be equal to x. That is:
log B N = x
What the definition means is that the log of a number (N) is just the power to which the base (B) is
raised to produce N. (This idea is fairly subtle and elusive. It is crucial you grasp this concept.)
Examples
Logarithmic notation and exponential notation are just alternative ways of stating the same
relationship between the three terms (N, B and x).
32 = 2 5
is equivalent to log 2 32 = 5
1000 = 10 3
is equivalent to log 10 1000 = 3
2 = 10 0.301030
is equivalent to
log 10 2 = 0.301030
In the first example, the log has a base of 2. The commonly encountered logarithm uses the base of
10, as shown in the other two examples. If you come across logarithms without a base, such as log 2,
it actually uses the base of 10. The usual notation of logarithm to the base of 10 is just log.
Note: There is an important logarithm, called the natural logarithm (Ln), in statistics. It uses a base of
e which is a number with non-recurring decimals and its value is approximately 2.7183. It is also
written as log e. and you will see a lot of equations involving natural logs in statistics materials and
econometrics.
FINDING THE LOG OF A NUMBER
To find the value of log of a number N with base B, log B N , is equivalent to solving for the unknown
(x) in the following equation:
x
N= B
That is, we ask: To which power must B be raised in order to get N?
Examples
(a)
To find the value of log 3 81 , we ask: What is the index (x), such that 3 to the power x is equal
to 81?
That is,
3x = 81
x=4
Therefore,
(b)
[What is value of x?]
[34 = 3 * 3 * 3 * 3 = 81]
log 3 81 = 4
log 7 49 = ?
We ask: To which power must 7 (base) be raised to obtain 49?
That is,
7x = 49
x=2
Therefore,
(c)
log 7 49 = 2
log10 10 = ?
We ask: To which power must 10 (base) be raised to obtain 10?
10x = 10
That is,
x = ??
Therefore,
log10 10 = ??
X=1
(d)
log10 5 = ?
Now we must use the calculator. Using the calculator, we get ‘0.698970’. It means:
10 0.698970 = 5
The table below shows the log of some selected numbers at base 10:
N
Log N
0
1 (10 )
2
3
5
10 (101)
20
30
50
100 (102)
200
300
500
1000 (103)
0
0.301030
0.477121
0.698970
1
1.301030
1.477121
1.698970
2
2.301030
2.477121
2.698970
3
Did you observe any interesting pattern?
BASIC LOGARITHMS RULES
1.
2.
3.
log
M
*
N

log
M

log
N
B
B
B
M
l
o
g
l
o
g
M

l
o
g
N
B 
B
B
N
p
lo
g
M

P
lo
g
M
B
B
4.
lo
gM
lo
g
 a
BM
lo
g
aB
Solving Power Equations (I)
The first type of power equation does not require logs. If you remember, we have done similar
problems before.
1.
P5 = 16
P  5 16
[Take the fifth root on both sides.]
P = 1.741
[Use your calculator.]
Solving Power Equations (II)
When the index is the unknown, we have to use logarithms to solve the equation.
Examples
1.
Solve the following equation for x:
3x = 2
Solutions:
3x = 2
log 3x = log 2
x*log 3 = log 2
x =
log 2
log 3
x = 0.63093
[Step 1: Take log on both sides.]
[Step 2: Use Rule 3.]
[Step 3: Divide both sides by log3.]
[Use your calculator.]
2.
Solve the equation for n:
460(1.08)n = 925
Solution:
460(1.08)n = 925
(1.08)n
[Step 1: Divide both sides by 460.]
(1.08)n = 2.0108696
log (1.08)n = log 2.0108696
[Step 2: Take log on both sides.]
n*log 1.08 = log 2.0108696
[Step 3: Use Rule 3.]
n=
log
2.0108696
log
1.08
n = 9.0768945
[Step 4: Divide both sides by log1.08.]
APPENDIX
Basic Rules in Logarithms
1.
log
M
*
N

log
M

log
N
B
B
B
Proof:
Let M = Bx and N = By
It follows that log B M = x and log B N = y.
Now,
LHS
[Definition of logarithm]
= log B
= log B (Bx
y
)
= log B (Bx+y )
[B ? = Bx+y]
= x+y
= log B M + log B N
= RHS
2.
M
l
o
g
l
o
g
M

l
o
g
N
B 
B
B
N
The proof is very similar to (1). I leave the proof to the reader as self-practice.
3.
p
lo
g
M

P
lo
g
M
B
B
Proof:
Let M = Bx.
It follows that log B M = x
Then,
LHS
= log B Mp
= log B (Bx) p
= log B (Bx*p)
= x*p
= p log B M
= RHS
4.
lo
gM
lo
g
 a
BM
lo
g
aB
Let M = ax and B = ay.
It follows that log a M = x and log a B = y
So, RHS =
loga M x

loga B y
[By substitution]
On the other hand, B = ay can be re-written as: a = B1/ y
So, we have M = ax = = (B1/ y)x = Bx / y
[Index rule]
Therefore we have:
LHS
g

lo
g
= lo
BM
BB
xy
/
=
x
y
= RHS