Logarithms and Logarithmic Laws
3C Logarithms
10
2
= 100
the base 10 raised to the power 2 gives 100
2 is
the power
which the base 10 must be raised to, to give 100
the power
= logarithm
2 is the logarithm to the base 10 of 100
Logarithm is the number which we need to raise
a base to for a given answer
to what power must I raise 2 to give an answer of 64?
ans =
6
written as log
2
64 = 6
to what power must I raise 2 to give an answer of 64?
ans =
6
written as log
2
64 = 6
to what power must I raise 5 to give an answer of 625?
ans =
4
written as log
5
625 = 4
to what power must I raise 9 to give an answer of 3?
ans =
1
2
written as log
9
3 =
1
2
log
a
n = p a
p
= n
base
answer
= number inside
1.
log
a
m + log
a
n = log
a
mn
2.
log
a
m - log
a
n = log
a
m
n
3.
n
log
a
m = log
a
m
n
4.
log
n
m = log
a
m
change of base law
log
a
n
N.B. The log of a negative is impossible to find
Proofs:
Law 1
log
a
m + log
a
n = log
a
m n
Let log
a
m = p & log
a
n = q
a
p
= m
a
q
= n
a
p
. a
q
=
m . n
a
p + q
=
m . n
base
answer
= number inside
log
a
m . n = p + q
log
a
m n = log
a
m + log
a
n
Law 2
Let log
a
m = p & log
a
n = q
a
p
= m
a
q
= n
a
p
=
a
q
m
n
a
p
-
q
=
m
n
= p - q
= log
a
m - log
a
n
( )
n
( )
n
Law 4
n
p
=
log
n
m =
log
a
m
log
a
n
Let log
n
m = p
take logs of
both
sides
log
a
n
p
=
log
a
m
Law 3
Let log
a
m = p
a
p
=
m
n
log
a
m = log
a
m
n
We need m
n
a
p
=
m
a
pn
=
m
n
base
answer
= number inside
log
a
m
n
= pn
log
a
m
n
= (log
a
m) n
log
a
m
n
= n log
a
m
m
p log
a
n =
log
a
m
p =
log
a
m
log
a
n
log
n
m =
log
a
m
log
a
n
e.g.1 log
4
64 = x
base
answer
= number inside
4
x
=
64
4
x
=
4
3
x = 3
e.g.2 log
2
x = 5
2
5
= x
x =
32
e.g.3 log
4
(5x + 6) = 2
4
2
= 5x + 6
16
= 5x + 6
10
= 5x
10
5
= x
2
= x
e.g.4 log
3
(2x - 4) = 1 + log
3
(4x + 8)
log
3
(2x – 4) – log
3
(4x + 8)
= 1
12x + 24 = 2x - 4
10x = -28
x = -
28
10
x = -2.8
For an unknown power always
take logs of both sides
e.g.5 6
n
= 3200
log
10
6
n
= log
10
3200
n log
10
6 = log
10
3200
n = 4.5045
Calculations using log
10
log
10
1000 = 3
as 10
3
= 1000
If we want log
2
32 =
log
n
m =
log
a
m
change of base law
log
a
n
log
10
32
log
10
2
e.g.6 log
2
55 = log
10
x
log
10
55
log
10
2
= log
10
x
5.78 = log
10
x
base
answer
= number inside
10
5.78
= x
x = 604449
Logarithms help us determine the power to which a base must be raised to give a specific result. This content explains concepts like logarithmic laws, changing bases, solving logarithmic equations, and includes examples for better understanding.
Download Presentation
Please find below an Image/Link to download the presentation.
The content on the website is provided AS IS for your information and personal use only. It may not be sold, licensed, or shared on other websites without obtaining consent from the author.If you encounter any issues during the download, it is possible that the publisher has removed the file from their server.
You are allowed to download the files provided on this website for personal or commercial use, subject to the condition that they are used lawfully. All files are the property of their respective owners.
The content on the website is provided AS IS for your information and personal use only. It may not be sold, licensed, or shared on other websites without obtaining consent from the author.
E N D
Presentation Transcript
3C Logarithms the base 10 raised to the power 2 gives 100 102 = 100 2 is the power which the base 10 must be raised to, to give 100 the power = logarithm 2 is the logarithm to the base 10 of 100 Logarithm is the number which we need to raise a base to for a given answer to what power must I raise 2 to give an answer of 64? ans = 6 written as log2 64 = 6
to what power must I raise 2 to give an answer of 64? ans = 6 written as log2 64 = 6 to what power must I raise 5 to give an answer of 625? ans = 4 written as log5 625 = 4 to what power must I raise 9 to give an answer of 3? ans = 1 2 written as log9 3 = 1 2 loga n = p ap = n baseanswer = number inside
1. loga m + loga n = loga mn 2. loga m - loga n = loga m 3. n loga m = loga mn n 4. logn m = loga m loga n change of base law N.B. The log of a negative is impossible to find
Law 2 loga m - loga n = loga m Let loga m = p & loga n = q ap = m ap = aq ap - q =m n m n m n Proofs: Law 1 loga m + loga n = loga m n Let loga m = p & loga n = q ap = m ap . aq = m . n ap + q = m . n loga m . n = p + q loga m n = loga m + loga n n aq = n aq = n m n loga = p - q baseanswer = number inside loga = loga m - loga n
Law 3 n loga m = loga mn Law 4 logn m = loga m loga n Let loga m = p Let logn m = p ap = m We need mn take logs of both sides m np = ( )n = m ( )n ap apn = mn loga mn = pn loga mn = (loga m) n loga mn = n loga m loga np = loga m p loga n = loga m p = loga m baseanswer = number inside loga n logn m = loga m loga n
e.g.1 log4 64 = x e.g.3 log4 (5x + 6) = 2 baseanswer = number inside 42= 5x + 6 16= 5x + 6 10= 5x 10 5 2 = x 4x= 64 4x= 43 x = 3 = x e.g.2 log2 x = 5 25= x x = 32
e.g.4 log3 (2x - 4) = 1 + log3 (4x + 8) log3 (2x 4) log3 (4x + 8) 2 4 log 4 8 x + 2 4 3 4 8 x + 3(4 8) 2 4 x x + = For an unknown power always take logs of both sides e.g.5 6n = 3200 = 1 = x 1 log10 6n = log10 3200 3 n log10 6 = log10 3200 x = 1 n = log10 3200 log10 6 n = 4.5045 12x + 24 = 2x - 4 10x = -28 x = -28 x = -2.8 10
Calculations using log10 logn m = logam change of base law loga n log10 1000 = 3 as 103 = 1000 If we want log2 32 = log10 32 log10 2 e.g.6 log2 55 = log10 x log10 55 log10 2 = log10 x baseanswer = number inside 5.78 = log10 x 105.78 = x x = 604449
