Transcript “Teach A Level Maths” Vol. 1: AS Core Modules 46: Indices and Laws of Logarithms © Christine Crisp Indices and Laws of Logarithms Unknown Indices We have.

“Teach A Level Maths”
Vol. 1: AS Core Modules
46: Indices and Laws of
Logarithms
© Christine Crisp
Indices and Laws of Logarithms
Unknown Indices
We have met the graph of y  a x and seen that
it represents growth or decay.
Because of important practical applications of
growth and decay, we need to be able to solve
equations of the type
x
a
b
where a and b are constants.
Equations with unknown indices are solved using
logarithms. We will see what a logarithm is and
develop some rules that help us to solve equations.
Indices and Laws of Logarithms
e.g. How would you solve
Ans: If we notice that
10 x  1000 ?
1000  10 3
Indices and Laws of Logarithms
e.g. How would you solve
Ans: If we notice that
then, (1) becomes
10 x  1000 - - - - (1)
1000  10 3
10 x  10 3
 x3
We can use the same method to solve


3 x  81
3 x  34
x4
or


5 2 x  25
52x  52
2x  2  x  1
Indices and Laws of Logarithms
Suppose we want to solve
10 x  75
We need to write 75 as a power ( or index ) of 10.
Tip: It’s useful to notice that, since 75 lies
This index is called a logarithm
( or log 2) and 10 is
1
between 10 and 100 ( or 10 and 10 ), x lies
the base.
between 1 and 2.
Our calculators give us the value of the logarithm
of 75 with a base of 10.
The button is marked
log
The value is
1 875 ( 3 d.p. ) so,
10 x  101  875
 x  1 875
Indices and Laws of Logarithms
A logarithm is just an index.
To solve an equation where the index is unknown, we
can use logarithms.
e.g. Solve the equation 10 x  4 giving the answer
correct to 3 significant figures.
x is the logarithm of 4 with a base of 10
We write
10 x  4  x  log10 4
 0  602 ( 3 s.f. )
In general if
x
10  b then x  log10 b
index  log
Indices and Laws of Logarithms
In general if
a b
x
x  log b
then
a
Memory aid
In general if
log b  x
a
then
a b
a to the power of x = b
x
Indices and Laws of Logarithms
Exercise
Solve the following equations giving the answers
correct to 2 d.p.
(a) 10 x  230
(b) 10 2 x  0  5
Solution:
(a) 10 x
 230
( Notice that 2  x  3 )
x  log 10 230
 x  2  36 ( 2 d.p. )
(b) 2 x  log10 0  5
 2 x   0  301  x   0 15 ( 2 d.p. )
Indices and Laws of Logarithms
In the exercise, we saw that
x
10  230
Generalizing this,
10 x  b


x  log 10 230
x  log 10 b
In general if
log b  x
10
then
10  b
x
This relationship is also true changing from the log
form to the index form,
Indices and Laws of Logarithms
The equation
ax  b
When the base, a, is 10, we found the equation is
easy to solve.
e.g. Solve the equation 10 x  275
Solution: 10 x  275  x  log10 275

e.g. To solve
2x  5
x  2  44 ( 3 s.f. )
we could write x  log 2 5
BUT there are no values for logs with base 2 on our
calculators so we can’t find this as a simple number.
We need to develop some laws of logs to
enable us to solve a variety of equations
with unknown indices or logs
Indices and Laws of Logarithms
A law of logs for
e.g.
loga x k
log 10 2  0  301
( from the calculator )
Also,
log10 2 2  log10 4  0  602 ( from calculator )
 2  0  301
 2 log10 2
And,
log10 2 3  log10 8  0  903 ( from calculator )
 3  0  301
 3 log10 2
Indices and Laws of Logarithms
A law of logs for
e.g.
loga x k
log 10 2  0  301
( from the calculator )
Also,
log10 2 2  log10 4  0  602 ( from calculator )
 2  0  301
 2 log10 2
And,
log10 2 3  log10 8  0  903 ( from calculator )
 3  0  301
 3 log10 2
Indices and Laws of Logarithms
A law of logs for
e.g.
loga x k
log 10 2  0  301
( from the calculator )
Also,
log10 2 2  log10 4  0  602 ( from calculator )
 2  0  301
 2 log10 2
And,
log10 2 3  log10 8  0  903 ( from calculator )
 3  0  301
 3 log10 2
We get
log10 x k  k log10 x
Indices and Laws of Logarithms
A law of logs for
loga x k
The same reasoning holds for any base, a, so
loga x k  k loga x
( the “power to the front ” law of logs )
Indices and Laws of Logarithms
Solving
ax b
e.g.1 Solve
2x  5
( Notice that 2 < x < 3 since 2 2
 4 and 2 3  8 )
We “take” logs
Solution:
2x  5
We don’t actually
the
logs anywhere: we put
log10 2 xtake
 log
10 5
them in, but the process is always called taking logs!
Using the “power to the front” law, we can simplify
the l.h.s.
x log10 2  log10 5
We used logs with baselog
1010because
the values are
5

on the calculator. xHowever,
log10 2 any base could be
used. You could check the result using the “ln”
2  32 you
( 3will
s.f.meet
) in A2 ).
button ( which uses a base

Indices and Laws of Logarithms
Solving
ax b
e.g.2 Solve the equation 1000  100 ( 3) x
Solution: We must change the equation into the
form b  a x before we take logs.
Divide by 100:
x
x
1000  100 ( 3)
Take logs:
 10  3
log10  log 3 x
Using the “power to the front” law:

log 10  x log 3
log10
x
log 3
x  2  10 ( 3 s.f. )
Indices and Laws of Logarithms
SUMMARY
 The Definition of a Logarithm
ax  b

x  log a b
 The “Power to the Front” law of logs:
loga x k  k loga x
x
 Solving the equation
na
b
•
Divide by n
•
“Take” logs
•
Use the power to the front law
•
Rearrange to find x.
Indices and Laws of Logarithms
Exercises
1. Solve the following equations giving the answers
correct to 2 d.p.
(a) 3 x  14
(b) 12 2 x  15
(a)
log 10 3 x  log 1014
 x log 10 3  log 10 14
log10 14
 x
 2  40 ( 2 d.p. )
log10 3
(b) “Take” logs: log 10 12 2 x  log 1015
 2 x log 10 12  log 10 15
log10 15
 2x
 1  0898
log10 12
“Take” logs:
 x  0 54 ( 2 d.p. )
Indices and Laws of Logarithms
Exercises
2. Solve the equation 500  200 ( 2) x giving the
answer correct to 2 d.p.
Solution: Divide by 200:
x
500  200 ( 2)
Take logs:
Power to the front:
Rearrange:
 25  2 x
log 2  5  log 2 x
log 2  5  x log 2
log 2  5
x
log 2
 x  1 32 ( 2 d.p. )
Indices and Laws of Logarithms
Log laws for Multiplying and Dividing
We’ll develop the laws by writing an
example with the numbers in index form.
Indices and Laws of Logarithms
42  26  1092
A log is just an index, so to write this in index
form we need the logs from the calculator.
log10 42  1 623 and log10 26  1 415
1 623
1 415

42  26
 10
 10
 101  623  1  415
 10 3  038 (  1092 )
 log10 (42  26)  3  038
So,
log10 ( 42  26) 
Indices and Laws of Logarithms
42  26  1092
A log is just an index, so to write this in index
form we need the logs from the calculator.
log10 42  1 623 and log10 26  1 415
1 623
1 415

42  26
 10
 10
 101  623  1  415
 10 3  038 (  1092 )
 log10 (42  26)  3  038
So,
log10 ( 42  26)  log10 42 
Indices and Laws of Logarithms
42  26  1092
A log is just an index, so to write this in index
form we need the logs from the calculator.
log10 42  1 623 and log10 26  1 415
1 623
1 415

42  26
 10
 10
 101  623  1  415
 10 3  038 (  1092 )
 log10 (42  26)  3  038
So,
log10 ( 42  26)  log10 42  log10 26
Indices and Laws of Logarithms
42  26  1092
A log is just an index, so to write this in index
form we need the logs from the calculator.
log10 42  1 623 and log10 26  1 415
1 623
1 415

42  26
 10
 10
 101  623  1  415
 10 3  038 (  1092 )
 log10 (42  26)  3  038
log10 ( 42  26)  log10 42  log10 26
log10 ( xy)  log10 x  log10 y
In general,
So,
Indices and Laws of Logarithms
Any positive integer could be used as a base instead
of 10, so we get:
loga ( xy)  loga x  loga y
A similar rule holds for dividing.
 x
loga    loga x  loga y
 y
If the base is missed out, you should assume it
could be any base e.g. log 2 might be base 10 or
any other number.
Indices and Laws of Logarithms
SUMMARY
 The Laws of Logarithms are:
•
loga xy  loga x  loga y
1. Multiplication law
•
2. Division law
•
3. Power law
x
loga  loga x  loga y
y
loga x k  k loga x
 The definition of a logarithm:
x
a
b
leads to 4.
log

a
x  log a b
1 0
k
log
a
k
6.
a
5.
loga a  1
Indices and Laws of Logarithms
e.g. 1 Express the following in terms of
log 2, log 3 and log 5
1
(a) log 15 (b) log 16
(c) log 
3

Solution:
(a) log 15  log 3  5  log 3  log 5 ( Law 1 )
 log 2 4  4 log 2 ( Law 3 )
1
(c) Either log   log 1  log 3 ( Law 2 )
 3
( Law 4 )
 0  log 3
  log 3
1
Or
log   log 3 1
 3    log 3 ( Law 3 )
(b)
log 16
Indices and Laws of Logarithms
e.g. 2 Express log(a b 2 ) in terms of
log a and log b
Solution:
We can’t use the power to the front law directly!
( Why not? )
There is no bracket round the ab, so the square
ONLY refers to the b.
So,
log(a b 2 )  log a  log b 2
( Law 1 )
 log a  2 log b
( Law 3 )
Indices and Laws of Logarithms
e.g. 3 Express each of the following as a single
logarithm in its simplest form:
(a) log 5  log 2  log 3 (b) 2 log10 4  1 log10 25  1
2
5

2


log 5  log 2  log 3  log

 3 
 10 
 log 
 3
(b) 2 log10 4  1 log10 25  1
Solution: (a)
2
1
25 2
 log10 4  log10
 log10 10
 2

2
 16  10 
 4  10 
 log10 
 log10 
  log10 32
1 
 51 


2
 25 
5
This could be simplified to log10 2  5 log10 2
2
Indices and Laws of Logarithms
Exercise
1. Express the following in terms of
log 2, log 3 and log 5
1
(c) log
(a) log 25 (b) log 6
10
Ans: (a) 2 log 5 (b) log 2  log 3 (c)  log 2  log 5
2
2. Express log a b in terms of
Ans:
log a and log b
2 log a  log b
3. Express the following as a single logarithm in
its simplest form:
(a) log 3  log 2  log 5 (b) 3 log10 2  1 log10 16  1
2
15
16
Ans: (a) log
(b) log10
2
5
Indices and Laws of Logarithms
Indices and Laws of Logarithms
The following slides contain repeats of
information on earlier slides, shown without
colour, so that they can be printed and
photocopied.
For most purposes the slides can be printed
as “Handouts” with up to 6 slides per sheet.
Indices and Laws of Logarithms
A logarithm is just an index.
To solve an equation where the index is unknown, we
can use logarithms.
e.g. Solve the equation 10 x  4 giving the answer
correct to 3 significant figures.
x is the logarithm of 4 with a base of 10
We write
10 x  4  x  log10 4
 0  602 ( 3 s.f. )
( from the calculator )
In general if
10 x  b then x  log10 b
index  log
Indices and Laws of Logarithms
10 x  230
Generalizing this,
10 x  b


x  log 10 230
x  log 10 b
This relationship is also true changing from the log
form to the index form,
so,
10 x  b
 x  log 10 b
In the exercise we used logs with a base of 10 but
the definition holds for any base, so
ax  b

Base
x  log a b
Indices and Laws of Logarithms
SUMMARY
 The Definition of a Logarithm
ax  b

x  log a b
 The “Power to the Front” law of logs:
loga x k  k loga x
x
 Solving the equation
na
b
•
Divide by n
•
“Take” logs
•
Use the power to the front law
•
Rearrange to find x.
Indices and Laws of Logarithms
Solving
ax b
e.g.1 Solve
Solution:
2x  5
2x  5
We “take” logs
We don’t actually take the logs anywhere: we put
them in, but the process is always called taking logs!
 log10 2 x  log10 5
Using the “power to the front” law, we can simplify
the l.h.s.
log10 5
 x log10 2  log10 5  x 
log10 2
 2  32 ( 3 s.f. )
We used logs with base 10 because the values are
on the calculator. However, any base could be
used. You could check the result using the “ln”
button ( which uses a base you will meet in A2 ).
Indices and Laws of Logarithms
e.g.2 Solve the equation
1000  100 ( 3) x
Solution: We must change the equation into the
form b  a x before we take logs.
Divide by 100:
1000  100 ( 3) x  10  3 x
Take logs:
log10  log 3 x
Using the “power to the front” law:

log 10  x log 3
log10
x
log 3
x  2  10 ( 3 s.f. )