MATHEMATICS
SS2 LESSON : WEEK 1.
TOPIC: REVISION ON APPROXIMATION, ROUNDING OFF OF figures, DECIMAL
PLACES, SIGNIFICANT figures and standard form.
1.
REVISION ON APPROXIMATION.
(a).What
is
Approximation?-This is art of getting a
figure reduced in such a way that only the quantity of the digit in the
figure is reduced but the value of the
shortened figure still remain almost the same as the original figure e.g.
135.6 (4
digit number) is shortened to 136 (3 digit)
(b)
How to Approximate?
To approximate any
figure, the last digit in the figure is adjusted to become 1 or 0 depending on
the value of the last digit and also on the number of the digit to be left
after the approximation processes is completed.
If
the last digit is less than 5 i.e. less or equal to 4, it will be reduced to 0
(zero) but if it 5 and above (5-9) it will become 1 e.g.
Approximate the following to the value stated
in front of each figure;
(i)
18.57…………….3 digit: Answer = 18.6 (7 is more than 5 so
becomes 1 and is added to the preceding digit which is 5 to becomes 6).
(ii) 200725…………..4 digit.: Answer = 2007 (5 is reduced to 1 and is
added to 2 to become 3 but we are to reduced the digit to 4 then the 3 is
reduced to 0 since it is not up to 5 hence the answer is 2007).
(c) Different system of approximation
available.
(i)
Significant figures
(ii)
Decimal places
(iii)
Rounding off.
SIGNIFICANT
FIGURES
This
is approximation to a particular number of the given digit. We can have 2
significant figures, 3 significant figures etc e.g.
Take
the following to the requested number of significant figures
(i)
2356………..3
significant figures : answer = 236 (6 is more than 5 so it’s reduced to 1
and added to the 5 that is before it to becomes 6)
(ii)
17895.357…………3 significant figures: answer
= 17900.00 (7 is more than 5 and so is reduced to 1 and added to 5 to
become 6. 6 is also reduced to 1 and added to 3 to become 4. 4 is less than 5
and is reduced to 0. The 0 is added to 5 and this gives 5and this is reduced to
1 and added to 9 to become 10. 10 is also reduced to 1 and added to 8 to become
9 )
(iii)
0.0130……….2 significant figures: answer = 0.013 (0 is not up to 5 and so is removed.
If the negative number is to be taken to significant figures, we start counting
from the real number after the decimal point not minding the number of zeros
before the real number .In this place, we start counting the two significant
figures to be left in the expected answer from 1 and not from 0 that is after
the decimal point. We now start the
reduction from the last digit of the given number as we have done in the case
of taking positive numbers to specified significant figures)
(iv)
0.00809……..2 significant figures: answer = 0.0081 (following the step in iii
above)
(v)
0.001118…….3 significant figures: answer = 0.00112 (following the step in iii
above)
Work
to do
(a) Take the following to 3 significant
figures
(i)
728.8
(ii)
0.0586
(iii)
80.96
(iv)
0.662
(v)
46.896
(b)
Take the following to 2 significant figures
(i)
0.0206
(ii) 2.347
(iii) 0.407
(iv) 56.999 (v) 0.99876
ROUNDING OFF
This means taking the
number to the nearest whole unit. It can be to the nearest ten, hundred, or
even thousand as the examiner wishes. Round
off the following to the corresponding term before it e.g.
(i)
215………..nearest ten: answer = 220 (5 is reduced to 1 and
added to the 1 before it to become 2)
(ii)
378.289………..nearest ten: answer = 380 (9 is more than 5, reduced to
1 added to8 to become 9 again which is reducd to 1 again added to 2 to become 3
which is not up to 5 and so becomes 0 and added to 8 to become 8 which is more
than 5 and reduced to 1 which is added to 7.Hence we have 380 to the nearest
ten)
(iii)
280.25…………..nearest hundred: answer = 300 (5 is reduced to 1 and added
to 2 to become 3. Since we are rounding off to the nearest hundred, 8 is
reduced to 1 and added to 2 to become 3. Hence we have 300 to the nearest
hundred)
(iv)
109.72…………nearest hundred:
answer = 100 (2 is reduced to 0 and added to 7. 7 is reduced to 1 and
added to 9 to become 10. 10 is also reduced to 1 and added to 0 to become 1 and
1 is reduced to 0 . Hence we have 100 to the nearest hundred )
(v)
399.62…………nearest ten: answer = 400 (2 is reduced to 0 added to
6 to become 6 which is reduced to 1 and added to 9 to become 10. 10 is reduced
to 1 and added to 9 to become 10. We are supposed to stop at the second 9 but
since it has become 10 automatically, it is reduced to 1 and added to 3 to
become 4.Hence we have 400 instead of 390)
(vi)
39962……….nearest hundred: answer= 400 (2 is reduced to 0 added to
6 to become 6 which is reduced to 1 and added to 9 to become 10. 10 is reduced
to 1 and added to 9 to become 10. We are supposed to stop at the second 9 but
since it has become 10 automatically, it is reduced to 1 and added to 3 to
become 4.Hence we have 400)
Work
to do
Round off the following into the
specified value stated in their front.
(i)
110
……………………...nearest hundred
(ii)
278……………………… nearest hundred
(iii)
884,45……………………nearest ten
(iv)
76321.087…......................nearest
ten
(v)
1989………………………nearest thousand
(vi)
4090………………………nearest thousand
(vii)
701.17……………………..nearest ten
(viii)
6999.98 ……………………nearest ten
(ix)
999.999……………………nearest hundred
(x)
5899.9978…………………nearest thousand
DECIMAL PLACES
This is when the figure given is to be
rounded off to some exact numbers in which the number of digits after decimal
is taken into consideration. When there is 1 digit after decimal, it is 1
decimal place. When it is 2 digits after decimal it is 2 decimal places etc.
e.g.
(1)
Round off the following to 1 decimal
place
(a)
325.63 : answer = 325.6 (3 is not up to 5 so
it is reduced to 0
and added to 6 to become 6. Since we are to have 1 digit after decimal,
hence the answer is 325.6)
(b)
28.5631: answer = 28.6 (1 is reduced to 0 and
added to 3 to become 3 which is also
reduced to 0 and added to 6 to become 6.
6 is also reduced to 1 and added to 5 to become 6 which gives us 1 digit after
decimal. Hence the answer is 28.6)
(c)
0.781 : answer = 0.9 (1 is reduced to 0 and
added to 8 to become 8 which is reduced to 1 and added to 7 to become 8 and
since we are to have 1 digit after decimal, the answer is 0.8)
(2)
Round off the following to 2 decimal
places (a) 3.4789 : answer = 3.48 (9 is reduced to 1
and added to 8 to become 9 which is added to 7 to become 8. this gives us 2
digits after decimal, the answer is 3.48)
(b)
362.440
: answer = 362.44 (0is added
to 4 to become 4 which gives us two digits after decimal, hence the answer is
362.44)
(c)
0.04329 answer = 0.04 (9 is reduced to 1 and
added to 2 to become 3 which is reduced to 0 and added to 3 to become 3. 3 is
reduced to 0 and added to 4 to become 4 which gives us two digits after
decimal, hence the answer is 0.04)
(d)
0.0951 : answer = 0.10 (1 is reduced to 0 and
added to 5 to become 5 which is reduced to 1 and added to 9 to become 10 and
since we are to have two digits after decimal, we leave the answer to remain at
0.10)
Work
to do
Take the figures in
the table below to the specified places of decimal
|
S/N
|
FIGURES
|
1
PLACES OF DECIMAL
|
2
PLACES OF DECIMAL
|
|
I
|
0.0795
|
|
|
|
2
|
1.4537
|
|
|
|
3
|
12.000975
|
|
|
|
4
|
432.5310
|
|
|
|
5
|
0.1523
|
|
|
|
6
|
0.989
|
|
|
|
7
|
399.99549
|
|
|
|
8
|
79.998
|
|
|
|
9
|
0.009998
|
|
|
|
10
|
602.985
|
|
|
STANDARD
FORM
This is the art of expressing numbers in
the form: A × 10n
A
= Digit from 1-9
n = the index or power to which
10 is raised and this depends on the numbers of times the decimal point in the
given figure moves either to the right or to the left. If the decimal point
moves to the left ‘n’ will be positive
but if the point moves to the right ‘n’
will be negative.
Stating
if a number is in standard form or not.
(i)
5.1×10 (ii)
70.0 × 102 (iii)
9.00 (iv) 1.20 ×
103 (v) 9.7 ×
10-1
(vi) 0.105 ×
10-2 (vii) 20.1 ×
103 .
SOLUTION
(i)
5.1 × 10 is not in the standard form
because the value of the power ‘n’ of the 10 is not stated though it is
known that ordinary 10 = 101 or ordinary 100 =1001 but in
the standard form it has to be stated i.e. the power to which 10 is raised must
be visibly put.
Correct
answer = 5.1 × 101
(ii)
70.0 × 102 is not in standard form because the figure
before the decimal point must be in unit
so it should not be 70 but
7.0.and to make 70 becomes 7.0 the decimal point have to move 1 time to the
left and so we have 7.0 × 101 × 102 and if we apply the
law of indices ( to be learnt fully later; am × an = am + n) then we
have 7.0
× 101+2 = 7.0 × 103
Correct
answer= 7.0 × 103
(iii)
9.00 is not n standard form because it does not gives us the value
of 10n though 9.00= 9 and ordinary 9 = 91 or 9.0 × 101
also ordinary 64 =641 or ordinary 28= 281 but in the
standard form the 10 and its power must be stated
Correct
answer = 9.00 × 101
(iv)
1.20 × 103 is in standard form since we have just one
digit before decimal point and 10 is also raised to the appropriate value of
n=3.
(v)
9.7 × 10-1 is in standard form since
there is only one digit before decimal point and10 is also raised to the
appropriate value of n = -1 showing that this is a negative number.
(vi)
0.105 × 10-2 is not in
standard form because 0 is not a real number and the digit that should precede
the decimal number in the standard form should start from 1-9.due to this 0.105
will become 1.05 × 10-1 (since the decimal point has moved to the
right just one time). So, the actual number is now 1.05 × 10-1 × 10-2
and in applying the law of indices (to be leant later; am × an
= am + n) 1.05 × 10-1 × 10-2 =1.05 × 10-1
+ -2
Correct
answer = 1.05 × 10-3.
(vii)
20.1×
103 is not in standard form ( for the reason check no (ii)
above)
Correct
answer = 2.01 × 104
work to do
State
if the following are in standard form or not
(i)
70.0
(ii) 6.5 × 104 (iii)
256 × 102 (1v) 0.00875 × 10-1 (v) 20.4 × 10-3
Expressing number in
standard form
Express
the following in standard form
(i)
510.0 = 5.1 × 102 (the decimal point moves 2 times to the left
before it can made the figure to have one digit before the point so, the 10 is
raised to power of 2 or in another way
just count the number of the digit before the decimal and subtract 1
from it. Here there are 3 digits before the decimal point, so, 3 - 1 = 2 the
decimal is completely removed from the place where it is and put after the
first digit. The 2 got is now taken to be the power of 10 )
(ii)
547.32 = 5.47 × 102 (as in (i) above)
(iii)
47.52 = 4.752 × 10 1( here
the point moves just once to the left before it can make the figure to have one
digit before the point so the 10 is
raised to power of 1 or in another
way the number of digit before the
decimal is 2 then 2-1 =1; the points is put after the first digit and the 1 got
is taken to be the power of 10 )
Expressing negative numbers in standard form
(i)
0.0072 = 7.2 ×10-3 (the point
is moved to the right 3 times before we
can have 1 digit before the point and since it moves to the right the power on
10 will be negative i.e -3 or in another way just count the number of 0 (zero)
not minding the point and this is 3 put the point before the first digit and
raised the 10 to power of -3)
(ii)
0.123 = 1.23 × 10-1 ( apply
the method above)
(iii)
0.0005 = 5.0 × 10 -4 (as in above)
work
to do
take the following into
standard form
(i)
0.00876
(ii) 0.765 (iii) 123.87 (iv) 2.98
(v) 900.0097 (vi) 87.098 × 102 (vii) 0.00985 × 101 (viii)
00000.00675 × 10-1
(ix) 12.86 × 10-3 (x) 9.08×
102
ARITHMETIC OPERATIONS IN STANDARD FORM
Addition in standard
form
Add up the following in standard form
(a) 34.785 and 1111.234
34.785
+1111.234
1146.019
So, 1146.019 = 1.146019 × 103 (add
the numbers together and thereafter take the answer to the standard form)
(b) 2.03214 × 102 + 1.730×101
2.03214
× 102= 203.214 and1.730 × 101 = 1.730 therefore we are to
add 203.214 and 1.730
203.214
+ 1.730
204.944
So, 204.944 = 2.04944 × 102 ( we cannot add the two number in standard
form directly like that so we convert to natural number before we now add
together and thereafter treat as in (a) above)
Work to do
Add
up the following and leave your answer in standard form
(i)
55.29 and 0.021
(ii)
0.000781 and 0.0002561
(iii)
678.087 and 4321.1
Subtraction in standard form
(a) Subtract 3.728 × 101 from
4.083 × 101
3.728 × 101 = 37.2
4.083 × 101 = 40.83
40.83
- 37.28
3.55
: 3.55 = 3.55× 100 (we
changed the number into natural
number then subtracted the number from
each other and the resulting answer is then taken into standard form)
(b) 483.25 – 21.07
483.25 - 21.07 462.18 : 462.18= 4.6218 × 102 (we
subtracted the number from each other
directly and the resulting answer is taken into standard form)
Work to do
Subtract the
following leaving your answer in standard form
(i)
Subtract 44.32 from 301.22
(ii)
0.6217 – 0.00502
(iii)
0.000784 – 0.00005643
Multiplication in standard form
Evaluate
the following in standard form:
(a) 250×40
250 = 2.50×102
40 = 4.0× 101
250× 40 = 2.5 ×102× 4.0 ×101
= 2.5×
4.0× 102 ×101
= 10.0× 102× 101
= 10.0× 102+1
= 10.0× 103
= 1.0 ×101 ×103
= 1.0× 101+3
= 1.0× 104 (each of the number is taken into
standard form then the
like term is collected multiply together and if the number is more than one
before the decimal point it is re–taken into standard form as accurately as
possible note that the power of 10 is added together considering the rule of
indices that am + an =
am+n)
OR
250× 40 = 10,000
= 1.0 ×104 (the number is
multiply together directly and
thereafter taken into standard form)
(b) 72.28× 100.02× 2.13
72.28 = 7.228× 101
100.02 = 1.0002× 102
2.13 = 2.13 ×100
72.28× 100.02 ×2.13 = 7.228 ×101×
1.0002 ×102× 2.13 ×100
=
7.228× 1.0002 ×2.13× 101× 102 ×100
= 15.39871913× 101+2+0
= 15.39871913× 103
= 1.539871913× 101 ×103
= 1.539871913 ×101+3
= 1.539871913 ×104
(c) 0.0667× 0.0081
0.0667 = 6.667× 10-2
0.0081 = 8.1 ×10-3
0.0667× 0.0081 = 6.67×
10-2× 8.1× 10-3
=
6.67× 8.1 ×10-2× 10-3
= 54.027 ×10 -2+ -3
= 54.027× 10-5
= 5.4027 ×101× 10-5
= 5.4027× 101+-5
= 5.4027× 101-5
= 5.0427× 10-4
Work
to do
Evaluate the following leaving your
answer in standard form.
(i)
1200 ×
0.0012
(ii)
23.74
× 541.26
(iii)
0.081 ×
0.0542
(iv)
75.48
× 201.357
(v)
0.000345 × 0.0289
Division
in standard form
(a) 471.8
÷ 17.31
=
(4.718 102) ÷ (1.73 101)
=
4.718 ÷1.73 ×102÷ 101
=
2.7256 ×102-1
=
2.7256× 101 (take each of the number into standard form and collect the
like term and divide them by each other
and for the power of 10 mind the rule of
indices of am ÷ an
= am-n)
(b) 14.40
÷ 0.012
=
1.440× 101÷ 1.2× 10-2
=
1.440÷ 1.2× 101÷ 10-2
=1.2×
101-2
=
1.2 ×101+2
=
1.2× 103 (as in (b) above)
(c) 0.0345÷
0.0024
=3.45×
10-2÷ 2.4× 10-3
=3.45
÷2.4 ×10-2÷ 10-3
=1.4375×
10-2+3
=1.4375×
101 (as in (a) and (b) above)
(d) 0.016÷
64
=1.16× 10-2÷ 6.4× 101 =1.16 ÷6.4× 10-2 ÷101
=1.8125× 10-2-1 =1.8125× 10-3 (apply the above method)
(e) 0.00667÷ 0.081
= 6.67 × 10-4 ÷ 8.1× 10-2 = 6.67 ÷ 8.1× 10-4
÷ 10-2
=5.4002 ×10-4- -2 =5.4002×102 (as in above)
Work
to do
Evaluate the following and leave the
answer in standard form
(i)
0.402
÷ 78.54
(ii)
20.667 ÷
0.081
(iii)
0.000674 ÷ 0.876
(iv)
0.00543 ÷ 0.000087655
(v)
78.421 ÷
9.604
(vi)
304.78
÷ 1678.098
(NOTE FOR MORE
EXERCISES CHECK NEW CONCEPT MATHEMATICS
SS2 CHAPTER2 PAGES 16-18)
SS2
LESSON:WEEK TWO
TOPIC:
REVISION ON INDICES AND LOGARITHM.
1.
REVISION ON
INDICES
Indices is the art of using index
/power to do mathematic operation such as addition, subtraction, multiplication
division e.t.c. 23 ,72, 5a2, are some examples
of indices.
to be able to operate in indices
we need to briefly study the rule of indices .
RULES OF INDICES
Rule
1:
.am × an = am
+ n
E.g..
(i) 52
×54
= 52+4
=
56
(ii)
64 ×128
(express the number in indices you do this by finding the prime factor
of the number)
= 26× 27
=26+7
=
213
Rule
2:
. am
an
=am
÷an
= am-n
E.g.
(i) 33
32
= 33÷ 32
= 33-2
= 31
=3
(iii)
64 ÷ 24
=
26÷ 24
= 22
Rule
3:
(Zero index):
a0 = 1
E.g.
(i) 1000 = 1
(ii)
20 = 1
Rule
4.
(Fractional index 1):
a 1/n
= n√a
E.g.
(i) Simplify 641/3
= 3√64
= 4
Rule
5:
(Fractional index 2):
am/n
= (n√a)m
E.g
. (i): (1024)3/10
= (10√1024)3
= 23
=8
(ii): (16,807)2/5
=
(5√16,807)2
= 72
=49
Rule
6:
(Power index 1)
(am)n
= am × n
= amn
E.g.
(i) (23)2
= 23× 2
= 26
(ii)Simplify (43)2
= 43×2
=46
Rule
7: (Power
Index 2)
(i)
(ab)m = am× bm
e.g
(i)
simplify 62
=
(2× 3)2
=
22× 32
=
4× 9
=
36
(ii)
Simplify (18)2
=
(2×3×3)2
=
22×32×32
=
499
=
324
Rule 8:(Power Index 3)
a m
b
=
am
bm
e.g.
(4/9)1/2
= 41/2
91/2
= (22)1/2
(32)1/2
= 2
3
Rule 9: (Negative index)
a-n =
1
an
E.g. (i) Simplify 2-3
= 1
23
= 1
8
(ii) Simplify 4-2
= 1
42
= 1
16
examples on general application
(i) 271/3×64-1/2
32-2/5
Solution
27=33 and 64=26
so also 32=25
Therefore,
(33)1/3 × (26)-1/2 (since amn = am×n)
(25)-2/5
=31 ×2-3
2-2
=
31
×2-3 ÷2-2 (since
am ÷ an = a m ̶
n)
= 31×2-3+ -2
= 31×2-3-2
= 31×2-5
(ii) 81/3 × 251/2 × 80
=(23)1/3 × (52)1/2 ×1 (since 80 =1)
= 21×51 ×1
= 2× 5× 1
=10
Work to do
(i)
Simplify
2563/4
(ii)
81 ×729
27
(iii)
251/3 ×125-1/2
5-1/6 ×5-2/3
(iv)
5n-5 n-2
53×
5n -125× 5n-2
(v)
√a×√ a3
√a-3
(vi)
35× 3× 27
81× 243
(vii)
125-1/3
× 251/2
49-1/2× 7
(viii)
81/2 ×
16-1/2
32-1/5
EXPONENTIAL EQUATION
Exponential
equation is the equation in which the unknown or one of the unknowns is in
index form. Examples are:
(i)
2x
= 8
(ii)
2x
= 23
x=3
(iii)
32x
= 81
32x = 34
2x
= 4
x = 2
(iv)
4x+3
= 128
22(x+3) = 27
2x+6 = 7
2x = 7-6
2x=1
x = ½
(v)
26(5x-1)
= 52x
26(5x÷5) = (5x)2 + 1
26(5x/5) = (5x)2
+1
Let 5x = p
26p/5 = p2
+1
26p = 5p2 +5
5p2-26p
+5 =0
+25p2 -25p × - p
-26p -25p + - p
5p2 – 25p – p +5 = 0
(5p2 - 25p) - (p - 5) = 0
5p (p-5) -1(p-5) = 0
(5p-1) (p -5) = 0
5p-1 = 0 ˃ p = 1/5
p- 5 = 0 i.e. p = 5
but p = 5x
so, 5x = 1/5
then 5x= 5-1
x = -1
also, p = 5
then 5x = 5
5x = 51
.
So, x = 1
.
. x
= 1 and -1
32x + 1 −18
(3x) −81 = 0
= (3)2x ×
3−18(3x) – 81=0
= (3x)2
× 3 -18(3x) -81 =0
Let 3x =P
= (p2 × 3) −18p
−81 =0
=3p2 −18p −81
= 0
-243p2 -27p × 9p
-18p
-27p + 9p
=3p2 −27p +9p−81 =0
(3p2−27p) + (p −81) =0
3p (p−9) +9(p−9)=0
3p (p−9) +9(p−9) =0
(3p+9) (p − 9) =0
3p+9 =0
=3p= −9
=p= −9/3
p−9 = 0
p = 9
but p = 3x
so, 3x=−3
3x = −31
No value for x here
If p = 9
Then 3x
=9
3x =32
So ,x = 2
. x=2
.
work to do
Find the value of x in question I − iv
(i)
22x
– 6(2x) +8 =0
(ii)
(6√243)2x =1/27 x + 3
(iii)
7
x + 1 + 7x =2744
(iv)
4x+1−
9(2x) = −2
(v)
(a).
Simplify 10 3n/2 ×15 n/2 × 6 n/6
45 n/3 × 20 2n/3
(b). if n = 2 evaluate (a) above
(NOTE
MORE EXERCISES CAN BE FOUND IN CAMBRIDGE IGCSE MATHS CHAPTER 18 PGS 280-283 (THIS IS ONE OF THE
TOPIC IN THE CAMBRIDGE EXAM RECNTLY INTRODUCED BY THE SCHOOL: IF YOU DO NOT SEE
THE TEXT THE SCHOOL WILLPROVIDE MORE INFORMATION LATER)
TOPIC: LOGARITHM
RULES
OF LOGARITHM
RULE 1: Logam
+ logan = loga (m×n)
e.g. evaluate log1050 if log102
= 0.3010, log103 = 0.4771, log105 = 0.6990 and log107 = 0.8451
(i)
log1050 = log10 (2×5×5)
=
log102 + log 105 + log105
=
0.3010 + 0.6990 + 0.6990
= 1.6990
(ii) Simplify log354
= log3 (27×2)
= log327 + log32 =
log333 + log32
=
3log33 + log32
= (3 × 1)+ log32
= 3+ log32
RULE
2: loga m – loga n = log a (m/n)
e.g. if log103 = 0.4771, log102
= 0.3010, log105 = 0.6990, log107 = 0.8451
evaluate log10(4.5)
= log10 (4.5)
=log10(45/10)
= log10(9/2)
= log109 – log102
= log10(3×3) – log102
= log103 + log103 log102
= 0.4771 + 0.4771 - 0.3010
= 0.9542 – 0.3010
= 0.6532
RULE 3: logaa=1
e.g (i)
log1010=1 (ii) log100100=1
RULE 4: loga ms = s logam
e.g. Simplify log327
= log333
= 3log33
= 3×1
= 3
If
log102 = 0.3010, log 103 = 0.4771, log105 =
0.6990, log107 = 0.8451, evaluate
(i)
Log108
= log10(2×2×2)
= log1023
= 3log102
= 3×0.3010
= 0.9030
(ii)
Log
1036
=
log1062
= 2log106
= 2log10(2×3)
= 2(log102 + log103)
=
2(0.3010 + 0.4771)
= 2(0.7782)
= 1.5562
RULE 5: loga√m = logam1/n = 1/n logam
e.g.
Simplify log3√9
= log391/2
= 1/2log39
= ½(log332)
= ½(2log33)
= ½(2×1)
= ½×2×1
= 1
(ii)
log√8
log8
= log81/2
= log8
= ½log 8
Log8
=½×1
=½
RULE 6: logab × logba = 1
e.g. evaluate log32×
log29
= log32× log29
= log32× log232
= log32×2 ×log23
= 2×log23×log32 =2 ×
1(since logab × logba=1) So,2 ×1=2
Change of base
y = logba
take
log to base 10 and convert base b to 10 to base 10
y
= log10a
log10b
e.g. evaluate log927
= log327
log39
= log333
log332
= 3log33
2log33
=
3
2
(ii)
evaluate log48 + log816 – log24
=
log28 + log216 – log24
log24 log28 1
=
log223 + log224 – log222
log222 log223 1
= 3log22 +4log22
– 2log22
2log22 3log22 1
= 3 + 4 – 2
2
3 1
= 9 + 8 – 12
6
=
5
6
LOGARITHM EQUATION
Logab = c ≡ ac
= b (this is to show the interconvertibility of logarithm and indices and it is
very helpful when trying to solve logarithm equation in most cases)
Log2100 ≡ 102 = 100
(i) if log2x = 3 find x
23 = x
(convert to indices)
x = 8
(ii)
if 2logm3 = 2
logm32
= 2
m2 = 32 (converting to indices)
m = 3
(iii)
6 log(n+4) = 2log8
Log (n+4)6 = log82 (
taking 8 to index form)
log (n+4)6
= log(23)2
log (n+4)6
= log26 (since the two sides are in the same log base then thebase
cut each other)
n+4 = 2
n = 2-4
n = -2
(iv)
log10 (x2 + 4) = 2 + log10x – log1020
log10(x2
+ 4) = log10100 + log10x – log1020
log10(x2
+ 4) = log10(100x/20)
x2 + 4 = 5x
x2
+ 4 – 5x = 0
x2
– 5x + 4 = 0
x2 + -4x + -x + 4 = 0
x2
– 4x – x + 4 = 0
x2 -4x – x + 4 = 0
( x2 – 4x) – (x – 4) = 0
x (x –
4) – 1(x – 4) = 0
(x – 1)
(x – 4) = 0
x – 1 =
0
x = 1
x – 4 = 0 x = 4
x = 1 or 4
Work
to do
(i)
logx27+ logy4=5……..(i) find x and y
Logx27 – logy4=1……(ii)
(ii)
log3X +log9X – log27X =7/3 find x
(iii)
23x +1= 5x+1 find x
(iv) solve for x and y in this equation
2 x + y=6
3 x− y=4
(v)
log45.83 =x
(NOTE
MORE EXERCISES CAN BE FOUND IN CAMBRIDGE IGCSE MATHS (THIS IS ONE OF THE TOPIC IN THE CAMBRIDGE
EXAM RECNTLY INTRODUCED BY THE SCHOOL: IF YOU DO NOT SEE THE TEXT THE SCHOOL
WILLPROVIDE MORE INFORMATION LATER)
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