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LOGARITHMS
LOGARITHMS

STANDARD NOTATIONS
Standard notation form is written in form of A x 10n whereby 1≤ A< 10 and n is any integers

Example

Write the following in standard form
(i) 2380

Solution:
2380 = 2.38 x 103


(ii) 97

Solution:
97 = 9.7 x 101


(iii) 100000

Solution:
100000 = 1 x 105


(iv) 8
Solution:
8 = 8 x 100

Example

Write the following in standard form

(i) 0.00056
= 5.6 x 10-4

(ii) 0.001
= 1 x 10-3

(iii) 0.34
= 3.4 x 10 -1

(iv) 2. 0001
= 2. 0001 x 100
EXERCISE 1:

i). Write the following in standard form

17000
= 1.7 x 104


ii) 0.00998
= 9.98 x 10-3
iii). Write in standard form

0.000625
= 6.25 x 10-4

8/300 correct to four significant figure

8/300 = 0.02666

Now 2.667 x 10-2

2.667 x 10-2
iv) If a = br and a = 8.4 x 104 , b = 7.0 x 102 Find r.

solution:

a = 84 000

b = 700

Now

br = a

(700) (r) = 84000

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r=120
r = 1.2 x 102
DEFINITION OF LOGARITHMS

Consider

3 x 3 x 3 x 3 then

3 x 3 x 3 x 3 = 34 = 81, the number 3 is the base ,and 4 is the exponent.

Now we say;

Logarithm of 81 to base 3 is equal to exponent 4

log381 = 4
In short bn = a

logba= n

Example 1.
Write the following in logarithmic form

i) a5 = 10
loga10 = 5
ii)10-3 = 0.001
10-3= 0.001
log100.001 = -3
iii) 2-1 = ½
log21⁄2 = -1
iv) 3 = 91/2
log39 = 1⁄2
Example 2

Write the following in exponential form
(i) log3729= 6
36 = 729
(ii) log31⁄3 = -1
3-1 = 1/3
(iii) log100.01 = -2
10-2 = 0.01
(iv)1⁄2 = log42
4(1/2) = 2
Example 3

If log100.01= y. Find y

Solution:

log100.01 = y

10y = 0.01

10y =1×10-2

10y=100×10-2

10y=10-2

y= -2


If log10x=-3 find x

Solution:
log10x = -3

10-3 = x

x=0.001


EXERCISE 1
1. Write in standard form
i) 405.06
ii) 0.912
Solution:
i) 405.06 = 4.0506 x 102
ii) 0.912 = 9.12 x 10-1
2. Write in logarithimic form
i)5-1 = 1⁄5
ii) 0.0001 =1 × 10-4

Solution:
i) 5-1 = 1⁄5
log5(1⁄5) = -1

ii) 0.0001 = 10-4
log100.0001 = -4
3. Write in exponential form
i) logax = n
ii)-3 = log100.001
iii) log2(1⁄64) = -6

Solution:
i) logax = n
an = x

ii)-3 =log100.001
10-3 = 0.001

iii) log2(1⁄64) = -6
2-6 = 1⁄64
4. To solve for x
i) log6x= 4
64 = x
x = 1296

ii) x = log36561
3x = 6561
x = 8

iii) logx10= 1
x1 = 10
x = 10

iv) log42 = x
4x = 2
22x = 21
2x= 1
x = 1⁄2


BASE TEN LOGARITHM
– Is an logarithm of a number to base 10. Also known as common logarithm
example i) log105= log5
ii) log1075 = log75
iii) log10p = log p

SPECIAL CASES
(1). logaa = x
ax = a1
x = 1
C:thlbcrtz__i__images__i__a1410.png

Generally logaa = 1

Example
i) log66 = 1
ii) log10 = 1

(2) loga(an) = x
ax = an
x = n
C:thlbcrtz__i__images__i__ee51.png

Example i) log4(45) = 5
ii) log10-3 = -3

Example 1

If log55 = log2m Find m

Solution:

log55 = log2m

But log55 = 1

1 = log2m

21 = m1

m = 2


Example 2

Given log525 + log4x = 6, Find x

Solution:

log525 + log4x = 6
log5(52)+ log4x = 6

2log55+log4x = 6

2 +log4x = 6

log4x = 4
x= 44

x = 256
EXERCISE 2.

Evaluate

i) log24096

ii) log0.0001


solution

i) log24096

let x = log24096
2x = 4096

2x = 212

x = 12

log24096=12


ii) log0.0001

Solution:

Let x = log0.0001
10x = 1/10000
10x = 1/(104)
10x = 10-4
x = -4

log0001=1


2) If logk81 – log232= -1

Solution:
logk81 – 5log22 = -1
logk81 = -1 + 5
logk81= 4
k4 = 81
k4 = 34
k = 3



3. Given log6y = log7343. Find y

Solution:

log6y = 3log77

log6y = 3

63 = y

216 = y

y = 216
4) Solve for m

i) log81 = m
8m = 1 since aº=1 then

8m=80

m=0


ii) log5m + log327 = 8
log5m + log333= 8
log5m +3 = 8
log5m = 5
m= 55

m = 3125
LAWS OF LOGARITHMS

MULTIPLICATION LAW

Suppose, logax = p and logay = q then
logax = p….(i)
logay = q….(ii)
Write equation (i) and (ii) into exponential form.
ap = x………(iii)
aq = y……..(iv)
Multiply equation (iii) and (iv)

xy = ap x aq
xy = a(p + q) …….(v)

In equation (v) apply loga both sides
loga (xy) = logaa(p + q)
logaxy = (p + q) logaa
logaxy = p + q
But p = logax
q = logay
C:thlbcrtz__i__images__i__ee61.png

Example

i) log6(8 ×12) = log88 + log612
ii) log49 +log43 = log4(9 ×3)

Example 1

i) Find x , If log3x = log315 + log312

Solution:

log3x = log315 + log312
log3x = log3( 15 ×12)
log3x = log3180
x = 180

Example 2

Given log520 = log54 + log5x .Find x

Solution:

log520 = log54 + log5x

log520 = log5(4 × x)
log520 = log54x
20 = 4x

X = 5

Example 3

If log80.01= log8(m ×2). Find m

solution

log80.01 = log8( 2m)

0.01 = 2m

m = 0.01/2

m = 0.005


QUOTIENT LAW

Suppose, logax= p and
logay = q then
logax = p……..(i)
logay = q……..(ii)

Write equation (i) and (ii) into exponent form
ap = x……(iii)
aq = y…..(iv)

Divide equation (iii) and (iv)
x/y = ap/aq
x/y = a(p – q) ….. (v)

In equation (v) apply log a both sides
loga(x/y) = logaa(p-q)
loga(x/y) = (p – q) logaa
But logaa = 1
loga(x/y) = p – q ,where p= logax and q=logay

C:thlbcrtz__i__images__i__ee41.png
i) log6( 8/12) = log68 – log612
ii) log49 – log43 = log4(9/3)

Example

If log220 = log2x – log28.Find x

Solution:

log220 = log2x – log28
log220 = log2(x/8)
Now, 20 = x/8
X = 20 x 8
X = 160
EXERCISE 3
1. Evaluate

i) log63 + log62

Solution:

= log63 + log62
= log6( 2 ×3)= log66
= 1

ii) log 40 + log 5 + log40

Solution:

= log1040 + log105 + log1040
= log10( 40 ×5 ×40)
=log108000

iii) log1025 – log109 + log10360
Solution:

log1025 – log109 + log10360
log10( (25 ×360 )/9)
=log101000
=log10103
=3log1010
=3

2. If log5a⁡x = log5a⁡9 + log5a⁡12. Find x

Solution:

log5a⁡x = log5a⁡9 + log5⁡a12
log5a⁡x = log5a⁡( 9 ×12)
log5a⁡x = log5a⁡108
x = 108
3. If log2a⁡5 = log2ay + log2a⁡0.001.Find Y

Solution:

i) log2a⁡5 = log2a⁡(y×0.001)

5⁡ = ⁡0.001y

y= 5/0.001

Y = 5000


ii)Find y if log6⁡100 = log6⁡5 + log6⁡80 – log6⁡y

Solution:

log6⁡100 = log6 (5 × 80)/ y

100 = ⁡400/y

y = 4
4. If log a = 0.9031, log b = 1.0792 and log c = 0.6990. Find log⁡ ac⁄b

Solution


log ⁡ac⁄b =log10⁡ a + log10⁡c – log10⁡b
=0.9031 + 1.0792 -0.6990

log ⁡ac⁄b = 1.2833


LOGARITHM OF POWER

If loga⁡x = p then
X = ap
Multiply by power in both sides xn = anp

Apply log a both sides
logaxn = logaanp
logaxn = np

But p = logax

logaxn= nlogax
C:thlbcrtz__i__images__i__ee71.png

Example(1)

Evaluate

i) log2 (128)6
ii) log7 (343)8

Solution
i) log2 (128)6 = 6log2 27

= (7 x 6) log22

= 42 x 1

= 42

ii) Log7 (343)8
Solution:
log7 3438 = 8log7 343
= 8log7 73
= (8 x 3) log7 7
= 24


Example (2)

If log5 625y = log3 7292 .Find y.

Solution:
log5 625y = log3 7292
log5 625y = 2log3 729
ylog5 54 = 2log3 36
(y x 4) log5 5 = (2 x 6) log3 3
4y log5 5 = 12 log3 3
4y = 12

y=2/4

y = 3
LOGARITHM OF ROOTS

C:thlbcrtz__i__images__i__ee81.png
Example (1)

C:thlbcrtz__i__images__i__a1510.png

C:thlbcrtz__i__images__i__ee101.png
EXERCISE 4:

1. Evaluate

i) log 60 + log 40 – log 0.3
ii) log3 √(1⁄27)

Solution:
i) Log60 + log40 – log0.3
log10 60 + log10 40 – log10 0.3
log10 (60 x 40/0.3) = log10 (2400/0.3)
= log10 8000
=3.9031

C:thlbcrtz__i__images__i__d15.png
C:thlbcrtz__i__images__i__a1610.png
3. Given log2 x = 1 – log2 3. Find x
Solution:

log2 x = 1 – log2 3
log2 x = log2 2-log2 3
log2 x = log2 (2/3)
x =2/3

4. Simplify

i) 2log5 + log36 – log9
ii) (log⁡8-log⁡4)/(log 4-log2)

Solution:

i) 2log5 + log36 – log9

log52 + log36 – log9

log1025 + log1036 – log109

= log10 (25 x 36)/9

= log10 (900/9)

= log10100

=log10 102

=2 log10 10

=2



ii) (log⁡8-log⁡4)/(log 4-log2)

Solution:

(log⁡8-log⁡4)/(log 4-log2)

= log10 (8/4) ÷log10 (4/2)

= log102 ÷ log102
= 1




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EcoleBooks | MATHEMATICS O LEVEL(FORM TWO) NOTES - LOGARITHMS

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2 Comments

  • EcoleBooks | MATHEMATICS O LEVEL(FORM TWO) NOTES - LOGARITHMS

    Elijah, April 15, 2024 @ 3:45 am Reply

    Thanks for the good work you have done.

  • EcoleBooks | MATHEMATICS O LEVEL(FORM TWO) NOTES - LOGARITHMS

    Elijah, April 15, 2024 @ 3:44 am Reply

    Thanks for the good work

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