MATHEMATICS O LEVEL(FORM TWO) NOTES – LOGARITHMS

LOGARITHMS

STANDARD NOTATIONS

Standard notation form is written as A × 10ⁿ whereby 1 ≤ A < 10 and n is any integer.

Example

Write the following in standard form:

1. 2380

Solution:
2380 = 2.38 × 10³

2. 97

Solution:
97 = 9.7 × 10¹

3. 100000

Solution:
100000 = 1 × 10⁵

4. 8

Solution:
8 = 8 × 10⁰

Example

Write the following in standard form:

1. 0.00056 = 5.6 × 10^-4
2. 0.001 = 1 × 10^-3
3. 0.34 = 3.4 × 10^-1
4. 2.0001 = 2.0001 × 10⁰

EXERCISE 1:

i) Write the following in standard form:

1. 17000 = 1.7 × 10⁴
2. 0.00998 = 9.98 × 10^-3

iii) Write in standard form:

0.000625 = 6.25 × 10^-4

8/300 correct to four significant figures:

8/300 = 0.02666

Now 2.666 × 10^-2

iv) If a = br and a = 8.4 × 10⁴, b = 7.0 × 10², find r.

Solution:
a = 84000
b = 700
br = a
(700)(r) = 84000

r = 120
r = 1.2 × 10²

DEFINITION OF LOGARITHMS

Consider 3 × 3 × 3 × 3 then

3 × 3 × 3 × 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

log_ba = n

Example 1

Write the following in logarithmic form:

1. a⁵ = 10
   log_a10 = 5
2. 10^-3 = 0.001
   log₁₀0.001 = -3
3. 2^-1 = ½
   log₂½ = -1
4. 3 = 9^1/2
   log₃9 = 1/2

Example 2

Write the following in exponential form:

1. log₃729 = 6
   3⁶ = 729
2. log₃1/3 = -1
   3^-1 = 1/3
3. log₁₀0.01 = -2
   10^-2 = 0.01
4. 1/2 = log₄2
   4^1/2 = 2

Example 3

If log₁₀0.01 = y, find y.

Solution:
log₁₀0.01 = y
10^y = 0.01
10^y = 1 × 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 × 10²
ii) 0.912 = 9.12 × 10^-1

2. Write in logarithmic 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) log_ax = n
   ii) -3 = log₁₀0.001
   iii) log₂(1/64) = -6

Solution:
i) aⁿ = x
ii) 10^-3 = 0.001
iii) 2^-6 = 1/64

4. Solve for x:
   i) log₆x = 4
   ii) x = log₃6561
   iii) log_x10 = 1
   iv) log₄2 = x

Solution:
i) 6⁴ = x = 1296
ii) 3^x = 6561, x = 8
iii) x¹ = 10, x = 10
iv) 4^x = 2, 2x = 1, x = 1/2

BASE TEN LOGARITHM

Is a logarithm of a number to base 10. Also known as common logarithm.

Examples:

• log₁₀5 = log 5
• log₁₀75 = log 75
• log₁₀p = log p

SPECIAL CASES

1. log_aa = 1

Because a¹ = a

Generally, log_aa = 1

Example

i) log₆6 = 1
ii) log 10 = 1

2. log_a(aⁿ) = n

Because a^x = aⁿ implies x = n

Example

i) log₄(4⁵) = 5
ii) log 10^-3 = -3

Example 1

If log₅5 = log₂m, find m.

Solution:
log₅5 = log₂m
But log₅5 = 1
So 1 = log₂m
2¹ = m
m = 2

Example 2

Given log₅25 + log₄x = 6, find x.

Solution:
log₅(5²) + log₄x = 6
2 log₅5 + log₄x = 6
2 + log₄x = 6
log₄x = 4
x = 4⁴ = 256

LAWS OF LOGARITHMS

MULTIPLICATION LAW

Suppose, log_ax = p and log_ay = q then

log_ax = p …(i)
log_ay = q …(ii)

Write equations (i) and (ii) into exponential form:

a^p = x …(iii)
a^q = y …(iv)

Multiply equations (iii) and (iv):

xy = a^p × a^q
xy = a^{p + q} …(v)

Apply log_a to both sides of (v):

log_a(xy) = log_a(a^{p + q})
log_a(xy) = (p + q) log_aa
log_a(xy) = p + q

But p = log_ax and q = log_ay

Example

i) log₆(8 × 12) = log₆8 + log₆12

ii) log₄9 + log₄3 = log₄(9 × 3)

Example 1

Find x if log₃x = log₃15 + log₃12.

Solution:
log₃x = log₃(15 × 12)
log₃x = log₃180
∴ x = 180

Example 2

Given log₅20 = log₅4 + log₅x, find x.

Solution:
log₅20 = log₅(4 × x)
20 = 4x
x = 5

EXERCISE 2

Evaluate:

1. log₂4096
2. log 0.0001

Solution:

i) Let x = log₂4096
2^x = 4096
2^x = 2¹²
x = 12
∴ log₂4096 = 12

ii) Let x = log₁₀0.0001
10^x = 1/10000
10^x = 10^-4
x = -4
∴ log 0.0001 = -4

LOGARITHM OF POWER

If log_ax = p then

x = a^p

Multiply by power n on both sides:

xⁿ = a^np

Apply log_a to both sides:

log_a(xⁿ) = log_a(a^np)

log_a(xⁿ) = np log_aa

log_a(xⁿ) = np

But p = log_ax

Example (1)

Evaluate:

1. log₂(128)⁶
2. log₇(343)⁸

Solution:

i) log₂(128)⁶ = 6 log₂128 = 6 × 7 = 42

ii) log₇(343)⁸ = 8 log₇343 = 8 × 3 = 24

LOGARITHM OF ROOTS

Example (1)

EXERCISE 4:

1. Evaluate:

1. log 60 + log 40 – log 0.3
2. log₃√(1/27)

Solution:

i) log 60 + log 40 – log 0.3 = log₁₀(60 × 40 / 0.3) = log₁₀8000 = 3.9031

ii) log₃√(1/27) = (1/2) log₃(1/27) = (1/2)(-3) = -3/2

3. Given log₂x = 1 – log₂3, find x.

Solution:
log₂x = 1 – log₂3
log₂x = log₂2 – log₂3
log₂x = log₂(2/3)
x = 2/3

4. Simplify:

1. 2 log 5 + log 36 – log 9
2. (log 8 – log 4) / (log 4 – log 2)

Solution:

i) 2 log 5 + log 36 – log 9 = log 5² + log 36 – log 9 = log (25 × 36) / 9 = log 100 = 2

ii) (log 8 – log 4) / (log 4 – log 2) = log (8/4) ÷ log (4/2) = log 2 ÷ log 2 = 1

Example

If log₂20 = log₂x – log₂8, find x.

Solution:
log₂20 = log₂(x/8)
20 = x/8
x = 160

EXERCISE 3

1. Evaluate:

1. log₆3 + log₆2
2. log₁₀25 – log₁₀9 + log₁₀360

Solution:

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

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

2. If log_5ax = log_5a9 + log_5a12, find x.

Solution:
log_5ax = log_5a(9 × 12) = log_5a108
x = 108

3. If log_2a5 = log_2ay + log_2a0.001, find y.

Solution:
log_2a5 = log_2a(y × 0.001)
5 = 0.001y
y = 5000