LOGARITHMS

LOGARITHMS

The logarithm of a positive real number  'N' to the base 'b' is define as the power to which 'b' must be raised to obtain the number 'N'

Thus if N = aʸ, then y = logₐN ( y>1).
Note that logarithm is an inverse of indices.

Example1:

Simplify (a) log₁₂₅0.2
(b) log₀.₂₅16  
(c) log₈0.25

    Solution
(a)  let log₁₂₅0.2 = 
         
        0.2 = 125ᵐ ⇒ from definition
    1/5 = 125ᵐ
      5⁻¹ = 5³ᵐ
       -1 = 3m
        m = -1/3

(b)   let log₀.₂₅16 = k
        
       0.25ᵏ =16 ⇒         from definition
  4⁻ᵏ = 16
   4⁻ᵏ = 4²
    k = -2

(c)  let log₈0.25 =m
        8ᵐ = 0.25
        8ᵐ = 4⁻¹
         2³ᵐ = 2⁻²
         3m = -2⇒ divide both side by the coefficient of m
           m =- 2/3



         
 THEORY OF LOGARITHMS



1.      logₐm + logₐn 
         
          = logₐmn

2.      logₐm - logₐn =
        
         logₐ(m/n)

3.      logₐ(m)ⁿ = 
    
         nlogₐ(m)

4.      logₐa =1

5.      logₐ1 = 0

6.      logₙm =(logₐm)
          /(logₐn) ⇒     change of base

7.       logₓⁿ=1/logₙˣ ⇒ change of base

8.       logₙm = logₐm 
          x logₙa

9.      ₐlogₐk = k


Example 1: 

If logₓy = 100 and log₂x = 10. Find y


Solution 

x = 2¹⁰ = 1024 ⇒from definition

log₁₀₂₄y = 100

log₁₀₂₄y = log₁₀₂₄1024 x 100
y =1024 x 100
y =102400



Example 2:

Given that log₂(log₁₆x) = -2.  Find x

Solution
2⁻² = log₁₆x

1/4 =log₁₆x

16^¼ = x

⁴√(16) =x

2 = x



Example 3:

If logy + 3logx =2. Express y in term of x


Solution

logy × x³ = 2

10² = y x x³

100 = yx³

y = 100/x³

(N.B  A log without a written base is in base 10) 

Example 4:

If log√(x) + logy = 2logz express x in terms of y and z


Solution

log√(x)  x  y  =2logz

log√(x)y = logz² ⇒ take log

√(x)y =z²

√(x) =z²/y

add square to both side
x =(z²/y)²



Example 5:

₃-2log₃5 = 5⁻² 

=1/25

Example 6:

₂1/4log₂16 x ₃2log₃5

= 16^¼  x  5² 
=2 x 25 = 50


Example 6:

10ⁿ = 0.63 and log₁₀63 = 1.8, find n

log10ⁿ = log0.63 =

log63/100

log10ⁿ =log63 - log100

log10ⁿ = -1.8 + log10²

log10ⁿ = -1.8 + 2
nlog10 = - 1.8 + 2
n x1 =0.2
n = 0 2 = 1/5



Example 7:

Evaluate log₄0.3 - log₄0.48 + log₄0.05

log₄(0.3 ÷ 0.48 x 0.05)  

log₄(3/10 ÷ 48/100 × 5/100)

log₄(3/10 x 100/48 x 5/100)
log₄32 =log₄16 x2 =
log₄16 + log₄2 =
log₄4² + 1/2log₂2=
2 + 1/2 =5/2
 =2.5


Example 8:
-₉2log₃2 x ₂3log₂6
=₃-4log₃2 x ₂3log₂6
=2⁻⁴ x 6³
= 1/16 x 216
=13.5

Example

Simplify
log₁₀6 + log₁₀45 - log₁₀27 

log₁₀6 x 45 ÷ 27 =

log₁₀270/27
log₁₀10 = 1.