Logarithm – Solving Problems Based On Laws Of Logarithm

CONTENT

• Logarithm of numbers (Index & Logarithmic Form)
• Laws of Logarithm
• Logarithmic Equation
• Change of Base
• Standard forms
• Logarithm of numbers greater than one
• Multiplication and divisions of numbers greater than one using logarithm
• Using logarithm to solve problems with roots and powers (no > 1)
• Logarithm of numbers less than one.
• Multiplication and division of numbers less than one using logarithm
• Roots and powers of numbers less than one using logarithm

Logarithm of numbers (Index & Logarithmic Form)

The logarithm to base a of a number P, is the index x to which a must be raised to be equal to P.

Thus if P = a^x, then x is the logarithm to the base a of P. We write this as x = log a P. The relationship log_aP = x and

a^x =P are equivalent to each other.

a^x =P is called the index form and log_aP = x is called the logarithm form

Conversion from Index to Logarithmic Form

Write each of the following in index form in their logarithmic form

a)         2⁶ = 64                       b)         25^1/2 = 5                      c)  4⁴= 1/256

Solution

a)         2⁶ = 64

            Log₂ 64 = 6

b)         25^1/2 = 5

            Log₂₅5=1/2

c)         4^-4= 1/256

Log₄1/256 = -4

Conversion from Logarithmic to Index form

a)         Log₂128 = 7              b)           log₁₀ (0.01) = -2                  c)            Log_1.5 2.25 = 2

Solution

a)         Log₂128 = 7

            2⁷ = 128

b)         Log₁₀ (0.01) = -2

            10^-2= 0.01

c)         Log1.5 2.25 = 2

1.5² = 2.25

Laws of Logarithm

a)         let P = b^x, then log_bP = x

            Q = b^y, then log_bQ = y

            PQ = b^x X b^y = b^x+y (laws of indices)

            Log_b PQ = x + y

:.          Log_b PQ = log_bP + Log_bQ

b)         P÷Q = b^x÷by = b^x+y

            Log_bP/Q = x –y

:.          Log_bP/Q = logbP – logbQ

c)         Pⁿ= (b^x)ⁿ = b^xn

            Log_bpⁿ = nb^x

:.          LogPⁿ = logbP

d)         b = b¹

:.          Log_bb = 1

e)         1 = b⁰

            Logb1 = 0

Example 

Solve each of the following:

a)         Log₃27 + 2log₃9 – log₃54

b)         Log₃13.5 – log₃10.5

c)         Log₂8 + log₂3

d)         Given that log₁₀2 = 0.3010 log₁₀3 = 0.4771 and log₁₀5 = 0.699 find the log₁₀64 + log₁₀27

Solution

a)         Log₃27 + 2 log₃9 – log₃54

= log₃ 27 + log₃ 9² –log₃54

= log₃ (27 x 9²/54)

= log₃ (27¹ x 81/54) = log₃ (81/2)                  

= log₃ 3⁴/log32

= 4log₃ 3 – log₃ 2

= 4 x (1) – log₃ 2 = 4 – log₃ 2

= 4 – log₃ 2

b)         log₃ 13.5 – log₃ 10.5

            = log₃ (13.5) – Log310.5 = log₃ (135/105)

            = log₃ (27/21) = log₃ 27 – log₃ 21

            = log₃ 3³ – log₃ (3 x 7)

= 3log₃ 3 – log₃ 3 -log₃7

= 2 – Log₃ 7

c)         Log₂8 + Log₃3

            = log₂2³+ log₃3

            = 3log₂2 + log₃3

             = 3 +1 = 4

d)         log₁₀ 64 + log₁₀ 27

            log₁₀ 2⁶ + log₁₀3³

            6 log₁₀ 2 + 3 log₁₀ 3

            6 (0.3010) + 3(0.4771)

            1.806 + 1.4314 = 3.2373.

EVALUATION

1.         Change the following index form into logarithmic form.

            (a)  6³= 216   (b) 3³ = 1/27   (c) 9² = 81

2.         Change the following logarithm form into index form.

            (a)  Log₈8 = 1             (b) log _½¼ = 2

3.         Simplify the following

             a)  Log₅12.5 + log52               b)      ½ log₄8 + log₄32 – log₄2      c) log₃81

4.         Given that log 2 = 0.3010, log3 0.4770, log5 = 0.6990,   find the value of log6.25 + log1.44

Logarithmic Equation

Change of Base

Let log_bP = x and this means P = b^x

Log_cP = log_cb^x = x log_cb

If x log_cb = log_cP

          x = log_cP

                 log_c b

:.          log_cP = log_cP

                          log_cb

Example :

Shows that log_ab   x   log_ba  = 1

                 Log_ab = log_cb

                               log_ca

                 Log_ba = log_ca

                               log_cb

:.          log_ab   x log_ba  =  log_cb  x log_ca

            log_ca  +  log_cb = 1

Evaluation

Solve (i) Log₃ (x² + 7x + 21) = 2 (ii) Log₁₀ (x² – 3x + 12) = 1

 (iii) 5^2x+1 – 26(5^x) + 5 = 0 find the value of x

Logarithm of numbers greater than one   

Numbers such as 1000 can be converted to its power of ten in the form 10ⁿ where n can be term as the number of times the decimal point is shifted to the front of the first significant figure i.e. 10000 = 10⁴

Number                                                 Power of 10

1. 10²
2. 10¹
3. 10⁰ 
   1. 10^-3
   1. 10^-1

Note: One tenth; one hundredth, etc are expressed as negative powers of 10 because the decimal point is shifted to the right while that of whole numbers are shifted to the left to be after the first significant figure.

A number in the form A x 10ⁿ, where A is a number between 1 and 10 i.e. 1 < A < 10 and n is an integer is said to be in standard form  e.g. 3.835 x 10³ and 8.2 x 10^-5 are numbers in standard form.

Multiplication and Division of number greater than one using logarithm

To multiply and divide numbers using logarithms, first express the number as logarithm and then apply the addition and subtraction laws of indices to the logarithms. Add the logarithm when multiplying and subtract when dividing.