LOGARITHM – SOLVING PROBLEMS BASED ON LAWS OF LOGARITHM

Back to: Further Mathematics SS1

The logarithm of numbers greater than one   

The 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 log 6.25 + log1.44

Logarithmic Equation

Solve the following equation:

a)         Log10 (x² – 4x + 7) = 2

b)        Log₈ (r² – 8r + 18) = 1/3

Solution

a)         Log₁₀ (x² – 4x + 7) = 2

x² – 4x + 7 = 10² (index form)

x² – 4x + 7 = 100

x² – 4x + 7 – 100 = 0

x² – 4x – 93 = 0

Using quadratic formula

x          = – b ±√b²– 4ac

2a

a = 1, b = – 4, c = – 93

x = – (- 4) ± √(- 4) ² – 4 x 1 x (- 93)

2 x 1

= + 4 ± √16 + 372

2

= + 4 ± √388/2

= x = 4 +√ 388/2 or 4 – √388/2

x =  11.84 or x = – 7.85

2)         Log₈ (x² – 8x + 18) =1/3

x² – 8x + 18 = 8^1/3

x² – 8x + 18 = (2)^3X1/3

x² – 8x + 18 =2

x² – 8x 18 – 2 = 0

x² – 8x + 16 = 0

x² – 4x – 4x + 16 = 0

x(x – 4) -4 (x – 4) = 0

(x – 4) (x – 4) = 0

(x – 4) twice

x = + 4 twice

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

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 log 6.25 + log1.44

Logarithmic Equation

Solve the following equation:

a)         Log10 (x² – 4x + 7) = 2

b)        Log₈ (r² – 8r + 18) = 1/3

Solution

a)         Log₁₀ (x² – 4x + 7) = 2

x² – 4x + 7 = 10² (index form)

x² – 4x + 7 = 100

x² – 4x + 7 – 100 = 0

x² – 4x – 93 = 0

School Owner? Looking for ready-made content and tools to save time and grow easily? Book a free demo session now

Get more class notes, videos, homework help, exam practice on Android [DOWNLOAD]

Get more class notes, videos, homework help, exam practice on iPhone [DOWNLOAD]