LOGARITHM – SOLVING PROBLEMS BASED ON LAWS OF LOGARITHM

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 must be raised to be equal to P.

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

a=P are equivalent to each other.

a=P is called the index form and logaP = 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)         251/2 = 5                      c)  44= 1/256

Solution

a)         26 = 64

Log2 64 = 6

b)         251/2 = 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)            Log1.5 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 = bx, then logbP = x

Q = by, then logbQ = y

PQ = bx X by = bx+y (laws of indices)

Logb PQ = x + y

:.          Logb PQ = logbP + LogbQ

b)         P÷Q = bx÷by = bx+y

LogbP/Q = x –y

:.          LogbP/Q = logbP – logbQ

c)         Pn= (bx)n = bxn

Logbpn = nbx

:.          LogPn = logbP

d)         b = b1

:.          Logbb = 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 – log2

b)         log13.5 – log10.5

= log(13.5) – Log310.5 = log(135/105)

= log(27/21) = log27 – log21

= log33 – log(3 x 7)

= 3log3 – log3 -log37

= 2 – Log7

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) – 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 = 81/3

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 logbP = x and this means P = bx

LogcP = logcbx = x logcb

If x logcb = logcP

x = logcP

logc b

:.          logcP = logcP

logcb

Example :

Shows that logab   x   logba  = 1

Logab = logcb

logca

Logba = logca

logcb

:.          logab   x logba  =  logcb  x logca

logca  +  logcb = 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 = ax, then x is the logarithm to the base a of P. We write this as x = log a P. The relationship logaP = x and

a=P are equivalent to each other.

a=P is called the index form and logaP = 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)         251/2 = 5                      c)  44= 1/256

Solution

a)         26 = 64

Log2 64 = 6

b)         251/2 = 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)            Log1.5 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 = bx, then logbP = x

Q = by, then logbQ = y

PQ = bx X by = bx+y (laws of indices)

Logb PQ = x + y

:.          Logb PQ = logbP + LogbQ

b)         P÷Q = bx÷by = bx+y

LogbP/Q = x –y

:.          LogbP/Q = logbP – logbQ

c)         Pn= (bx)n = bxn

Logbpn = nbx

:.          LogPn = logbP

d)         b = b1

:.          Logbb = 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 – log2

b)         log13.5 – log10.5

= log(13.5) – Log310.5 = log(135/105)

= log(27/21) = log27 – log21

= log33 – log(3 x 7)

= 3log3 – log3 -log37

= 2 – Log7

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

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