LOGARITHMS

The logarithm of  a number z to a base b is the power to which b must be raised to give the number z

Specifically, the logarithms of a number to a base 2 is the power to which 2 must be raised to give the number

In general, then if z = bⁿ  the logarithm of Z to base b is

                                    n = log_bz

z = number, b = base, n = logarithm of z to base b

Note:  Logarithm to the base e written in x or log_ex are called natural or Naperian logarithm.

Logarithms to the base 10 written log x or log₁₀x are called common logarithms.

 

Basic Rules of logarithms

1.         Log_aM + log_aN = log_a(m×n)    (multiplication)

2.         Log_aM − log_aN =   (division)

3.         Log_aMⁿ = nlog_aM       (Raising to power)

4.         Log_a1 = 0 (Logarithms of unity)

5.         Log_aa = 1  (Logarithms of the base)

6.         Log_b^N =  (To change a logarithm   base)

In particular

7.         log _ba =

8.        

9.        

10.       log _am = log_aN   if and only M =N           

            11.     If a^x = z then log_a z  = x  

 Example

Simplify

     1.   log ₃ 81

      2.   log₃

      3.   log ₁₂₅25

 

 Solution

    1.     Let  log₃81 = a

                3^a  ₌81

                3^a= 3⁴

                a = 4

              log₃81 = 4

    2.     Let   log_{3  =} b

               3^b =

                 3^b =

                 3^b = 3^-2

                  b = −2

        

3          Let log ₁₂₅25 = c

               125^c = 25

                 5^{3 c}  = 5²

                  3c   = 2

                    c  =    

Example

     Evaluate  without using tables

1.  3log2  + log20 – log

2.  2log₆3 + log ₆12+ log₆8 -  log₆24

 

    Solution

1. Since the  base is not specified , you may assume

    base 10 all through

    3 log2 + log 20 – log             

     = log 2³ + log20 - log

     = log 8+ log20 – log    

       = log   = log

     =log₁₀100 =2

      

2. 2log₆3 + log₆12+ log₆8 – log₆24

   = log_6­3² + log ₆12 + log₆8 –log₆24

  = log₆9 + log₆12 + log₆8 –log₆24

  = log₆  =  log₆36

   = 2 log₆6 =   2

 

 Example

Solve for x in the following  equation

1. log₄  - log₄  = 2

      2. log₇  + log₇  =  log₇

      3. log₃ x+  3 log_x3  =  4

4. log_x3  + log_x27 = 2

 

Solution

1.        log₄  - log₄  =2

                          

                              = 4² =16

                              x– 3 = 16

                              x–3 = 16x + 48

                                     16x –x = –3 – 48

                              15x = – 51

                             x =  =   

 

2.  log₇3x + log₇  = log₇

                 log₇3x  =log₇

                      3x  =  16 –10

                     6x² – 3x = 16x –10

                     6x² –19x + 10 = 0

                 = 0

                      x =  or x =   

 

3.         log₃x + 3log_x3 = 4

            Note log_x3 =  (Using Rule 7)

            log₃x + 3

            log₃x +  = 4

            Let log₃x = a

            a +  = 4

            a² – 4a + 3 = 0

            (a –1) (a–3) = 0

            a = 1 or a = 3

            when a = 1

             log₃x = 1

x = 3¹ = 3

            or when a = 3

             log₃x = 3

            x = 3³ = 27

            x = 3 or x = 27

 

 

4.         log_x3 + log_x27 = 2

                        log_x 3×27 = 2

log_x 81 = 2

             x² = 81

                         x² = 9²

              x = 9

Example

Solve the equation                                                                                                                               

1.                                                      log₁₀ (a² + 6a + 28) = 2

2.                                                      log₃ (a² – a – 2) = 2log (a+1)

Solution

1.                  log₁₀ (a²+6a+28) = 2

            a² + 6a + 28 = 10²

            a² + 6a + 28 = 100

            a²+6a – 72 = 0

           (a – 6)(a+12) = 0

            a = 6 or a = –12

2.         log₃(a²– a –2) = 2log₃ (a+1)

            log₃ (a²– a–2) = log₃ (a+1)²  

  Since the logarithm base is the same

            a² – a – 2 = (a+1)²

            a² – a – 2 = a² + 2a +1

                 – a – 2a = 3

                       –3a = 3

                           a = –1

 

Example    Solve for x

            log₂(x²– 4x–16) – log₂ (x² – 3x +4) = –1

Solution

            log₂(x² – 4x – 16) – log₂ (x²-3x+4) = –1

            log₂

                 

            2(x²–4x–16) = x²–3x+4

            2x² – 8x – 32 = x² – 3x + 4

            x² – 5x – 36 = 0

            (x – 9) (x+4) = 0

            x = 9 or x = -4

Example

1.         Solve   log (x–1) + 2 logy = 2log3

                             logx + log y = log 6

            and find the value of (x+y)^½

2.         Solve the equation

    (i)     2(log₂a – 1 ) = –log₂b

                               b = a – 1

 

   (ii)                  ab = 10

            2(log₁₀ a-1) = –log₁₀b

Solution

1.         log (x–1) + 2log y = 2log 3

                        log x + log y = log 6

            log (x –1) + 2log y = 2log3

            log (x-1) +log y²  = log3²

                               log y² (x-1) = log 9

                                    y²(x-1) =  9 - - - - (1)

 

                         log x + log y = log 6

                 log xy = log 6

                                    xy = 6       -  - - - (2)

                             i.e  x =

            from (2) x =  and substituting  it into (1)

                        y²  = 9

                        y²  = 9

                        y (6 – y) = 9

                        6y – y² = 9

                        y² – 6y + 9 = 0

                        (y –3)² = 0

                          y = 3 twice

            Substituting y = 3 into x =  gives x = 2

                        x = 2 twice , y = 3 twice

            The value of (x + y)^½ = 5^½

                        (x + y)^½ =       

 

2i.        2(log₂a – 1) = –log₂b

                        2log₂a– 2 = – log₂b

                        log₂ a² +2 = log₂b

                        log₂ a² +log₂b = 2

                        log₂a² + log₂b = 2

                        log₂ a² b = 2

                        a²b = 4

            Substitutes b = a – 1 into a²b = 4

                        a²(a –1) = 4

                        a³–a²–4 = 0

                        (a+1)(a-2)(a–2) = 0

                        a = –1 or a = 2 twice

                        when a = –1, b = –1 – 1 = –2

                        when a = 2 twice  b = 2–1 = 1 twice

                     (a,b) = (–1, –2) or (a, b) = (2,1) twice

 

ii.          ab = 10, 2(log₁₀a –1) =– log₁₀b

                Solution

                        ab = 10  - - - - - - - - - - - - (i)

                        2(log₁₀a –1) = –log₁₀b - - - - - (ii)

                        2 log₁₀a –2 = –log₁₀b

                        2 log₁₀a + log₁₀ b = 2

             log₁₀ a² + log₁₀b = 2

             log₁₀ a²b = 2

             a²b = 10²

             a²b = 100 - - - - - - (iii)

from (i) a = , substituted into (iii)

             b = 100

            . b = 100

             = 100

               b = 1

when b = 1 then a =  = 10

            a = 10, b = 1

Example

Given that log_x a + log_x b = g and log_xa – log_xb = h Prove that a = x^½(g+h) and find the similar expression for b

Solution

log_xa + log_xb = g

    log_x^ab = g

         ab = x^g     

            b =  - - - - - - - - - - (i)

Also log_xa –log_xb = h

            log_x  = h

                  = x^h

                   b =  - - - - - - - - (ii)

Equating (i) and (ii)

                        b =

                        a² = x^g . x^h = x^g+h

                        a^. = ^{½ (g+h)}

Also from b = 

            a =   - - - -  - - - - - - - (iii)

From b =

            a = bx^h - - - - - - - - - - - - - (iv)

Equating (iii) and (iv)

                        a =   = bx^h

                            b² = 

                           b² = x^g-h

                           b = x^{½ (g-h)}

Example

If 2log_ba + 2log_ab = 5, show that log_b a is either ½ or 2.  Hence find all pairs of values of a and b which satisfies simultaneously the equation above and the equation ab = 27   

 

Solution

1.         2 log_ba + 2 log_ab = 5

            From rules of logarithms log_ab =

                        2 log_ba +  = 5

                        Let x = log_b a

                        2x +  = 5

                        2x² – 5x + 2 = 0

                        (2x–1) (x–2) = 0

                        x = ½ or x = 2

                         log_ba = ½ or 2

 

(ii)        2 log_ba + 2log_ab = 5 - - - - - -  - (i)

                                  ab = 27 -  -- -- - - (ii)

            Already obtained log_b a = ½ or 2

                         log_ba = ½ - - - - - - - -(a)

                             ab = 27 - - --- - - - - - (b)

            From (b) a = , substituted into (a)

                        log_b  = ½

                         = b^½

                        27 = b^½ . b = b^½+1       

                        b =

                        b= 9    

                If b = 9, then a =  = 3

            So (a,b) = (3,9)

Also   log_ba = 2 –  - - - - - - - - - (c)

            ab = 27 - - - - - - - - - - - (d)

from (d) a =  substituted in (c)

                        log_b  = 2

                         = b²

                        27 = b³

                          b = 3

If b = 3 then a =  = 9

            (a,b) = (9,3)

(a,b) = (3,9) or (a,b) = (9,3)

 

Example

1.         Show that

Solution

log_ab =  or log_ba =  (Using Rule 7)

            (Change the base to xyz)

 log_xyzx + log_xyzy + log_xyzz = log_xyzxyz

                        log_xyzxyz = 1

Example

If a²+b² = 7ab, show that

log (a+b) = log 3 +

Solution

a² + b² = 7ab

(a+b)² – 2ab = 7ab

(a+b)² = 9ab

Take log of both sides

               log (a+b)² = log 9ab

              2 log (a+b) = log 9ab

              2 log (a+b) = log 9 + log a + log b

            2 log (a+b) = log 3² + log a + log b

            2 log (a+b) = 2 log 3 + log a + log b

               log (a+b) =     

             log (a+b) = log 3 +

Example

   If a² + b² = 47ab show that

    log a + log b = 2 log

Solution

            a² +b² = 47ab

            (a+b)² – 2ab = 47ab

            (a+b)² = 49ab

            (a+b)² = 7² ab

             =ab

Take logarithms of both sides

     log  = log ab

              2log  = log a + log b

 log a + log b = 2 log

Example

             If log_2x x = a, log_3x 2x = b and log_4x 3x = c

             Show that abc +1 = 2bc

Solution

            abc = log_2x x. log_3x2x. log_4x3x

                  changing the base

            abc =

            abc =

            abc+1 = log_4xx+1

                        = log_4x x +log_4x4x

                        = log_4x4x² - - - - - - (1)

            2bc = 2log_3x^2x.log_4x^3x

            = 2  

            =

            = log_4x(2x)² = log_4x4x²

            2bc = log_4x4x² - - - - (2)

                abc +1 = 2bc = log_4x4x²

Example

Prove the definition of logarithms that if m, n, x are positive numbers then

                        log_mnx =

                        Let y = log_mn^x - - - - - (1)

                        w = log_nx - - - - - - - - (2)

                        z = log_nm - - - - - - - - (3)

Using the Rule (11)

            x = (mn)^y = m^yn^y - - - - - - - (a) from (1)

            x = n^w                                   (b) from (2)

            m = n^z                                  (c) from (3)

Equating (a) and (b)

            m^yn^y = n^w----------------------(d)

Substitute m = n^z into (c)

            (n^z)^y n^y = n^w

            (n^zy) n^y = n^w

            n^zy+y = n^w

Equating the powers since the base are equal

            zy+y = w

            y(z+1) = w

 

Example

Solve the equation 2^x.3^1−x = 6

Solution

Take log of both side

              log 2^x.3^1−x = log 6

            log2^x +log3^1-x  = log 6

            x log 2 + (1−x) log 3 = log 6

x log2 + log3 – x log 3 = log 6

            x log 2 – x log3 = log 6 – log 3

            x (log 2 – log 3) = log                       

            x (log 2 – log 3) = log 2

            x =

            x =

            x =

 

Example

            Solve the equation 5^2x – 5^x+1 + 4 = 0

Solution

            5^2x – 5^x+1+4 = 0

            (5^x)² – 5^x.5¹+4 = 0

            Let b = 5^x

            b² – b×5+4 = 0

            b² – 5b + 4 = 0

            (b−4) (b−1) = 0

               b = 4 or b = 1

when b = 4

            5^x = 4

Taking the log of both sides

            log 5^x = log 4

            x log5 = log4

            x =

x = 0.8613

when b = 1

            5^x = 1

            5^x = 5⁰

            x = 0

            x = 0 or x = 0.8613

 

Example

Solve the Simultaneous equations  2^x+y= 6,

3^x−y  = 4

 

Solution

From the firs equation , take the logarithm of both sides

            log2^{x+ y} =log 6

            (x+y)log2=log6
            x + y =   

            x+ y =

            x+ y = 2. 5854………{1}

         for the second equation, take the logarithm

         of both sides

log 3^x−y  = log 4

 (x – y)log 3 = log 4

              x–y =

             x–y =

             x–y =1.2612……{2}

Adding equation {1} and {2} together

            2x = 3.8460

            x = 1.9233

Substituting  x = 1.9233 into equation(1)

                     y = 0.6621

         x = 1.9233  and  y = 0.6621

 

Example: Show without using tables

1.      2 log₁₀6 + 3log₁₀2 =log₁₀288

2.      log₁₀19  =   { 2 log₁₀6 + 1}¹⁰

            Hence deduce from these result

           = .

Solution

i           L. H.S = 2 log₁₀6 + 3 log ₁₀2

             = log₁₀6²   + log₁₀2³

             =   log₁₀36 +log ₂8

             = log₁₀36x8

             = log₁₀288 = R.H.S

ii           2log₁₀19=log₁₀19²
                       = log₁₀361   

                        log₁₀360

                        log₁₀36x10

                        log₁₀36+ log₁₀10

                        log₁₀6² +1

       2log₁₀19  2 log₁₀6 +1

        log₁₀19

iii.         2 log₁₀ 17 = log₁₀ 17²

                             = log₁₀ 289

                              log₁₀ 288

            2 log₁₀17  2 log₁₀⁶ + 3log₁₀2

               log₁₀17  ½ (2 log₁₀6 + 3log₁₀2)

log₁₀19- log₁₀17  [2 log₁₀6 + 1] –

                                       ½[2log₁₀6 + 3 log₁₀2]

 ½ [2 log₁₀6+1 – 2 log₁₀6 – 3 log₁₀2]

log₁₀   ½ [1 – 3log₁₀2]

    ½ [1-log₁₀⁸]           

log₁₀  ½ [log₁₀10 – log₁₀8]

               ½ [log₁₀  ]  ½

            log₁₀  = log₁₀

                     log₁₀  = log₁₀         

                       

                         

Example

Show that  log_a (a+b)² = 2+log_a  

Solution

            Log_a (a+b)² = log_a (a²+2ab+b²)

            Factorize a² from a²+2ab+b²

            log_a (a+b)² = log_a a²

                                   

Example

Without using tables or calculator, evaluate

Solution

           

            = 

 

=

=

           

 

Example

Without tables or calculator, evaluate

                          

Solution

           

           

Example

a.      If x log₁₀x = log₂x, Find the value of x

b.      Find N if log₅ N + log₂₅ N = 6

Solution