Classwork Series and Exercises {Mathematics- SS3}: Theory of Logarithm

Theory of Logarithm

The logarithm of a number X to a base A is the power (index) of A in X

If log_AX =Y, then X =A^Y

Example: log ₁₀1000 = 3, then 10³ = 1000

This rule holds for all bases, therefore if 2⁴ = 16, then log₂16 = 4

TRY THIS:

Express the following in logarithmic form;

a)    25^1/2 = 5 [fusion_builder_container hundred_percent=”yes” overflow=”visible”][fusion_builder_row][fusion_builder_column type=”1_1″ background_position=”left top” background_color=”” border_size=”” border_color=”” border_style=”solid” spacing=”yes” background_image=”” background_repeat=”no-repeat” padding=”” margin_top=”0px” margin_bottom=”0px” class=”” id=”” animation_type=”” animation_speed=”0.3″ animation_direction=”left” hide_on_mobile=”no” center_content=”no” min_height=”none”][log₂₅5= ½ ]

b)    1.2³ =1.728 [log_1.21.728 = 3]

Express in index form:

a)    log₄64 = 3 [4³ = 64]

b)    log ₂₅0.2 = – ½  [25^{– ½}  = 0.2]

Laws of Logarithm

1. log (MN) = log M + log N 

Example:  log (2 x 3) = log 2 + log 3

2. Log (^M/N) = log M – log N

Example: log 2.5 = log ⁵/2 = log 5 – log 2

3. Log_M M =1 (from M¹ = M)

Example: Log₁₀₀100 = 1

4. Log M¹ = 0 (from M⁰ = 1)

Example: log₅₅₆₇1 =0

5. LogM^p = plogM

Example: If log 2 =0.3010, find log 8

Solution: log 8 =log 2³ = 3log2 = 3 x 0.3010 = 0.9030

Examples:

1. Simplify log 16 – log 8 = log (¹⁶/8) = log 2

2. Evaluate log₁₀5 + log₁₀20 = log₁₀ (5 X 20) = log₁₀100 = log₁₀10² =2log₁₀10 =2

3. Given that log₁₀­2 = 0.3010 and log₁₀3 = 0.4771, calculate without the use mathematical tables or calculator (a) log₁₀54 (b) log₁₀0.24 (c) log₁₀15

Solutions

(a)    Log₁₀54 = log₁₀(2X27) = log₁₀(2 x3 x 3 x3) = log₁₀(2 x3³) = log₁₀2 + log₁₀3³

Log₁₀54 = log₁₀2 + 3log₁₀10 = 0.3010 + 3(0.4771) = 1.7323

(b) Log₁₀0.24 = log₁₀ (²⁴/100) = log₁₀24 – log₁₀100

          Log₁₀24 = log₁₀ (2 x2 x 2 x3) = log₁₀ (2³X 3) = log₁₀2³ + log₁₀3 = 3log₁₀2 + log₁₀3

         log₁₀100 =2log_10­10

        log 0.24 = 3log₁₀2 + log₁₀3 – 2log₁₀10

       = 3(0.3010) + 0.4771 – 2(1)

è 0.9030 + 0.4771 -2 = -0.6199

(c) Log_10­15 = log₁₀(³⁰/2) = log₁₀(3 x10)= log₁₀3 + log₁₀10 – log₁₀2

      = 0.4771 + 1 – 0.3010 = 1.1761

Further Examples

Solve for x if (i) log₁₀X = 3 (ii) log₁₀X = -2 (iii) log_X81 = 4 (iv) 2log_X (3³/8) =6 (v) log x – log (2x -1) = 1

Solutions:

(i)            Log₁₀X = 3; 10³ = X,  è X = 1000

(ii)          Log₁₀X = -2; 10^-2 = X, è X = 0.01

(iii)         Log_X81 = 4; X⁴ = 81, X⁴ = 3⁴ è X = 3 (by comparison of powers and bases)

(iv)         2log_X(3³/8) = 6; log_X(3³/8)² = 6

= (3³/8)² è X⁶; (²⁷/8)² = X⁶

= [(³/2)³]² = X⁶; (³/2)⁶ = X⁶

X = ³/2 = 1½           

(v)          Log X – log (2X -1) = 1

    Log  (   X          )= 1                

               (2X -1)

Take Log 10 to both sides

Log 10 (   X          )= log ₁₀10; from log₁₀10 = 1                      

                (2X -1)

             (   X          )= 10              

                (2X -1)

è X = 10(2X -1);  X =  20X -10

     X – 20X = -10; -19X = -10

       èX = ¹⁰/19

EXERCISES

Lets see how much you’ve learnt, attach the following answers to the comment below

1.    Evaluate log₁₀6 + log₁₀45 – log₁₀270 (a) 0 (b) 1 (c) 1.1738 (d) 1.3802 (e) 10

Guideline:log₁₀6 + log₁₀45 – log₁₀270 = log₁₀ (6 x 45) = log₁₀ (270) = log₁₀1 = 0. Answer: 0

                                                                                       270                270

2.    If log₁₀a is 4, what is a? (a) 0.4 (b) 40 (c) 400 (d) 1000 (e) 10,000

Guideline: log₁₀a = 4, then a = 10⁴ = 10,000. Answer: 10,000

3.    What is the value of log₂8 – log₃¹/9? (a) – 1½ (b) -1 (c) -5 (d) 1½ (e) 5

Guideline: log₂8 = log₂2³ = 3log₂2 = 3 (log₂2 = 1)

           Log₃¹/9 = log₃9^-1 = log₃3^{2 x -1} =log₃3^-2 = -2log₃3 = -2

è log₂8 – log₃¹/9 = 3 – (-2) = 3 + 2 = 5. Answer: 5

4.    X = log₁₀√35 + log₁₀√2 – log√7, find x (a) ½ (b) 1 (c)2 (d) 3 (e) √10

Guidelines: Use same process as in number 1; log₁₀√(35 x 2) = log₁₀√70 =log₁₀√10 = ½ . Answer: ½

                                                                                                   √7

5.    Given that log₁₀3 = 0.477, evaluate log ₁₀2.7 correct to 2 significant figures

(a)  0.43 (b) 0.55 (c) 1.4 (d) 1.43 (e) 2.52

Guideline: log 2.7 = log₁₀(27/10) = log₁₀27 – log₁₀10 = log₁₀3³ – log₁₀10

è 3log₁₀3 – log₁₀10 = 3(0.477) – 1 = 1.431 – 1 = 0.431 è 0.43 (2 sig.fig.). Answer: 0.43

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