Theory of Logarithm
The logarithm of a number X to a base A is the power (index) of A in X
If logAX =Y, then X =AY
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) 251/2 = 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 [log1.21.728 = 3]
Express in index form:
a) log464 = 3 [43 = 64]
b) log 250.2 = – ½ [25– ½ = 0.2]
Laws of Logarithm
- log (MN) = log M + log N
Example: log (2 x 3) = log 2 + log 3
- Log (M/N) = log M – log N
Example: log 2.5 = log 5/2 = log 5 – log 2
- LogM M =1 (from M1 = M)
Example: Log100100 = 1
- Log M1 = 0 (from M0 = 1)
Example: log55671 =0
- LogMp = 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 =2log1010
log 0.24 = 3log102 + log103 – 2log1010
= 3(0.3010) + 0.4771 – 2(1)
è 0.9030 + 0.4771 -2 = -0.6199
(c) Log1015 = 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) logX81 = 4 (iv) 2logX (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) LogX81 = 4; X4 = 81, X4 = 34 è X = 3 (by comparison of powers and bases)
(iv) 2logX(33/8) = 6; logX(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 = log332 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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