1. LOGARITHMS

2. Definition: The “Log” of a number, to a given base, is the power to which the base must be raised in order to equal the number. e.g.your calculator tells you that log 100= 2 This is because the log button on the calculator uses base 10. …and 10 must be raised to the power 2, to equal 100. i.e 10 2 = 100 We write:log 10 100 = 2( Since 10 2 = 100 ) n.b. log to base 10 is sometimes written as lgi.e. lg100 = 2

3. Note that any positive value can be chosen as the base number: e.g. In base 2; log 2 8= 3( Since 2 3 = 8 ) In general: If log a x = p, then a p = x e.g. In base 7; log 7 49= 2( Since 7 2 = 49 ) Note also:log 3 3= 1 log 5 5= 1 i.e. For any base: log a a = 1 Also:log 3 1= 0 log 8 1= 0 i.e. For any base: log a 1 = 0

4. Example 1:Find the value of: a) log 2 32 b) log 16 4 c) log 2 a) log 2 32= 5( Since 2 5 = 32 ) b) log 16 4 = Example 2:Solve the following equations: a) log 4 x = 2 b) log 3 (2x + 1) = 2 a) log 4 x = 2Hence, x = 16 b) log 3 (2x + 1) = 2 9 = 2x + 1 Hence, x = 4 ( Since 16 ½ == 4 ) c) log 2 1818 = –3 ( Since 2 –3 = 123123 1818 = ) 4 2 = x 3 2 = 2x + 1 1212

5. Laws of Logarithms As logarithms are themselves indices the laws of logs are closely related to the laws of indices. If log a X = p,then a p = X.If log a Y = q, then a q = Y i) XY = a p a q i.e XY = a p + q So, log a XY = p + q log a XY = log a X + log a Y iii) X n = (a p ) n i.e. X n = a pn So, log a X n = pn log a X n = nlog a X So, log a X Y = p – q ii) X Y apap aqaq = a p – q X Y = i.e.

6. Example 1: Express the following as a single logarithm: a) log4 + 2log3 b) logP – 2logQ + 3logR a)log4 + 2log3 = log4 + log3 2 = log 36 b) logP – 2logQ + 3logR = logP – logQ 2 + logR 3 Example 2: Express log in terms of log x and log y. y2y2 = log PR 3 Q 2 log y2y2 = – log y 2 1212 = log x – 2 log y

7. Example 3: Given log 2 3 = p, and log 2 5 = q, find the following in terms of p and q: a) log 2 15 b) log 2 60 c) log 2 2.7. a) log 2 15 =log 2 (3)(5)= log 2 3 + log 2 5= p + q b) log 2 60 =log 2 (3)(5)(2 2 )= log 2 3 + log 2 5 + log 2 2 2 = log 2 3 + log 2 5 + 2log 2 2 = p + q + 2 c) log 2 2.7 == log 2 27 – log 2 10= log 2 3 3 – log 2 (2)(5) = 3log 2 3 – ( log 2 2 + log 2 5 ) = 3p –1 – q or just log 2 4 log 2 27 10

8. Example 4: Solve log 4 x + log 4 (3x + 2) = 2. (3x + 8)(x – 2) = 0 Since x must be positive, log 4 x (3x + 2) = 2 3x 2 + 2x – 16 = 0 Using: logX + logY = logXY x = 2 4 2 = x (3x + 2 )

9. Equations of the form a x = b. These can be solved using logarithms: Example 5: Solve the equation 2 = 7. x Taking logs of both sides: log 2 = log 7 x logX n = nlogX Using: x log 2 = log 7 x = log 7 log 2 = 2.81( 3 sig.figs.)

10. Summary of key points: This PowerPoint produced by R.Collins ; Updated Feb.2012 The “Log” of a number, to a given base, is the power to which the base must be raised in order to equal the number. If log a x = p, then a p = x log a a = 1 log a 1 = 0 Laws of Logs: logX + logY = logXY logX n = n logX