Logarithms Laws of logarithms

Product Rule loga xy = loga x + loga y loga x + loga y = loga xy Examples: log5 (35) = log5 3 + log5 5 log3 5 + log3 4 = log3 (5  4) = log3 20

Quotient Rule = loga x – loga y loga x – loga y = log5 20 – log5 4 = =1 = log3 16 – log3 5

Power Rule loga xm = m loga x m loga x = loga xm logx 53 = 4 log9 3 = log9 34 = log9 81 = log9 92 = 2 log9 9 = 2 (log9 9 = 1) 3 logx 5

loga 3 + loga 4 – loga5 = loga (3  4) – loga 5 (Rule 1) Express the following as a single logarithms loga 3 + loga 4 – loga5 = loga (3  4) – loga 5 (Rule 1) = loga12 – loga 5 = (Rule 2) = loga 2.4

= log4 x5 – log4 y2 + log4 z3 (Rule 3) Express the following as a single logarithms 5 log4 x – 2 log4 y + 3 log4 z = log4 x5 – log4 y2 + log4 z3 (Rule 3) (Rule 2) = (Rule 1) + log4 z3

Question: Given that log2 3 = 1.58 and log2 5 = 2.32, (a) log2 75 Find value of each of the following. (a) log2 75 (b) log2 0.3 (c) log2 √5

SOLUTION (a) log2 75 = log2 [325] = log2 3 + log2 25 Rule 1 Given that log2 3 = 1.58 and log2 5 = 2.32 (a) log2 75 = log2 [325] = log2 3 + log2 25 Rule 1 = log2 3 + log2 52 = log2 3 + 2 log2 5 = 1.58 + 2(2.32) = 6.22

Solution (b) log2 0.3 = log2 (3÷10) = log2 3  log2 10 Rule 2 = 1.58  (2.32 + 1) =  1.74 (c) log2 √5 = (1/2) log2 5 = (1/2)(2.32) = 1.16

CHANGE OF BASE Change of base-a to base c is as follows: loga b = For example, to change log4 8 to base-2 log4 8 =

EXAMPLE Evaluate log5 12. log5 12 = Use calculator Use at least 4 significant figures

CHANGE OF BASE Change of base-a to base b is as follows: loga b = For example, to change log32 2 to base-2 log32 2 =

EXERCISE Given that log2 5 = 2.32 find the value for each of the following without using calculator. ( without changing to base-10) (a) log5 4 (b) log5 2 (c) log4 50

Given that log2 5 = 2.32 (a) log5 4 Change to base-2 Rule 3 =0.8621

(b) log5 2 = (c) log4 50 =0.431 =2.82