1. Logarithms
Laws of logarithms

2. 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

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

4. 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
= (log9 9 = 1)
3 logx 5

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 (Rule 1)
= loga12 – loga 5
= (Rule 2)
= loga 2.4

6. = 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

7. 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

8. 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 + log Rule 1
= log2 3 + log2 52
= log log2 5
= (2.32)
= 6.22

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

10. 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 =

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

12. 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 =

13. 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

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

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