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# Laws of Logarithm

1. a) let P = bx, then logbP = x

Q = by, then logbQ = y

PQ = bx X by = bx+y (laws of indices)

LogbPQ = x + y

:.             LogbPQ = logbP + LogbQ

1. b) P÷Q = bx÷by = bx+y

LogbP/Q = x –y

:.             LogbP/Q = logbP – logbQ

1. c) Pn= (bx)n = bxn

Logbpn = nbx

:.             LogPn = logbP

1. d) b = b1

:.             Logbb = 1

1. e) 1 = b0

Logb1 = 0

#### EXAMPLE – Solve each of the following:.

1. a) Log327 + 2log39 – log354
2. b) Log35 – log310.5
3. c) Log28 +  log23
4. d) given that log102 = 0.3010 log103 = 0.4771 and log105 = 0.699 find the log1064 + log1027

Solution

1. a) Log327 + 2 log39 – log354

= log3 27 + log3 92 –log354

= log3 (27 X 92/54)

= log3 (271 X 81/54) = log3 (81/2)

= log3 34/log32

= 4log3 3 – log3 2

4 X (1) – log3 2 = 4 – log3 2

= 4 – log2

1. b) log5 – log10.5

= log(13.5)- Log310.5 = log(135/105)

= log(27/21) = log27 – log21

= log3 – log(3 X 7)

= 3log3  – log3 -log37

= 2 – Log7

1. c) Log28 + Log33

= log223+ log33

= 2log22 + log33

2+1=3

1. d) log10 64 + log10 27

log10 26 + log1033

6 log10 2 + 3 log10 3

6 (0.3010) + 3(0.4771)

1.806 + 1.4314 = 3.2373.

EVALUATION ( USE THE DISCUSSION BOX AT THE BOTTOM TO SUBMIT YOUR ANSWER FOR DISCUSSION AND APPRAISAL)

1. Change the following index form into logarithmic form.

(a)          63= 216 (b) 33 = 1/27        (c) 92 = 81

1. Change the following logarithm form into index form.

(a)          Log88 = 1              (b) log ½¼ = 2

1. Simplify the following
2. a) Log55 + log52
3. b) ½ log48 + log432 – log42
4. c) Log381
5. Given that log 2 = 0.3010, log3 0.477

Log105 = 0.699, find the log10 6.25 + log10

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