8/9/2019 LawsOf Logarithms

1/15

Logarithms

Laws of logarithms

8/9/2019 LawsOf Logarithms

2/15

log a xy = loglog aa x + log x + log aa yyloglog aa x + log x + log aa yy = log a xy

Examples:log 5 (3v 5) = log 5 3 + log 5 5

log 3 5 + log 3 4 = log 3 (5 v 4)= log 3 20

P roduct Rule

8/9/2019 LawsOf Logarithms

3/15

Quotient Rule

= log 3 16 log 3 5

)y x

( loga

3

165

log ( )

)( log4

205

)y x ( log a == loglog aa x x loglog aa yy

loglogaa x x loglog

aayy ==

loglog 5 5 2020 loglog 5 5 44 === log= log

5 5 5 5

=1=1

8/9/2019 LawsOf Logarithms

4/15

log x m = m log x m log x = log x m

log x 5 3 =

4 log 9 3 = log 9 34

= log 9 81= log 9 92

= 2 log 9 9

= 2 (log 9 9 = 1)

P ower Rule

3 log 3 log x x 5 5

8/9/2019 LawsOf Logarithms

5/15

E xpress the following s single log rithms

loga 3 + log a 4 log a5

log 3 + log 4 log 5 = log (3 v 4) log 5 (Rule 1)= log 12 log 5

= (Rule 2)

= log 2.4

125al

8/9/2019 LawsOf Logarithms

6/15

8/9/2019 LawsOf Logarithms

7/15

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

(c) log2 5

8/9/2019 LawsOf Logarithms

8/15

SOLUTION

G iv en that log 2 3 = 1.58 and log 2 5 = 2 .32

(a) log2 75 = log2 [3v25]= log2 3 + log2 25 Rule 1= log2 3 + log2 52= log2 3 + 2 log2 5= 1.58 + 2(2.32)= 6.22

8/9/2019 LawsOf Logarithms

9/15

(b) log2 0.3 = log2 (3 10)= log2 3 log2 10 Rule

2 = log2 3 [log2 (5v2)]= log2 3 [log2 5 + log2 2]

= 1.58 (2.32 + 1)= 1.74(c) log2 5 = (1/2) log2 5

= (1/2)(2.32)

S olutionS olution

8/9/2019 LawsOf Logarithms

10/15

8/9/2019 LawsOf Logarithms

11/15

EX AM PLEE valuat e log5 12 .

log 5 12 =10

10

12

5

log log

1 07920 6990 . .

!

544 1 .!

Use calculato r

U se at l east 4 significant figu res

8/9/2019 LawsOf Logarithms

12/15

CHAN GE OF B AS EC hange of base-a to base b is as follows:

loga b =

For example, to change log32 2 to base-2

log32 2 =

b

b

l og bl og a

2

132log

1

5!

1

bl og a

!

52

1

2log !

8/9/2019 LawsOf Logarithms

13/15

EXERCIS E

G iv en that log 2 5 = 2.32 find th e valu e

for each of th e following without u s ingcalculato r . ( without changing to ba se- 10)(a ) log5 4

(b ) log5 2(c) log4 50

8/9/2019 LawsOf Logarithms

14/15

G iv en that log2

5 = 2 .32

(a ) log5 4

22 32

!.

5

4

2

2

loglog

!

5 2

2

22

loglog!

=0. 862 1

Change to base-2

2

25

!log

Rul e 3

8/9/2019 LawsOf Logarithms

15/15

(b ) log5 2 =

(c) log4 50

2

15log

12 32

!

.

=0.431

2

2

50

4!

l l

2

22

2 25

2

v!

log ( )log

22 2

2 5

2!

l l

5 2

12l !

21 2 5

2!

log

3 225 0 ..!

=2.82