Upload
sforazzura-suhaimi
View
219
Download
0
Embed Size (px)
Citation preview
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