Let’s Learn
MATH
Andrea Fauzian
, ALGEBRAIC OPERATIONS
LOGARITHM
1. Understanding logarithms
The logarithm is the inverse of the power. Suppose a is a positive number (a > 0) and g is
a positive number not equal to 1 (g > 0, g≠ 1), then:
glog a = x if only if gx = a
or it can be written:
a. to glog a = x a = gx
b. to gx = a x = glog a
2. The properties of logarithms are as follows:
a. glog (a × b) = glog a + glog b
b. glog (ab ) = log a – log b
g g
glog
c. an = n × glog a
p
g
log a
d. log a =
p
log g
glog 1
e. a=
a
log g
glog
f. a × alog b = glog b
gn
g. log a m = m g
log a
n
g
log a
h. g =a
QUESTION SETTLEMENT
1. Value of
5 3 To solve it, first convert all numbers into rank numbers
log 9∙81log 625+ 5log 125 5 81 5 3
( 6log 216− 6log36 ) =… log 9∙ log 625+ log 125
( 6log 216− 6log36 ) .
A. 625
B. 125
2 3
5
C. 25 log 9∙9 log54+ 5log53
( 6 )
D. -25 log63− 6log62
E. -125
MATH
Andrea Fauzian
, ALGEBRAIC OPERATIONS
LOGARITHM
1. Understanding logarithms
The logarithm is the inverse of the power. Suppose a is a positive number (a > 0) and g is
a positive number not equal to 1 (g > 0, g≠ 1), then:
glog a = x if only if gx = a
or it can be written:
a. to glog a = x a = gx
b. to gx = a x = glog a
2. The properties of logarithms are as follows:
a. glog (a × b) = glog a + glog b
b. glog (ab ) = log a – log b
g g
glog
c. an = n × glog a
p
g
log a
d. log a =
p
log g
glog 1
e. a=
a
log g
glog
f. a × alog b = glog b
gn
g. log a m = m g
log a
n
g
log a
h. g =a
QUESTION SETTLEMENT
1. Value of
5 3 To solve it, first convert all numbers into rank numbers
log 9∙81log 625+ 5log 125 5 81 5 3
( 6log 216− 6log36 ) =… log 9∙ log 625+ log 125
( 6log 216− 6log36 ) .
A. 625
B. 125
2 3
5
C. 25 log 9∙9 log54+ 5log53
( 6 )
D. -25 log63− 6log62
E. -125
