12.4 Properties of Logarithmic Functions
Topics:
Logarithms of Products
Logarithms of Powers
Logarithms of Quotients
Using the Properties Together
Logarithms of Products
The Product Rule for Logarithms
For any positive numbers M, N and a ( a ≠ 1) ,
log a MN = log a M + log a N .
(The logarithm of a product is the sum of the logarithms of the factors)
A Proof of the Product Rule:
Let log a M = x and log a N = y . Converting to exponential equations, we have
a x = M and a y = N .
Now we multiply the last two equations, to obtain
MN = a x a y , or MN = a x + y
Converting back to a logarithmic equation, we get
log a MN = log a a x + y = x + y = log a M + log a N .
Example of the product rule
(2) Express as a sum of logarithms: log 2 (16 ⋅ 32 )
Solution: log 2 (16 ⋅ 32 ) = log 2 16 + log 2 32
(6) Express as a sum of logarithms: log t 3ab
Solution: log t 3ab = log t 3 + log t a + log t b
(8) Express as a single logarithm: log b 65 + log b 2
Solution: log b 65 + log b 2 = log b ( 65 ⋅ 2 ) = log b (130 )
(10) Express as a single logarithm: log t H + log t M
Solution: log t H + log t M = log t ( HM )
Logarithms of Powers
The Power Rule for Logarithms
For any positive numbers M and a, ( a ≠ 1) , and any real number p,
log a M p = p log a M
(The logarithm of a power of M is the exponent times the logarithm of M.)
A Proof of the Power Rule:
let x = log a M . We then write the equivalent exponential equation a x = M . Raising
both sides to the pth power, we get
(a )
x
p
= M p , or a xp = M p
Converting back to a logarithmic equation gives us
log a M p = xp
But x = log a M , so substituting, we have
log a M p = ( log a M ) p = p log a M
Homework examples:
Express as a product:
(12) log b t 5
Solution: log b t 5 = 5log b t
(16) log c M −5
Solution: log c M −5 = −5log c M
Logarithms of Quotients
The Quotient Rule for Logarithms
For any positive numbers M, N, and a, ( a ≠ 1) ,
M
= log a M − log a N .
N
(The logarithm of a quotient is the logarithm of the dividend minus the logarithm of the
divisor.)
log a
A Proof of the Quotient Rule:
M
log a
= log a MN −1 =
N
log a M + log a N −1 = log a M + (−1) log a N =
log a M − log a N
Homework examples:
Express as a difference of logarithms:
23
(18) log 3
9
23
Solution: log 3
= log 3 23 − log 3 9
9
(20) log a
y
x
y
= log a y − log a x
x
Express as a single logarithm (# 21 – 26)
Solution: log a
(22) log b 42 − log b 7
Solution: log b 42 − log b 7 = log b
(26) log b 5 − log b 13
Solution: log b 5 − log b 13 = log b
42
= log b 6
7
5
13
Using the Properties Together
The Logarithm of the Base to a Power
For any base a,
log a a k = k
(The logarithm, base a, of a to a power is the power.)
SUMMARY OF PROPERTIES
For any positive numbers M, N and a, ( a ≠ 1) :
log a MN = log a M + log a N ;
M
log a
= log a M − log a N ;
N
log a M p = p log a M ;
log a a k = k
Caution! Keep in mind that in general,
log a ( M + N ) ≠ log a M + log a N ;
log a MN ≠ ( log a M )( log a N )
log a ( M − N ) ≠ log a M − log a N ;
log a
M log a M
≠
N log a N
Homework examples:
#27 – 38 Express in terms of the individual logarithms of w, x, y, and z.
(28) log a xy 4 z 3
Solution: log a xy 4 z 3 = log a x + log a y 4 + log a z 3 = log a x + 4 log a y + 3log a z
(30) log b
x2 y5
w4 z 7
x2 y5
= log b x 2 y 5 − log b w4 z 7 = log b x 2 + log b y 5 − ( log b w4 + log b z 7 ) =
4 7
wz
2 log b x + 5log b y − 4 log b w − 7 log b z
Solution: log b
(36) log c
3
x4
y3 z 2
Solution: log c
3
 x4 
x4
log
=
c 3 2 
y3 z 2
y z 
(
1
3
 x4  1
1
= log c  3 2  = log c x 4 −  log c y 3 z 2  =
3
y z  3
(
)
1
1
log c x 4 − log c y 3 + log c z 2  = ( 4 log c x − 3log c y − 2 log c z ) =
3
3
4
2
log c x − log c y − log c z
3
3
#39 – 46 Express as a single logarithm and, if possible, simplify:
1
(40) 2 log b m + log b n
2
1
1
Solution: 2 log b m + log b n = log b m 2 + log b n 2 = log b m 2 n
2
a
− log a ax
(42) log a
x
a
a
a
a
Solution: log a
− log a ax = log a
= log a
= log a
2
x
x ax
x a
ax
Rationalizing the denominator
So the solution is log a
a
x a
=
a
x a
⋅
a a a a a
a
=
=
=
2
ax
x
a x a
a
a
− log a ax = log a
x
x
)