Download Properties of Logarithms

Properties of Logarithms
Since the exponential and logarithmic functions with base a are inverse functions, the
Properties of Exponents give rise to the Properties of Logarithms.
Properties of Exponents: (a and b are real numbers, m and n are integers)
(ab) n = a n b n
b0 = 1
b1 = b
am
= a m−n
n
a
an
⎛a⎞
⎜ ⎟ = n
b
⎝b⎠
⎛a⎞
⎜ ⎟
⎝b⎠
a −n b m
=
b −m a n
(a m ) n = a mn
a m a n = a m+ n
Properties of Logarithms:
n
−n
⎛b⎞
=⎜ ⎟
⎝a⎠
n
Let b be a positive number, with b ≠ 1. Let M > 0,
N > 0, p and x be any real numbers.
log b 1 = 0
log b b = 1
log b bx = x
b log b x = x ; x > 0
log b MN = log b M + log b N
⎛M ⎞
log b ⎜ ⎟ = log b M – log b N
⎝N⎠
The logarithm of a product of numbers
is the sum of the logarithms of the
numbers.
The logarithm of a quotient of numbers
is the difference of the logarithms of the
numbers.
Log b Mp = p log b M
The logarithm of a power of a number
is the exponent times the logarithm of
the number.
Example 1:
Use the Properties of Logarithms to rewrite the expression in a form with
no logarithm of a product, quotient, or power.
⎛ 5 x3 ⎞
(a) log 3 ⎜ 2 ⎟
⎝ y ⎠
(b) ln 5 x 2 ( z + 1)
Solution (a):
⎛ 5 x3 ⎞
log 3 ⎜ 2 ⎟ = log 3 ( 5 x3 ) − log 3 y 2
⎝ y ⎠
= log 3 5 + log 3 x 3 − log 3 y 2
= log 3 5 + 3log 3 x − 2 log 3 y
Law 2
Law 1
Law 3
Solution (b):
ln 5 x 2 ( z + 1) = ln ( x 2 ( z + 1) )
Example 2:
1
5
= 15 ln ( x 2 ( z + 1) )
Law 3
= 15 ⎡⎣ ln x 2 + ln ( z + 1) ⎤⎦
Law 1
= 15 ln x 2 + 15 ln ( z + 1)
Distribution
= 15 ( 2 ln x ) + 15 ln ( z + 1)
Law 3
= 52 ln x + 15 ln ( z + 1)
Multiplication
Rewrite the expression as a single logarithm.
(a) 3log 4 5 + log 4 10 − log 4 7
(b) log x − 2 log y + 12 log ( 9 z 2 )
Example 2 (Continued):
Solution (a):
3log 4 5 + log 4 10 − log 4 7 = log 4 53 + log 4 10 − log 4 7
Law 3
= log 4 125 + log 4 10 − log 4 7
Because 53 = 125
= log 4 (125 ⋅10 ) − log 4 7
Law 1
= log 4 (1250 ) − log 4 7
Simplification
⎛ 1250 ⎞
= log 4 ⎜
⎟
⎝ 7 ⎠
Law 2
Solution (b):
log x − 2 log y + 12 log ( 9 z 2 ) = log x − log y 2 + log ( 9 z 2 )
1
2
Law 3
= log x − log y 2 + log ( 3z )
Simplification
= log
Law 2
( ) + log (3z )
= log ⎡( ) ( 3 z ) ⎤
⎣
⎦
= log ( )
x
y2
x
y2
Law 1
3 zx
y2
Example 3:
Simplification
Evaluate the expression
(a) log 2 5 16
(b) log 3 189 − log 3 7
Solution (a):
1
log 2 5 16 = log 2 16 5
= 15 log 2 16
Law 3
= 15 log 2 24
Because 16 = 24
= 15 ( 4 )
Property 3 of Logarithms
=
Multiplication
4
5
Example 3 (Continued):
Solution (b):
log 3 189 − log 3 7 = log 3 189
7
Law 2
= log 3 27
Simplification
= log 3 3
Because 27 = 33
=3
Property 3 of Logarithms
3
Change of Base:
A calculator can be used to approximate the values of common logarithms (base 10) or
natural logarithms (base e). However, sometimes we need to use logarithms to other
bases. The following rule is used to convert logarithms from one base to another.
Change of Base Formula:
log b x =
Example 4:
log a x
log a b
Use the Change of Base Formula and a calculator to evaluate the
logarithm, correct to six decimal places. Use either natural or common
logarithms.
(a) log 6 17
(b) log 5 2.33
Solution (a):
The Change of Base Formula says
log b x =
log a x
.
log a b
Thus, if we let the new base a = 10
log 6 17 =
Example 4 (Continued):
log10 17
≈ 1.581246
log10 6
Solution (b):
Again we will use the Change of Base Formula. This time we will let the
new base be a = e.
log 5 2.33 =
Example 5:
Simplify:
ln 2.33
≈ 0.525568
ln 5
( log8 12 )( log12 7 )
Solution:
Using the Change of Base Formula with the new base a = 10 :
log8 12 =
log12
log 8
and
log12 7 =
log 7
log12
Thus,
⎛ log12 ⎞ ⎛ log 7 ⎞
⎟⎜
⎟
⎝ log 8 ⎠ ⎝ log12 ⎠
( log8 12 )( log12 7 ) = ⎜
log 7
log 8
= log 8 7
=
Simplification
Change of Base Formula