Logarithms

5.5 Properties and Laws of Logarithms
Do Now: Solve for x .
( ) ( )
1. log 64 = x 2. log 3 =
x
4 27
x = 1/3
3. log(10 6 ) = x 4. ln(e 12
) =
x
x = 3
x = 6 x = 12

Consider some more examples…
Without evaluating log (678), we know the
expression “means” the exponent to which
10 must be raised in order to produce 678.
log (678) = x  10x = 678
If 10x = 678, what should x be in order to
produce 678?
x = log(678) because 10log(678) = 678

And with natural logarithms…
Without evaluating ln (54), we know the
expression “means” the exponent to which e
must be raised in order to produce 54.
ln (54) = x  ex = 54
If ex = 54, what should x be in order to
produce 54?
x = ln(54) because eln(54) = 54

Basic Properties of Logarithms
Common Logarithms Natural Logarithms
1. log v is defined only when v
> 0.
1. ln v is defined only when v > 0.
2. log 1 = 0 and log 10 = 1 2. ln 1 = 0 and ln e = 1
3. log 10k = k for every real
number k.
3. ln ek = k for every real number k.
4. 10logv=v for every v > 0. 4. elnv=v for every v > 0.
** NOTE: These properties hold for all bases –
not just 10 and e! **

Example 1: Solving Equations Using
Properties
Use the basic properties of logarithms to
solve each equation.
1. log(x - 3) = 5 2. ln(2x +1) = 7
10 10
log(x 3) 5
- =
x 3 10
5
- =
x = 10 5
+
3
x =
100,003
e e
ln(2x 1) 7
7
=
2x + 1 =
e
2x = e 7
-
1
7 x
e -
1
2
+
=

Laws of Logarithms
Because logarithms represent exponents, it is helpful to
review laws of exponents before exploring laws of
logarithms.
When multiplying like bases, add the exponents.
aman=am+n
When dividing like bases, subtract the
exponents.
a = -
m n
m
n
a
a

Product and Quotient Laws of Logarithms
For all v,w>0,
log(vw) = log v + log w
ln(vw) = ln v + ln w
For all v,w >
0,
( )
ln( ) ln v ln w
log log v log w
w v
w v
= -
= -

Using Product and Quotient Laws
1. Given that log 3 = 0.4771 and log 4 = 0.6021,
find log 12.
log 12 = log (3•4) = log 3 + log 4 = 1.0792
2. Given that log 40 = 1.6021 and log 8 = 0.9031,
find log 5.
log 5 = log (40 / 8) = log 40 – log 8 = 0.6990

Power Law of Logarithms
For all k and v > 0,
log vk = k log v
ln vk = k ln v
For example…
log 9 = log 32 = 2 log 3

Using the Power Law
1. Given that log 25 = 1.3979, find log .
4 25
log (25¼) = ¼ log 25 = 0.3495
2. Given that ln 22 = 3.0910, find ln Ö22.
ln (22½) = ½ ln 22 = 1.5455

Simplifying Expressions
Logarithmic expressions can be simplified using logarithmic
properties and laws.
Example 1:
Write ln(3x) + 4ln(x) – ln(3xy) as a single logarithm.
ln(3x) + 4ln(x) – ln(3xy) = ln(3x) + ln(x4) – ln(3xy)
= ln(3x•x4) – ln(3xy)
= ln(3x5) – ln(3xy)
=
( ) 3xy
=
ln
3x5 ln
( x4 ) y

Simplifying Expressions
Simplify each expression.
1. log 8x + 3 log x – log 2x2
2. ( ) (4 2 )
ln x + ln ex ¼
x
log 4x2

Logarithms

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Logarithms