Download Properties of Laws and Logarithms

5.5
Properties and Laws
of Logarithms
Objective:
1. Use properties and laws of logarithms to
simplify and evaluate expressions.
Basic Properties of Logarithms
Example #1
Solving with the Basic Properties
 Use the basic properties of logarithms to solve the equations
below.
A.
log( x  3)  1
log  x  3 
1
10
 10
x  3  10
x  13
A power with base 10 is the
inverse of the common log.
Since they are inverses they
“undo” each other and we are
left with the resulting
expression.
Don’t forget when something is
done to one side of an equation,
it must also be done to the
other.
Example #1
Solving with the Basic Properties
 Use the basic properties of logarithms to solve the equations
below.
B.
 3  ln( 2 x)  2
ln 2 x   1
ln 2 x 
e
e
2x  e
e
x
2
1
Just like common logs,
natural logs have an inverse
as well, but with powers of
base e.
Product Law of Logarithms
Example #2
 Use the Product Law of Logarithms to evaluate each
logarithm.
A.
Given that log 3  0.4771 and
log 4  0.6021, find log 12.
log 12
 log 3  4
 log 3  log 4
 0.4771  0.6021
 1.0792
Example #2
 Use the Product Law of Logarithms to evaluate each
logarithm.
B.
Given that log 3  0.4771 and
log 4  0.6021, find log 27.
log 27
 log 3  3  3
 log 3  log 3  log 3
 3 log 3
 3(0.4771)
 1.4313
Example #2
 Use the Product Law of Logarithms to evaluate each
logarithm.
C.
Given that ln 3  1.0986 and
ln 6  1.7918, find ln 54.
ln 54
 ln 3  3  6 
 ln 3  ln 3  ln 6
 2 ln 3  ln 6
 2(1.0986 )  1.7918
 3.989
Quotient Law of Logarithms
Example #3
 Use the Quotient Law of Logarithms to evaluate each
logarithm.
A.
Given that log 40  1.6021 and
log 8  0.9031, find log 5.
log 5
 40 
 log  
 8 
 log 40  log 8
 1.6021  0.9031
 0.699
Example #3
 Use the Quotient Law of Logarithms to evaluate each
logarithm.
B.
Given that ln 32  3.4657 and
ln 8  2.0794, find ln 4.
ln 4
 32 
 ln  
 8
 ln 32  ln 8
 3.4657  2.0794
 1.3863
Power Law of Logarithms
Example #4
 Use the Power Law of Logarithms to evaluate each
logarithm.
A.
Given that log 25  1.3979,
find log 4 25.
log 4 25
 log 25 
1
4
 14  log 25
 14  1.3979
 0.349475
Example #4
 Use the Power Law of Logarithms to evaluate each
logarithm.
B.
Given that ln 22  3.0910,
find ln 22 .
ln 22
 ln 22
1
2
 12  ln 22
 12  3.0910
 1.5455
Example #5
Simplifying Expressions
 Write each expression as a single logarithm.
A.
log 8x  3 log x  log 2 x
When condensing logarithms into a
single logarithm, you must use the
properties of logarithms backwards.
Always bring coefficients up as
exponents first, then combine
addition and subtraction by
multiplication and division.
2
 log 8 x  log x 3  log 2 x 2
 8x  x3 

 log 
2 
2
x


 8x 4 
 log  2 
 2x 
 log 4 x 2
Example #5
Simplifying Expressions
 Write each expression as a single logarithm.
B.
4 ln( ex )  4
You cannot change the
subtraction into division
because they both must
be logarithms.
2
   4 ln e
 ln e x   ln e
 ln ex
2 4
4 8
This implies that
 e 4 x8 
 ln  4 
 e 
4  ln e  4
 ln x 8
4  1  4 & ln e  1
4
Example #5
Simplifying Expressions
 Write each expression as a single logarithm.
B.
4 ln( ex )  4
2

Alternatively, this problem can also
be solve by expanding the
expression first before condensing.

 4 ln e  ln x 2  4
 4 ln e  4 ln x 2  4
 4  4 ln x 2  4
 4 ln x 2
 
 ln x
2 4
 ln x 8
Example #5
Simplifying Expressions
 Write each expression as a single logarithm.
C.
 2  log 2000 x  log 10 y
Like the previous problem,
this one can also be solved by
introducing a logarithm first,
but this time use the fact
that:
log 10  1
 2 log 10  log 2000 x  log10 y
  log 10 2  log 2000 x  log 10 y
 2000 x 

 log 2

 10  10 y 
 2000 x 

 log
 1000 y 
 2x 
 log 
 y 
Example #5
Simplifying Expressions
 Write each expression as a single logarithm.
C.
 2  log 2000 x  log 10 y
This problem could have been solved by expanding first as well:
 2  log 20 x  100   log 10 y
 2  log 20 x  log 100  log 10 y
 2  log 20 x  2  log 10 y
 log 20 x  log 10 y
 20 x 

 log 
 10 y 
 2x 
 log  
 y 
Example #6
Simplifying by Substitution
 Write the expression in terms of u and v given
u  ln x and v  ln y
A.
4
3
ln( x y )
Expand the expression first, the
substitute.
 ln x  ln y
 4 ln x  3 ln y
 4u  3v
4
3
Example #6
Simplifying by Substitution
 Write the expression in terms of u and v given
u  ln x and v  ln y
B.
3 x 
ln  2   ln 3 x  ln y 2
 y 
1
2
3
 ln x  ln y


1
ln
x

2
ln
3
1
u

2
v
3
y