Transcript section 11.4

11.4 Notes
Solving logarithmic equations
11.4 Notes
In this unit of study, you will learn several
methods for solving several types of
logarithmic equations.
In a previous lesson, you learned how to solve
logarithmic equations that have only one
logarithm term.
In this lesson, you will learn how to solve
logarithmic equations that have more than one
logarithm term.
11.4 Notes
The objective when solving logarithmic equations
with more than one term in them is to try to get one
logarithm term, with matching bases, on each side
of the equal sign: log  8 x  20   log  x  6 
7
7
Once this has been achieved, the logarithm terms
can be ignored, the arguments of the terms set
equal to each other and solved: 8 x  2 0  x  6
x  2
11.4 Notes
Before answers to logarithmic equations can be
declared the solutions, it must be determined
whether they satisfy the original equation.
Because the range of y  b is  0,  , the
domain of log b y  x is  0,   . Therefore,
when answers to logarithmic equations are
substituted back into the original equation, the
arguments of the equation must be in the interval
x
 0,   .
11.4 Notes
log 7  8 x  20   log 7  x  6 
8 x  20  x  6
x  2
When -2 is substituted back into the original
equation, the argument of the logarithm terms on
each side of the equal sign is +4; therefore, -2 is the
solution to the equation.
11.4 Notes – Example 1
log 3  9 x  1   log 3  4 x  16 
9 x  1  4 x  16
x  3
When -3 is substituted back into the original
equation, the argument of the logarithm term on the
left side of the equal sign is -28; therefore, -3 is not
a solution. This equation has no solution.
11.4 Notes – Practice 1
lo g 1 2  x  9   lo g 1 2  3 x  1 3 
x  9  3 x  13
x 2
When 2 is substituted back into the original
equation, the argument of the logarithm term on the
left side of the equal sign is -7; therefore, 2 is not a
solution. This equation has no solution.
11.4 Notes
The objective when solving logarithmic equations
with more than one term in them is to try to get one
logarithm term, with matching bases, on each side
of the equal sign.
log 2 3  log 2 7  log 2 x
When a logarithmic equation contains more than
one logarithm term on either side of the equation, it
is necessary to get only one logarithm term, with
matching bases, on each side of the equal sign.
Doing so requires the use of properties of
logarithms.
11.4 Notes
Product property: log m n  log m  log n
b
b
b
Quotient property:
Power property:
log b
m
n
 log b m  log b n
log b m  n log b m
n
11.4 Notes – Example 2
log 2 3  log 2 7  log 2 x
log 2 3 7  log 2 x
3 7  x
x  21
11.4 Notes – Practice 2
lo g 7  lo g  x  2   lo g 6 x
lo g  7  x  2    lo g 6 x
7  x  2  6x
7 x  14  6 x
x  14
11.4 Notes – Example 3
log x 
1
log 81
2
1
lo g x  lo g 8 1 2
1
x  81 2
x
81
x  9
x9
11.4 Notes – Example 4
lo g 8  x  3   lo g 8 x  lo g 8 4
 x3
log 8 
  log 8 4
 x 
x3
4
x
x  3  4x
x 1
11.4 Notes – Practice 3
3 log 5 x  log 5 4  log 5 16
log 5 x  log 5 4  log 5 16
3
x
log 5
 log 5 16
4
3
x
 16
4
3
x  64
3
x
3
x 4
64
11.4 Notes – Example 5
lo g x  lo g  x  2 1   2
lo g  x  x  2 1    2 log 10
log  x  21 x   log 10
2
x  21 x  10
2
2
2
x  21 x  100
2
x  21 x  100  0
x   25,
2
 x  4   x  25   0
x40
x  25  0
x4
4
11.4 Notes – Practice 4
lo g 2  x  2   1  lo g 2  x  2 
lo g 2  x  2   1 log 2 2  lo g 2  x  2 
log 2  x  2   log 2 2  log 2  x  2 
x2
log 2 
  log 2  x  2 
 2 
x2
 x2
x6
2
x  2  2 x  2
x  2  2x  4