Download logarithmic equation

Warm Up
Solve.
1. log16x =
3
2
2. logx8 = 3
3. log10,000 = x
64
1.1
4
7-4 Logarithmic
Equations and
Inequalities
TEXTBOOK PAGE 478
An exponential equation is an equation containing
one or more expressions that have a variable as an
exponent. To solve exponential equations:
• Try writing them so that the
bases are all the same.
• Take the logarithm of both
sides.
A logarithmic equation is an equation with a
logarithmic expression that contains a variable.
You can solve logarithmic equations by using
the properties of logarithms.
Solving by Rewriting as an Exponential
Solve log4(x+3) = 2
42 = x+3
16 = x+3
13 = x
Example 1
Solve for x: log 6 x  2
Solution:
Let’s rewrite the problem
in exponential form.
6 x
2
We’re finished !
Example 2
1
Solve for y: log 5
y
25
Solution:
Rewrite the problem in
exponential form.
1
5 
25
y
2
5 5
y
y  2
 1

Since   5 2 
25

Example 3
Evaluate log3 27.
Solution:
Try setting this up like this:
log3 27  y Now rewrite in exponential form.
3  27
y
3 3
y
y3
3
Example 4
2
Evaluate: log7 7
Solution:
log7 7  y
2
7 7
y2
y
2
First, we write the problem with a variable.
Now take it out of the logarithmic form
and write it in exponential form.
Example 5
Evaluate: 4
log 4 16
Solution:
4
log 4 16
y
First, we write the problem with a variable.
log4 y  log4 16
Now take it out of the exponential form
and write it in logarithmic form.
Just like 2  8 converts to log2 8  3
3
y  16
Finally, we want to take a look at
the Property of Equality for
Logarithmic Functions.
Suppose b  0 and b  1.
Then logb x1  log b x 2 if and only if x1  x 2
Basically, with logarithmic functions,
if the bases match on both sides of the equal
sign , then simply set the arguments equal.
Example 1
Solve:
log3 (4x 10)  log3 (x 1)
Solution:
Since the bases are both ‘3’ we simply set
the arguments equal.
4x 10  x 1
3x 10  1
3x   9
x 3
Caution
Watch out for calculated solutions that are not
solutions of the original equation.
Example 2
Solve:
log8 (x 14)  log8 (5x)
2
Solution:
Since the bases are both ‘8’ we simply set the arguments equal.
2
x 14  5x
2
x  5x 14  0
(x  7)(x  2)  0
Factor
(x  7)  0 or (x  2)  0
x  7 or x  2 continued on the next page
Example 2
continued
Solve:
log8 (x 14)  log8 (5x)
2
Solution:
x  7 or x  2
It appears that we have 2 solutions here.
If we take a closer look at the definition of a
logarithm however, we will see that not only
must we use positive bases, but also we
see that the arguments must be positive as
well. Therefore -2 is not a solution.
Let’s end this lesson by taking a closer look
at this.
Use a table and graph to solve 2x = 4x – 1.
Use a graphing calculator. Enter 2x as Y1 and
4(x – 1) as Y2.
In the table, find the x-values
where Y1 is equal to Y2.
The solution is x = 2.
In the graph, find the x-value
at the point of intersection.
Use a table and graph to solve 2x > 4x – 1.
Use a graphing calculator. Enter 2x as Y1 and
4(x – 1) as Y2.
In the table, find the x-values
where Y1 is greater than Y2.
The solution is x < 2.
In the graph, find the x-value
at the point of intersection.
Use a table and graph to solve log x2 = 6.
Use a graphing calculator. Enter log(x2) as Y1
and 6 as Y2.
In the table, find the x-values
where Y1 is equal to Y2.
The solution is x = 1000.
In the graph, find the x-value
at the point of intersection.