Transcript + x - s3.amazonaws.com

Objectives
Solve exponential and logarithmic
equations and equalities.
Solve problems involving exponential
and logarithmic equations.
Vocabulary
exponential equation
logarithmic equation
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.
Helpful Hint
When you use a rounded number in a check, the
result will not be exact, but it should be
reasonable.
Example 1A: Solving Exponential Equations
Solve and check.
98 – x = 27x – 3
(32)8 – x = (33)x – 3 Rewrite each side with the same
base; 9 and 27 are powers of 3.
To raise a power to a power,
316 – 2x = 33x – 9
multiply exponents.
16 – 2x = 3x – 9 Bases are the same, so the
exponents must be equal.
x=5
Solve for x.
Example 1A Continued
Check
98 – x = 27x – 3
98 – 5 275 – 3
93 272
729 729 
The solution is x = 5.
Example 1B: Solving Exponential Equations
Solve and check.
4x – 1 = 5
5 is not a power of 4, so take the
log 4x – 1 = log 5
log of both sides.
Apply the Power Property of
(x – 1)log 4 = log 5
Logarithms.
log5
x –1 = log4
Divide both sides by log 4.
log5
x = 1 + log4 ≈ 2.161
Check Use a calculator.
The solution is x ≈ 2.161.
Check It Out! Example 1a
Solve and check.
32x = 27
(3)2x = (3)3
Rewrite each side with the same
base; 3 and 27 are powers of 3.
32x = 33
To raise a power to a power,
multiply exponents.
2x = 3
Bases are the same, so the
exponents must be equal.
x = 1.5
Solve for x.
Check It Out! Example 1a Continued
Check
32x =
27
32(1.5) 27
33 27
27 27 
The solution is x = 1.5.
Check It Out! Example 1b
Solve and check.
7–x = 21
log 7–x = log 21
(–x)log 7 = log 21
log21
–x = log7
log21
21 is not a power of 7, so take the
log of both sides.
Apply the Power Property of
Logarithms.
Divide both sides by log 7.
x = – log7 ≈ –1.565
Check It Out! Example 1b Continued
Check Use a calculator.
The solution is x ≈ –1.565.
Check It Out! Example 1c
Solve and check.
23x = 15
log23x = log15
(3x)log 2 = log15
log15
3x = log2
x ≈ 1.302
15 is not a power of 2, so take the
log of both sides.
Apply the Power Property of
Logarithms.
Divide both sides by log 2,
then divide both sides by 3.
Check It Out! Example 1c Continued
Check Use a calculator.
The solution is x ≈ 1.302.
Example 2: Biology Application
Suppose a bacteria culture doubles in size
every hour. How many hours will it take for
the number of bacteria to exceed 1,000,000?
At hour 0, there is one bacterium, or 20 bacteria. At
hour one, there are two bacteria, or 21 bacteria,
and so on. So, at hour n there will be 2n bacteria.
Solve 2n > 106
Write 1,000,000 in scientific
annotation.
log 2n > log 106
Take the log of both sides.
Example 2 Continued
nlog 2 > log 106
Use the Power of Logarithms.
nlog 2 > 6
log 106 is 6.
6
n > log 2
Divide both sides by log 2.
6
n > 0.301
Evaluate by using a calculator.
n > ≈ 19.94
Round up to the next whole number.
It will take about 20 hours for the number of bacteria
to exceed 1,000,000.
Example 2 Continued
Check In 20 hours, there will be 220 bacteria.
220 = 1,048,576 bacteria.
Check It Out! Example 2
You receive one penny on the first day, and
then triple that (3 cents) on the second day,
and so on for a month. On what day would
you receive a least a million dollars.
$1,000,000 is 100,000,000 cents. On day 1, you
would receive 1 cent or 30 cents. On day 2, you
would receive 3 cents or 31 cents, and so on. So, on
day n you would receive 3n–1 cents.
Solve 3n – 1 > 1 x 108
Write 100,000,000 in scientific
annotation.
log 3n – 1 > log 108
Take the log of both sides.
Check It Out! Example 2 Continued
(n – 1) log 3 > log 108 Use the Power of Logarithms.
(n – 1)log 3 > 8
8
n – 1 > log 3
8
log 108 is 8.
Divide both sides by log 3.
n > log3 + 1
Evaluate by using a calculator.
n > ≈ 17.8
Round up to the next whole number.
Beginning on day 18, you would receive more than
a million dollars.
Check It Out! Example 2
Check On day 18, you would receive 318 – 1 or
over a million dollars.
317 = 129,140,163 cents or 1.29 million dollars.
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.
Remember!
Review the properties of logarithms from Lesson
7-4.
Example 3A: Solving Logarithmic Equations
Solve.
log6(2x – 1) = –1
6
log (2x –1)
6
= 6–1
2x – 1 = 1
6
7
x = 12
Use 6 as the base for both sides.
Use inverse properties to remove
6 to the log base 6.
Simplify.
Example 3B: Solving Logarithmic Equations
Solve.
log4100 – log4(x + 1) = 1
100
log4(x + 1 ) = 1
log4( x + 1 )
100
4
= 41
100
=4
x+1
x = 24
Write as a quotient.
Use 4 as the base for both sides.
Use inverse properties on the
left side.
Example 3C: Solving Logarithmic Equations
Solve.
log5x 4 = 8
4log5x = 8
log5x = 2
x = 52
x = 25
Power Property of Logarithms.
Divide both sides by 4 to isolate log5x.
Definition of a logarithm.
Example 3D: Solving Logarithmic Equations
Solve.
log12x + log12(x + 1) = 1
log12 x(x + 1) = 1
12 log
x(x +1)
12
= 121
x(x + 1) = 12
Product Property of Logarithms.
Exponential form.
Use the inverse properties.
Example 3 Continued
x2 + x – 12 = 0
(x – 3)(x + 4) = 0
Multiply and collect terms.
Factor.
x – 3 = 0 or x + 4 = 0 Set each of the factors equal to zero.
x = 3 or x = –4
Solve.
Check Check both solutions in the original equation.
log12x + log12(x +1) = 1
log12x + log12(x +1) = 1
log123 + log12(3 + 1)
log123 + log124
log1212
1
1
1
1
1
log12( –4) + log12(–4 +1) 1 x
log12( –4) is undefined.
The solution is x = 3.
Check It Out! Example 3a
Solve.
3 = log 8 + 3log x
3 = log 8 + 3log x
3 = log 8 + log x3
3 = log (8x3)
103 = 10log (8x3)
1000 = 8x3
125 = x3
5=x
Power Property of Logarithms.
Product Property of Logarithms.
Use 10 as the base for both sides.
Use inverse properties on the
right side.
Check It Out! Example 3b
Solve.
2log x – log 4 = 0
x
2log( 4
log
2(10
)=0
Write as a quotient.
x
4
Use 10 as the base for both sides.
) = 100
2( x ) = 1
4
x=2
Use inverse properties on the
left side.
Caution
Watch out for calculated solutions that are not
solutions of the original equation.
Example 4A: Using Tables and Graphs to Solve
Exponential and Logarithmic Equations and
Inequalities
Use a table and graph to solve 2x + 1 > 8192x.
Use a graphing calculator. Enter 2^(x + 1) as Y1
and 8192x as Y2.
In the table, find the x-values
where Y1 is greater than Y2.
In the graph, find the x-value
at the point of intersection.
The solution set is {x | x > 16}.
Example 4B
log(x + 70) = 2log( x )
3
Use a graphing calculator. Enter log(x + 70) as Y1
and 2log( x ) as Y2.
3
In the table, find the x-values
where Y1 equals Y2.
The solution is x = 30.
In the graph, find the x-value
at the point of intersection.
Check It Out! Example 4a
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.
Check It Out! Example 4b
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.
Check It Out! Example 4c
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.
Lesson Quiz: Part I
Solve.
1. 43x–1 = 8x+1
x= 5
3
2. 32x–1 = 20
x ≈ 1.86
3. log7(5x + 3) = 3
x = 68
4. log(3x + 1) – log 4 = 2
x = 133
5. log4(x – 1) + log4(3x – 1) = 2
x=3
Lesson Quiz: Part II
6. A single cell divides every 5 minutes. How long
will it take for one cell to become more than
10,000 cells? 70 min
7. Use a table and graph to solve the equation
23x = 33x–1.
x ≈ 0.903