Holt Algebra 2
7-5 Exponential and Logarithmic Equations and Inequalities7-5 Exponential and Logarithmic Equations and Inequalities
Holt Algebra 2
Warm Up
Lesson Presentation
Lesson Quiz
Holt Algebra 2
7-5 Exponential and Logarithmic Equations and Inequalities
Warm UpSolve.
1. log16
x =
2. logx1.331 = 3
3. log10,000 = x
64
1.1
4
32
Holt Algebra 2
7-5 Exponential and Logarithmic Equations and Inequalities
Solve exponential and logarithmic equations and equalities.
Solve problems involving exponential and logarithmic equations.
Objectives
Holt Algebra 2
7-5 Exponential and Logarithmic Equations and Inequalities
exponential equationlogarithmic equation
Vocabulary
Holt Algebra 2
7-5 Exponential and Logarithmic Equations and Inequalities
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.
Holt Algebra 2
7-5 Exponential and Logarithmic Equations and Inequalities
When you use a rounded number in a check, the result will not be exact, but it should be reasonable.
Helpful Hint
Holt Algebra 2
7-5 Exponential and Logarithmic Equations and Inequalities
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.
316 – 2x = 33x – 9 To raise a power to a power, multiply exponents.
Example 1A: Solving Exponential Equations
16 – 2x = 3x – 9 Bases are the same, so the exponents must be equal.
x = 5 Solve for x.
Holt Algebra 2
7-5 Exponential and Logarithmic Equations and Inequalities
Example 1A Continued
Check 98 – x = 27x – 3
98 – 5 275 – 3
93 272
729 729
The solution is x = 5.
Holt Algebra 2
7-5 Exponential and Logarithmic Equations and Inequalities
Solve and check.4x – 1 = 5
log 4x – 1 = log 5 5 is not a power of 4, so take the log of both sides.
(x – 1)log 4 = log 5 Apply the Power Property of Logarithms.
Example 1B: Solving Exponential Equations
Divide both sides by log 4.
Check Use a calculator.
The solution is x ≈ 2.161.
x = 1 + ≈ 2.161log5log4
x –1 = log5log4
Holt Algebra 2
7-5 Exponential and Logarithmic Equations and Inequalities
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.
Check It Out! Example 1a
2x = 3 Bases are the same, so the exponents must be equal.
x = 1.5 Solve for x.
Holt Algebra 2
7-5 Exponential and Logarithmic Equations and Inequalities
Check 32x = 27
32(1.5) 2733 27
27 27
The solution is x = 1.5.
Check It Out! Example 1a Continued
Holt Algebra 2
7-5 Exponential and Logarithmic Equations and Inequalities
Solve and check.
7–x = 21
log 7–x = log 21 21 is not a power of 7, so take the log of both sides.
(–x)log 7 = log 21 Apply the Power Property of Logarithms.
Check It Out! Example 1b
Divide both sides by log 7.
x = – ≈ –1.565log21log7
–x = log21log7
Holt Algebra 2
7-5 Exponential and Logarithmic Equations and Inequalities
Check Use a calculator.
The solution is x ≈ –1.565.
Check It Out! Example 1b Continued
Holt Algebra 2
7-5 Exponential and Logarithmic Equations and Inequalities
Solve and check.
23x = 15
log23x = log15 15 is not a power of 2, so take the log of both sides.
(3x)log 2 = log15 Apply the Power Property of Logarithms.
Check It Out! Example 1c
Divide both sides by log 2, then divide both sides by 3.
x ≈ 1.302
3x = log15 log2
Holt Algebra 2
7-5 Exponential and Logarithmic Equations and Inequalities
Check Use a calculator.
The solution is x ≈ 1.302.
Check It Out! Example 1c Continued
Holt Algebra 2
7-5 Exponential and Logarithmic Equations and Inequalities
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?
Example 2: Biology Application
Solve 2n > 106
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.
Write 1,000,000 in scientific annotation.
Take the log of both sides.log 2n > log 106
Holt Algebra 2
7-5 Exponential and Logarithmic Equations and Inequalities
Example 2 Continued
Use the Power of Logarithms.
log 106 is 6.nlog 2 > 6
nlog 2 > log 106
6log 2n > Divide both sides by log 2.
60.301n > 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.
Holt Algebra 2
7-5 Exponential and Logarithmic Equations and Inequalities
Example 2 Continued
Check In 20 hours, there will be 220 bacteria.
220 = 1,048,576 bacteria.
Holt Algebra 2
7-5 Exponential and Logarithmic Equations and Inequalities
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.
Solve 3n – 1 > 1 x 108
$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.
Write 100,000,000 in scientific annotation.
Take the log of both sides.log 3n – 1 > log 108
Check It Out! Example 2
Holt Algebra 2
7-5 Exponential and Logarithmic Equations and Inequalities
Use the Power of Logarithms.
log 108 is 8.(n – 1)log 3 > 8
(n – 1) log 3 > log 108
8log 3n – 1 > Divide both sides by log 3.
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 Continued
8log3n > + 1
Holt Algebra 2
7-5 Exponential and Logarithmic Equations and Inequalities
Check On day 18, you would receive 318 – 1 or over a million dollars.
317 = 129,140,163 cents or 1.29 million dollars.
Check It Out! Example 2
Holt Algebra 2
7-5 Exponential and Logarithmic Equations and Inequalities
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.
Holt Algebra 2
7-5 Exponential and Logarithmic Equations and Inequalities
Review the properties of logarithms from Lesson 7-4.
Remember!
Holt Algebra 2
7-5 Exponential and Logarithmic Equations and Inequalities
Solve.
Example 3A: Solving Logarithmic Equations
Use 6 as the base for both sides.
log6(2x – 1) = –1
6 log6
(2x –1) = 6–1
2x – 1 = 1 6
7 12
x =
Use inverse properties to remove 6 to the log base 6.
Simplify.
Holt Algebra 2
7-5 Exponential and Logarithmic Equations and Inequalities
Solve.
Example 3B: Solving Logarithmic Equations
Write as a quotient.
log4100 – log4(x + 1) = 1
x = 24
Use 4 as the base for both sides.
Use inverse properties on the left side.
100 x + 1log
4( ) = 1
4log4 = 41
100x + 1( )
= 4 100 x + 1
Holt Algebra 2
7-5 Exponential and Logarithmic Equations and Inequalities
Solve.
Example 3C: Solving Logarithmic Equations
Power Property of Logarithms.
log5x 4 = 8
x = 25
Definition of a logarithm.
4log5x = 8
log5x = 2
x = 52
Divide both sides by 4 to isolate log5x.
Holt Algebra 2
7-5 Exponential and Logarithmic Equations and Inequalities
Solve.
Example 3D: Solving Logarithmic Equations
Product Property of Logarithms.
log12
x + log12
(x + 1) = 1
Exponential form.
Use the inverse properties.
log12
x(x + 1) = 1
log12
x(x +1) 12 = 121
x(x + 1) = 12
Holt Algebra 2
7-5 Exponential and Logarithmic Equations and Inequalities
Example 3 Continued
Multiply and collect terms.
Factor.
Solve.
x2 + x – 12 = 0
log12
x + log12
(x +1) = 1
(x – 3)(x + 4) = 0
x – 3 = 0 or x + 4 = 0 Set each of the factors equal to zero.
x = 3 or x = –4
log12
x + log12
(x +1) = 1
log12
3 + log12
(3 + 1) 1log
123 + log
124 1
log12
12 1
The solution is x = 3.1 1
log12
( –4) + log12
(–4 +1) 1
log12
( –4) is undefined.
x
Check Check both solutions in the original equation.
Holt Algebra 2
7-5 Exponential and Logarithmic Equations and Inequalities
Solve.
3 = log 8 + 3log x
Check It Out! Example 3a
3 = log 8 + 3log x
3 = log 8 + log x3
3 = log (8x3)
103 = 10log (8x3)
1000 = 8x3
125 = x3
5 = x
Use 10 as the base for both sides.Use inverse properties on the right side.
Product Property of Logarithms.
Power Property of Logarithms.
Holt Algebra 2
7-5 Exponential and Logarithmic Equations and Inequalities
Solve.
2log x – log 4 = 0
Check It Out! Example 3b
Write as a quotient.
x = 2
Use 10 as the base for both sides.
Use inverse properties on the left side.
2log( ) = 0 x 4
2(10log ) = 100
x 4
2( ) = 1 x 4
Holt Algebra 2
7-5 Exponential and Logarithmic Equations and Inequalities
Watch out for calculated solutions that are not solutions of the original equation.
Caution
Holt Algebra 2
7-5 Exponential and Logarithmic Equations and Inequalities
Use a table and graph to solve 2x + 1 > 8192x.
Example 4A: Using Tables and Graphs to Solve Exponential and Logarithmic Equations and Inequalities
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}.
Holt Algebra 2
7-5 Exponential and Logarithmic Equations and Inequalities
log(x + 70) = 2log( )
In the table, find the x-values where Y1 equals Y2.
In the graph, find the x-value at the point of intersection.
x 3
Use a graphing calculator. Enter log(x + 70) as Y1 and 2log( ) as Y2. x
3
The solution is x = 30.
Example 4B
Holt Algebra 2
7-5 Exponential and Logarithmic Equations and Inequalities
In the table, find the x-values where Y1 is equal to Y2.
In the graph, find the x-value at the point of intersection.
Check It Out! Example 4aUse a table and graph to solve 2x = 4x – 1.
Use a graphing calculator. Enter 2x as Y1 and 4(x – 1) as Y2.
The solution is x = 2.
Holt Algebra 2
7-5 Exponential and Logarithmic Equations and Inequalities
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.
Check It Out! Example 4bUse a table and graph to solve 2x > 4x – 1.
Use a graphing calculator. Enter 2x as Y1 and 4(x – 1) as Y2.
The solution is x < 2.
Holt Algebra 2
7-5 Exponential and Logarithmic Equations and Inequalities
In the table, find the x-values where Y1 is equal to Y2.
In the graph, find the x-value at the point of intersection.
Check It Out! Example 4cUse a table and graph to solve log x2 = 6.
Use a graphing calculator. Enter log(x2) as Y1 and 6 as Y2.
The solution is x = 1000.
Holt Algebra 2
7-5 Exponential and Logarithmic Equations and Inequalities
Lesson Quiz: Part I
Solve.
1. 43x–1 = 8x+1
2. 32x–1 = 20
3. log7(5x + 3) = 3
4. log(3x + 1) – log 4 = 2
5. log4(x – 1) + log
4(3x – 1) = 2
x ≈ 1.86
x = 68
x = 133
x = 3
x = 5 3
Holt Algebra 2
7-5 Exponential and Logarithmic Equations and Inequalities
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?
7. Use a table and graph to solve the equation 23x = 33x–1.
70 min
x ≈ 0.903
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