Embed Size (px)
Citation preview
Solve exponential and logarithmic equations and equalities.
Solve problems involving exponential and logarithmic equations.
Objectives
exponential equationlogarithmic equation
Vocabulary
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.
When you use a rounded number in a check, the result will not be exact, but it should be reasonable.
Helpful Hint
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.
Example 1A Continued
Check 98 – x = 27x – 3
98 – 5 275 – 3
93 272
729 729
The solution is x = 5.
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
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.
Check 32x = 27
32(1.5) 2733 27
27 27
The solution is x = 1.5.
Check It Out! Example 1a Continued
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
Check Use a calculator.
The solution is x ≈ –1.565.
Check It Out! Example 1b Continued
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
Check Use a calculator.
The solution is x ≈ 1.302.
Check It Out! Example 1c Continued
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
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.
Example 2 Continued
Check In 20 hours, there will be 220 bacteria.
220 = 1,048,576 bacteria.
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
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
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
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.
Review the properties of logarithms from Lesson 7-4.
Remember!
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.
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
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.
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
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.
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.
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
Watch out for calculated solutions that are not solutions of the original equation.
Caution
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}.
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
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.
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.
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.
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
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
