Slide 1 - Cape Fear Community College

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Transcript Slide 1 - Cape Fear Community College

MAT 171 Precalculus Algebra
Trigsted - Pilot Test
Dr. Claude Moore - Cape Fear Community College
CHAPTER 5:
Exponential and Logarithmic
Functions and Equations
5.1
5.2
5.3
5.4
5.5
5.6
Exponential Functions
The Natural Exponential Function
Logarithmic Functions
Properties of Logarithms
Exponential and Logarithmic Equations
Applications of Exponential and Logarithmic Functions
OBJECTIVES
1. Solving Exponential Equations
2. Solving Logarithmic Equations
Two Logarithmic properties
If u = v, then logb u = logb v.
Logarithm property of equality
logb ur = r logb u.
Power Rule for logarithms
To Solve Exponential Equations
1. If the equation can be written in the form bu = bv,
then solve the equation u = v
2. If the equation cannot easily be written in the form bu = bv.
a. Use the logarithm property of equality to “take the log of both
sides” (typically using the base 10 or e).
b. Use the product rule of logarithms to “bring down” any exponents.
c. Solve for the given variable.
Solve each equation. Round to four decimals places.
(3x - 2) (ln e) = ln 33
Properties of Logarithms
If b > 0,
b ≠ 1, u and v represent positive numbers and r is any real number,
then
Product rule:
Quotient rule:
Power rule:
Logarithm Property of Equality
If a logarithmic equation can be written in the form
then u = v.
Furthermore, if u = v, then
Solving Logarithmic Equations
1. Determine the Domain of the variable.
2. Use properties of logarithms to combine
all
logarithms, and write as a single logarithm, if needed.
3. Eliminate the logarithm by rewriting the equation in
exponential form.
4. Solve for the given variable.
5. Check for any extraneous solutions. Verify that each
solution is in the domain of the variable.
Domain is solution to
x – 6 > 0 or x > 6
Domain is solution to
5x+6 > 0 and 3x-2 > 0
x > -5/6 and x > 2/3
Which is x > 2/3
Since 33/43 ≈ 0.767 is greater than 2/3 ≈ 0.667, x = 33/43
is the solution.
Domain is solution
to x – 2 > 0 or x > 2.
13.
ex-3 e3x+7 = 24
e(x-3)+(3x+7) = 24
e4x+4 = 24
ln e4x+4 = ln 24
4x+4 = ln 24
4x = -4 + ln 24
x = 0.25(-4 + ln 24)
x ≈ -0.2055
14. 2(ex-1)2 e3-x = 80
2(e2x-2) e3-x = 80
e(2x-2)+(3-x) = 40
ex+1 = 40
ln ex+1 = ln 40
x+1 = ln 40
x = -1 + ln 40
x ≈ 2.6889
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