Table of contents
- 0. Functions7h 52m
- Introduction to Functions16m
- Piecewise Functions10m
- Properties of Functions9m
- Common Functions1h 8m
- Transformations5m
- Combining Functions27m
- Exponent rules32m
- Exponential Functions28m
- Logarithmic Functions24m
- Properties of Logarithms34m
- Exponential & Logarithmic Equations35m
- Introduction to Trigonometric Functions38m
- Graphs of Trigonometric Functions44m
- Trigonometric Identities47m
- Inverse Trigonometric Functions48m
- 1. Limits and Continuity2h 2m
- 2. Intro to Derivatives1h 33m
- 3. Techniques of Differentiation3h 18m
- 4. Applications of Derivatives2h 38m
- 5. Graphical Applications of Derivatives6h 2m
- 6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
- 7. Antiderivatives & Indefinite Integrals1h 26m
0. Functions
Properties of Logarithms
5:27 minutes
Problem 3.9.87.b
Textbook Question
Explain why or why not. Determine whether the following statements are true and give an explanation or counterexample.
b. ln(x + 1) + ln(x − 1) = ln(x² − 1), for all x.
Verified step by step guidance
1
Step 1: Recall the logarithmic property that states \( \ln(a) + \ln(b) = \ln(ab) \). This property allows us to combine the logarithms on the left-hand side of the equation.
Step 2: Apply the property from Step 1 to the left-hand side: \( \ln(x + 1) + \ln(x - 1) = \ln((x + 1)(x - 1)) \).
Step 3: Simplify the expression \((x + 1)(x - 1)\) using the difference of squares formula: \((x + 1)(x - 1) = x^2 - 1\).
Step 4: Substitute the simplified expression from Step 3 back into the equation: \( \ln((x + 1)(x - 1)) = \ln(x^2 - 1) \).
Step 5: Conclude that the original statement \( \ln(x + 1) + \ln(x - 1) = \ln(x^2 - 1) \) is true for all \( x \) such that \( x > 1 \) or \( x < -1 \), because the domain of the logarithmic function requires that the arguments \( x + 1 \) and \( x - 1 \) are positive.
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