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
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.
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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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