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
Exponential & Logarithmic Equations
Problem 87f
Textbook Question
Determine whether the following statements are true and give an explanation or counterexample.
(4x+1)ln x = xln(4x+1)
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1
Step 1: Start by examining the given equation: \((4x+1)^{\ln x} = x^{\ln(4x+1)}\). We need to determine if this equation holds true for all values of \(x\) or if there is a counterexample.
Step 2: Consider taking the natural logarithm of both sides of the equation to simplify the exponents. This gives us \(\ln((4x+1)^{\ln x}) = \ln(x^{\ln(4x+1)})\).
Step 3: Apply the logarithmic identity \(\ln(a^b) = b \cdot \ln a\) to both sides. This results in \(\ln x \cdot \ln(4x+1) = \ln(4x+1) \cdot \ln x\).
Step 4: Notice that both sides of the equation are identical, \(\ln x \cdot \ln(4x+1) = \ln x \cdot \ln(4x+1)\), which suggests that the original equation is true for all \(x > 0\) where the logarithms are defined.
Step 5: Conclude that the statement is true for all \(x > 0\) where both sides of the equation are defined, as the simplification shows both sides are equal.
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