Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![Here is the transcription of the image along with the necessary explanation for an educational website:
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### Logarithmic Equation Example
In this example, we are dealing with natural logarithms and their properties. The given equation for today’s practice is:
\[ 10) \ln (x - 6) + \ln (x + 1) = \ln (x - 15) \]
To solve this equation, we can use the properties of logarithms. Specifically:
1. **Logarithm Addition Property**: The sum of two logarithms is equivalent to the logarithm of the product of their arguments.
\[
\ln(a) + \ln(b) = \ln(ab)
\]
Applying this property to our equation:
\[
\ln((x - 6)(x + 1)) = \ln (x - 15)
\]
Since the logarithms are equal, their arguments must be equal:
\[
(x - 6)(x + 1) = x - 15
\]
Next, we'll solve for \(x\) by expanding and simplifying the equation.
Feel free to attempt the problem on your own before checking the solution!
---](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F629c87a2-ae5f-4a64-87c9-a8c68753814a%2F6cb96dcf-5929-4a08-9e63-056121785289%2F6ycdp1j_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Here is the transcription of the image along with the necessary explanation for an educational website:
---
### Logarithmic Equation Example
In this example, we are dealing with natural logarithms and their properties. The given equation for today’s practice is:
\[ 10) \ln (x - 6) + \ln (x + 1) = \ln (x - 15) \]
To solve this equation, we can use the properties of logarithms. Specifically:
1. **Logarithm Addition Property**: The sum of two logarithms is equivalent to the logarithm of the product of their arguments.
\[
\ln(a) + \ln(b) = \ln(ab)
\]
Applying this property to our equation:
\[
\ln((x - 6)(x + 1)) = \ln (x - 15)
\]
Since the logarithms are equal, their arguments must be equal:
\[
(x - 6)(x + 1) = x - 15
\]
Next, we'll solve for \(x\) by expanding and simplifying the equation.
Feel free to attempt the problem on your own before checking the solution!
---
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