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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Solve the following logarithmic equation for x.
![The image contains the logarithmic equation:
\[
\log_5(2x - 8) = \log_5(4x - 10)
\]
This equation demonstrates the concept of logarithmic equality. Since the bases of the logarithms are the same, you can set the arguments equal to each other:
\[
2x - 8 = 4x - 10
\]
Solving for \(x\) involves these steps:
1. Move all terms involving \(x\) to one side:
\[
2x - 4x = -10 + 8
\]
2. Simplify:
\[
-2x = -2
\]
3. Divide by -2:
\[
x = 1
\]
Thus, \(x = 1\) is the solution to the equation. Note that it's important to check that the solution doesn't result in a negative or zero argument inside the logarithms, which would make them undefined. Here, \(x = 1\) satisfies the condition since both arguments become negative, implying this solution doesn't satisfy the real conditions for logarithms.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F64f3aadb-8e18-4115-8198-8731f6be87fe%2Fff2ff7dc-f608-467b-bbaf-c22ca3135c9f%2Fcpk3dec_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The image contains the logarithmic equation:
\[
\log_5(2x - 8) = \log_5(4x - 10)
\]
This equation demonstrates the concept of logarithmic equality. Since the bases of the logarithms are the same, you can set the arguments equal to each other:
\[
2x - 8 = 4x - 10
\]
Solving for \(x\) involves these steps:
1. Move all terms involving \(x\) to one side:
\[
2x - 4x = -10 + 8
\]
2. Simplify:
\[
-2x = -2
\]
3. Divide by -2:
\[
x = 1
\]
Thus, \(x = 1\) is the solution to the equation. Note that it's important to check that the solution doesn't result in a negative or zero argument inside the logarithms, which would make them undefined. Here, \(x = 1\) satisfies the condition since both arguments become negative, implying this solution doesn't satisfy the real conditions for logarithms.
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