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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Equations and Inequations
Equations and inequalities describe the relationship between two mathematical expressions.
Linear Functions
A linear function can just be a constant, or it can be the constant multiplied with the variable like x or y. If the variables are of the form, x2, x1/2 or y2 it is not linear. The exponent over the variables should always be 1.
Question
![**Mathematical Problem: Logarithmic Equation**
Problem Statement:
Solve the equation:
\[ \log_2(x+1) - \log_2(x-4) = 3 \]
---
**Explanation:**
This problem requires solving a logarithmic equation involving base-2 logarithms. The equation can be simplified and solved using logarithmic properties, such as the quotient rule of logarithms:
\[ \log_b(a) - \log_b(c) = \log_b\left(\frac{a}{c}\right) \]
Applying this property, the equation becomes:
\[ \log_2\left(\frac{x+1}{x-4}\right) = 3 \]
To solve for \(x\), express the equation in exponential form:
\[ \frac{x+1}{x-4} = 2^3 \]
Calculate \(2^3\):
\[ \frac{x+1}{x-4} = 8 \]
Now, solve for \(x\) by cross-multiplying:
\[ x+1 = 8(x-4) \]
This leads to:
\[ x+1 = 8x - 32 \]
Rearrange the terms to isolate \(x\):
\[ 1 + 32 = 8x - x \]
\[ 33 = 7x \]
\[ x = \frac{33}{7} \]
Thus, the solution to the equation is \( x = \frac{33}{7} \).
**Important:**
Check the domain of the original logarithms to ensure the solution is valid. The expressions \((x+1)\) and \((x-4)\) must be positive:
1. \( x + 1 > 0 \Rightarrow x > -1 \)
2. \( x - 4 > 0 \Rightarrow x > 4 \)
Given \( x = \frac{33}{7} \approx 4.71 \), the solution meets both conditions. Therefore, the solution is valid within the domain of the logarithmic functions.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe0a10977-fbe8-44dd-8710-8a9db5a9a487%2F046c5731-d87e-443a-9e3f-cdf1c3ee61a8%2Fbtwfv2b_processed.png&w=3840&q=75)
Transcribed Image Text:**Mathematical Problem: Logarithmic Equation**
Problem Statement:
Solve the equation:
\[ \log_2(x+1) - \log_2(x-4) = 3 \]
---
**Explanation:**
This problem requires solving a logarithmic equation involving base-2 logarithms. The equation can be simplified and solved using logarithmic properties, such as the quotient rule of logarithms:
\[ \log_b(a) - \log_b(c) = \log_b\left(\frac{a}{c}\right) \]
Applying this property, the equation becomes:
\[ \log_2\left(\frac{x+1}{x-4}\right) = 3 \]
To solve for \(x\), express the equation in exponential form:
\[ \frac{x+1}{x-4} = 2^3 \]
Calculate \(2^3\):
\[ \frac{x+1}{x-4} = 8 \]
Now, solve for \(x\) by cross-multiplying:
\[ x+1 = 8(x-4) \]
This leads to:
\[ x+1 = 8x - 32 \]
Rearrange the terms to isolate \(x\):
\[ 1 + 32 = 8x - x \]
\[ 33 = 7x \]
\[ x = \frac{33}{7} \]
Thus, the solution to the equation is \( x = \frac{33}{7} \).
**Important:**
Check the domain of the original logarithms to ensure the solution is valid. The expressions \((x+1)\) and \((x-4)\) must be positive:
1. \( x + 1 > 0 \Rightarrow x > -1 \)
2. \( x - 4 > 0 \Rightarrow x > 4 \)
Given \( x = \frac{33}{7} \approx 4.71 \), the solution meets both conditions. Therefore, the solution is valid within the domain of the logarithmic functions.
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