Use the definition (p. 75) and properties of logarithmic functions (Theorem 1, p. 77) to solve the equation log5 (x? + 1) – log5(x - 1) = 1

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Author:James Stewart
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Chapter1: Functions And Models
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**Using Logarithmic Properties to Solve Equations**

In this section, we will explore how to solve the equation using the properties of logarithmic functions.

Given:  
\[ \log_5(x^2 + 1) - \log_5(x - 1) = 1 \]

Refer to the definition (p. 75) and properties of logarithmic functions as explained in Theorem 1 (p. 77) to solve this problem.

### Steps to Solve:

1. **Apply the Quotient Rule**:
   Using the quotient rule for logarithms: \(\log_b(m) - \log_b(n) = \log_b\left(\frac{m}{n}\right)\).
   
   \[
   \log_5 \left(\frac{x^2 + 1}{x - 1}\right) = 1
   \]

2. **Convert to Exponential Form**:
   If \(\log_b(a) = c\), then \(a = b^c\).

   \[
   \frac{x^2 + 1}{x - 1} = 5^1
   \]

3. **Solve the Equation**:
   \[
   \frac{x^2 + 1}{x - 1} = 5 
   \]

   Multiply both sides by \(x - 1\):
   \[
   x^2 + 1 = 5(x - 1)
   \]

   Simplify and solve for \(x\):
   \[
   x^2 + 1 = 5x - 5
   \]
   Rearrange:
   \[
   x^2 - 5x + 6 = 0
   \]

4. **Factor the Quadratic**:
   \[(x - 2)(x - 3) = 0\]

   Solve for \(x\):
   \[
   x = 2 \quad \text{or} \quad x = 3
   \]

5. **Check for Validity**:
   Both solutions need to satisfy the original equation conditions:

   - For \(x = 2\), calculate the log terms:
     \[
     \log_5((2)^2 + 1) - \log_5(2 - 1) = 1
     \]
     \[
     \log_5(5)
Transcribed Image Text:**Using Logarithmic Properties to Solve Equations** In this section, we will explore how to solve the equation using the properties of logarithmic functions. Given: \[ \log_5(x^2 + 1) - \log_5(x - 1) = 1 \] Refer to the definition (p. 75) and properties of logarithmic functions as explained in Theorem 1 (p. 77) to solve this problem. ### Steps to Solve: 1. **Apply the Quotient Rule**: Using the quotient rule for logarithms: \(\log_b(m) - \log_b(n) = \log_b\left(\frac{m}{n}\right)\). \[ \log_5 \left(\frac{x^2 + 1}{x - 1}\right) = 1 \] 2. **Convert to Exponential Form**: If \(\log_b(a) = c\), then \(a = b^c\). \[ \frac{x^2 + 1}{x - 1} = 5^1 \] 3. **Solve the Equation**: \[ \frac{x^2 + 1}{x - 1} = 5 \] Multiply both sides by \(x - 1\): \[ x^2 + 1 = 5(x - 1) \] Simplify and solve for \(x\): \[ x^2 + 1 = 5x - 5 \] Rearrange: \[ x^2 - 5x + 6 = 0 \] 4. **Factor the Quadratic**: \[(x - 2)(x - 3) = 0\] Solve for \(x\): \[ x = 2 \quad \text{or} \quad x = 3 \] 5. **Check for Validity**: Both solutions need to satisfy the original equation conditions: - For \(x = 2\), calculate the log terms: \[ \log_5((2)^2 + 1) - \log_5(2 - 1) = 1 \] \[ \log_5(5)
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