Section 7.3: Solving Logarithmic Equations

This section teaches techniques to solve logarithmic equations, including using properties of logs and converting to exponential form.

Example 1: Using the Same Base

Solve \( \log_2 x = 5 \).

Step 1: Convert to exponential form: \( x = 2^5 \)

Step 2: Evaluate: \( x = 32 \)

Solve \( \log_3 (x+1) - \log_3 2 = 2 \).

Step 1: Apply quotient property: \( \log_3 \frac{x+1}{2} = 2 \)

Step 2: Convert to exponential form: \( \frac{x+1}{2} = 3^2 = 9 \)

Step 3: Solve for x: \( x + 1 = 18 \) → \( x = 17 \)

Solve \( \log_5 x + \log_5 (x-4) = 2 \).

Step 1: Apply product property: \( \log_5 [x(x-4)] = 2 \)

Step 2: Convert to exponential form: \( x(x-4) = 5^2 = 25 \)

Step 3: Solve quadratic: \( x^2 - 4x - 25 = 0 \)

Step 4: Use quadratic formula: \( x = \frac{4 \pm \sqrt{16+100}}{2} = \frac{4 \pm \sqrt{116}}{2} = 2 \pm \sqrt{29} \)

Step 5: Check domain: \( x - 4 > 0 \) → \( x > 4 \), so solution: \( x = 2 + \sqrt{29} \)

Solve \( \log_2 (x-1) = 4 \)
Solve \( \log_4 x + \log_4 (x-3) = 2 \)
Solve \( \log_5 (2x+1) - \log_5 3 = 1 \)
Solve \( \log_3 (x+2) + \log_3 (x-1) = 2 \)
Solve \( \log_6 x - \log_6 2 = 1 \)