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Problem 10. Solve the equation for x: log56 x + log5 (x - 1) = 1.

Problem 10. Solve the equation for x: log56 x + log5 (x - 1) = 1.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Problem 10

### Problem 10: Solving Logarithmic Equations Objective: Solve the equation for \( x \): \[ \log_{56} x + \log_{56} (x - 1) = 1 \] Explanation: To solve this logarithmic equation, you can utilize properties of logarithms such as the product rule. The product rule states that \( \log_b (a) + \log_b (c) = \log_b (a \cdot c) \). Applying this property to the given equation: \[ \log_{56} x + \log_{56} (x - 1) = \log_{56} [x \cdot (x - 1)] \] So, the equation simplifies to: \[ \log_{56} [x (x - 1)] = 1 \] Since the base of the logarithm and the right side of the equation are the same, we can rewrite this as: \[ x (x - 1) = 56^1 \] Therefore: \[ x^2 - x = 56 \] \[ x^2 - x - 56 = 0 \] This is a quadratic equation. To solve for \( x \), you can use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Where \( a = 1 \), \( b = -1 \), and \( c = -56 \). Thus: \[ x = \frac{1 \pm \sqrt{1 + 224}}{2} \] \[ x = \frac{1 \pm \sqrt{225}}{2} \] \[ x = \frac{1 \pm 15}{2} \] So, the possible values for \( x \) are: \[ x = \frac{16}{2} = 8 \] and \[ x = \frac{-14}{2} = -7 \] Since the logarithm of a negative number is undefined, we discard \( x = -7 \). Hence, the solution for \( x \) is: \[ x = 8 \]
Transcribed Image Text:### Problem 10: Solving Logarithmic Equations Objective: Solve the equation for \( x \): \[ \log_{56} x + \log_{56} (x - 1) = 1 \] Explanation: To solve this logarithmic equation, you can utilize properties of logarithms such as the product rule. The product rule states that \( \log_b (a) + \log_b (c) = \log_b (a \cdot c) \). Applying this property to the given equation: \[ \log_{56} x + \log_{56} (x - 1) = \log_{56} [x \cdot (x - 1)] \] So, the equation simplifies to: \[ \log_{56} [x (x - 1)] = 1 \] Since the base of the logarithm and the right side of the equation are the same, we can rewrite this as: \[ x (x - 1) = 56^1 \] Therefore: \[ x^2 - x = 56 \] \[ x^2 - x - 56 = 0 \] This is a quadratic equation. To solve for \( x \), you can use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Where \( a = 1 \), \( b = -1 \), and \( c = -56 \). Thus: \[ x = \frac{1 \pm \sqrt{1 + 224}}{2} \] \[ x = \frac{1 \pm \sqrt{225}}{2} \] \[ x = \frac{1 \pm 15}{2} \] So, the possible values for \( x \) are: \[ x = \frac{16}{2} = 8 \] and \[ x = \frac{-14}{2} = -7 \] Since the logarithm of a negative number is undefined, we discard \( x = -7 \). Hence, the solution for \( x \) is: \[ x = 8 \]
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