Problem 10. Solve the equation for x: log56 x + log5 (x - 1) = 1.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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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:
\[ x = 8 \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe047ce83-e95d-4948-8d53-fb1226f895dd%2Fea7e1f7b-46dc-473d-8a99-76c69779df7e%2Fr13mesyj_processed.jpeg&w=3840&q=75)
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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