Solve for x. Check for extraneous solutions. Show your work using the Equation Editor tool. logs (10) + logg 8 = logg 16 Paragraph BI UVA AO =O GO + v ... KY
Solve for x. Check for extraneous solutions. Show your work using the Equation Editor tool. logs (10) + logg 8 = logg 16 Paragraph BI UVA AO =O GO + v ... KY
College Algebra (MindTap Course List)
12th Edition
ISBN:9781305652231
Author:R. David Gustafson, Jeff Hughes
Publisher:R. David Gustafson, Jeff Hughes
Chapter5: Exponential And Logarithmic Functions
Section5.3: Logarithmic Functions And Their Graphs
Problem 139E
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Question
![### Solving Logarithmic Equations
#### Task:
Solve for \( x \). Check for extraneous solutions. Show your work using the Equation Editor tool.
\[ \log_{8}(x - 10) + \log_{8} 8 = \log_{8} 16 \]
#### Solution:
First, use the properties of logarithms to combine the terms on the left side of the equation:
\[
\log_{8}(x - 10) + \log_{8} 8 = \log_{8}[(x - 10) \cdot 8]
\]
Since \(\log_{8} AB = \log_{8} A + \log_{8} B\), we can rewrite the equation as:
\[
\log_{8}[8(x - 10)] = \log_{8} 16
\]
Since the logarithms on both sides of the equation have the same base, we can set their arguments equal to each other:
\[
8(x - 10) = 16
\]
Next, solve for \( x \):
\[
8x - 80 = 16
\]
Add 80 to both sides:
\[
8x = 96
\]
Divide by 8:
\[
x = 12
\]
#### Check for Extraneous Solutions:
We need to check if \( x = 12 \) is an extraneous solution by substituting it back into the original equation and ensuring both sides are equal:
\[
\log_{8}(12 - 10) + \log_{8} 8 = \log_{8} 16
\]
Simplify inside the logarithms:
\[
\log_{8}(2) + \log_{8} 8 = \log_{8} 16
\]
We know that \(\log_{8} 8 = 1\) and \(\log_{8} 16 = 4/3\) since \(16 = 8^{4/3}\):
\[
\log_{8}(2) + 1 = 4/3
\]
If \(\log_{8}(2) + 1 = 4/3\) was indeed the simplified and correct interpretation of the expression, it confirms \( x = 12 \) is a valid solution.
Finally, we've verified that \( x = 12 \) satisfies the original](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc3692e55-09d2-4c7f-90e9-0ed90d8b0dce%2F0d0a393d-a412-4e5b-a48f-66ac650ee144%2Fdlks3n7_processed.png&w=3840&q=75)
Transcribed Image Text:### Solving Logarithmic Equations
#### Task:
Solve for \( x \). Check for extraneous solutions. Show your work using the Equation Editor tool.
\[ \log_{8}(x - 10) + \log_{8} 8 = \log_{8} 16 \]
#### Solution:
First, use the properties of logarithms to combine the terms on the left side of the equation:
\[
\log_{8}(x - 10) + \log_{8} 8 = \log_{8}[(x - 10) \cdot 8]
\]
Since \(\log_{8} AB = \log_{8} A + \log_{8} B\), we can rewrite the equation as:
\[
\log_{8}[8(x - 10)] = \log_{8} 16
\]
Since the logarithms on both sides of the equation have the same base, we can set their arguments equal to each other:
\[
8(x - 10) = 16
\]
Next, solve for \( x \):
\[
8x - 80 = 16
\]
Add 80 to both sides:
\[
8x = 96
\]
Divide by 8:
\[
x = 12
\]
#### Check for Extraneous Solutions:
We need to check if \( x = 12 \) is an extraneous solution by substituting it back into the original equation and ensuring both sides are equal:
\[
\log_{8}(12 - 10) + \log_{8} 8 = \log_{8} 16
\]
Simplify inside the logarithms:
\[
\log_{8}(2) + \log_{8} 8 = \log_{8} 16
\]
We know that \(\log_{8} 8 = 1\) and \(\log_{8} 16 = 4/3\) since \(16 = 8^{4/3}\):
\[
\log_{8}(2) + 1 = 4/3
\]
If \(\log_{8}(2) + 1 = 4/3\) was indeed the simplified and correct interpretation of the expression, it confirms \( x = 12 \) is a valid solution.
Finally, we've verified that \( x = 12 \) satisfies the original
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