Chapter6: Exponential And Logarithmic Functions
Section6.1: Exponential Functions
Problem 62SE: A scientist begins with 100 milligrams of aradioactive substance that decays exponentially. After 35...
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![### Solving Logarithmic Equations
To solve the logarithmic equation, follow these steps:
**Problem:**
Solve:
\[ \log_{3}(2x - 1) = \log_{3}(x + 8) \]
Given that both sides of the equation have the same base, we can equate the arguments:
\[ 2x - 1 = x + 8 \]
**Procedure:**
1. **Isolate the variable terms:**
\[ 2x - 1 = x + 8 \]
2. **Subtract \( x \) from both sides:**
\[ 2x - x - 1 = 8 \]
\[\ \boxed{x} - 1 = 8\ \]
3. **Add 1 to both sides to solve for \( x \):**
\[ x = 8 + 1 \]
\[ x = \boxed{9} \]
Thus, the value of \( x \) that satisfies the equation is \( 9 \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8585c01c-e95f-4e9e-ab09-9be83264bdef%2F8ee7bb5f-62a3-4dda-94ae-98067135b5dc%2Femdg86k_processed.png&w=3840&q=75)
Transcribed Image Text:### Solving Logarithmic Equations
To solve the logarithmic equation, follow these steps:
**Problem:**
Solve:
\[ \log_{3}(2x - 1) = \log_{3}(x + 8) \]
Given that both sides of the equation have the same base, we can equate the arguments:
\[ 2x - 1 = x + 8 \]
**Procedure:**
1. **Isolate the variable terms:**
\[ 2x - 1 = x + 8 \]
2. **Subtract \( x \) from both sides:**
\[ 2x - x - 1 = 8 \]
\[\ \boxed{x} - 1 = 8\ \]
3. **Add 1 to both sides to solve for \( x \):**
\[ x = 8 + 1 \]
\[ x = \boxed{9} \]
Thus, the value of \( x \) that satisfies the equation is \( 9 \).
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