To solve the equation \( \ln(3x - 11) = \ln(4 + 2x) \), we can use the property that if \( \ln(a) = \ln(b) \), then \( a = b \). Thus, we set the arguments of the logarithms equal to each other:
\[ 3x - 11 = 4 + 2x \]
Now, we will solve for \( x \). Let's first isolate \( x \) by rearranging the equation:
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Subtract \( 2x \) from both sides: \[ 3x - 2x - 11 = 4 \] This simplifies to: \[ x - 11 = 4 \]
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Now, add \( 11 \) to both sides: \[ x = 4 + 11 \] Thus, we have: \[ x = 15 \]
Now, we need to check if this solution is valid by substituting \( x = 15 \) back into the original arguments of the logarithm:
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For \( 3x - 11 \): \[ 3(15) - 11 = 45 - 11 = 34 \]
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For \( 4 + 2x \): \[ 4 + 2(15) = 4 + 30 = 34 \]
Both arguments are equal and positive, which means our solution is valid.
Therefore, the solution to the logarithmic equation is: \[ \boxed{15} \]