To find the zero or zeros of the function \( f(x) = \log_9(3x + 2) + \log_9(x - 3) \), we need to set the function equal to zero and solve for \( x \).
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Set the function equal to zero:
\[ \log_9(3x + 2) + \log_9(x - 3) = 0 \]
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Using the properties of logarithms, we can combine the logs:
\[ \log_9((3x + 2)(x - 3)) = 0 \]
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This means:
\[ (3x + 2)(x - 3) = 9^0 = 1 \]
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Expand the left side:
\[ 3x^2 - 9x + 2x - 6 = 1 \]
\[ 3x^2 - 7x - 6 = 1 \]
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Set the equation to zero:
\[ 3x^2 - 7x - 7 = 0 \]
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Now, we can solve this quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 3 \), \( b = -7 \), and \( c = -7 \).
\[ b^2 - 4ac = (-7)^2 - 4(3)(-7) = 49 + 84 = 133 \]
So, the solutions for \( x \) are:
\[ x = \frac{7 \pm \sqrt{133}}{6} \]
Therefore, the zeros of the function are:
\[ x = \frac{7 + \sqrt{133}}{6} \quad \text{and} \quad x = \frac{7 - \sqrt{133}}{6} \]
From the given responses, the correct answer is:
\[ x = \frac{7 \pm \sqrt{133}}{6} \]