To find the zeros of the function \( f(x) = \log_3(x - 1) + \log_3(2x + 3) \), we start by setting the function equal to zero:
\[ \log_3(x - 1) + \log_3(2x + 3) = 0 \]
Using the logarithmic property that states \( \log_b(a) + \log_b(c) = \log_b(ac) \), we can combine the logarithms:
\[ \log_3((x - 1)(2x + 3)) = 0 \]
This implies:
\[ (x - 1)(2x + 3) = 1 \]
Now we expand the left side:
\[ 2x^2 + 3x - 2x - 3 = 2x^2 + x - 3 \]
Setting this equal to 1 gives:
\[ 2x^2 + x - 3 = 1 \] \[ 2x^2 + x - 4 = 0 \]
Now we can apply the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2 \), \( b = 1 \), and \( c = -4 \):
\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-4)}}{2 \cdot 2} \] \[ x = \frac{-1 \pm \sqrt{1 + 32}}{4} \] \[ x = \frac{-1 \pm \sqrt{33}}{4} \]
Thus, the solutions for \( x \) are:
\[ x = \frac{-1 + \sqrt{33}}{4} \quad \text{and} \quad x = \frac{-1 - \sqrt{33}}{4} \]
The zero(s) of the function are:
\[ x = \frac{-1 \pm \sqrt{33}}{4} \]
This matches the last response option:
x = \(\frac{-1 \pm \sqrt{33}}{4}\)