Question

Find the zero(es) of the function f left parenthesis x right parenthesis equals log subscript 3 baseline left parenthesis x minus 1 right parenthesis plus log subscript 3 baseline left parenthesis 2 x plus 3 right parenthesis.
(1 point)
Responses

x = 2 and x equals negative 1
x = 2 and Image with alt text: x equals negative 1

x = 1 and x equals negative start fraction 2 over 3 end fraction
x = 1 and Image with alt text: x equals negative start fraction 2 over 3 end fraction

x equals start fraction negative 1 plus start root 33 end root over 4 end fraction
Image with alt text: x equals start fraction negative 1 plus start root 33 end root over 4 end fraction

x equals start fraction negative 1 plus or minus start root 33 end root over 4 end fraction
Image with alt text: x equals start fraction negative 1 plus or minus start root 33 end root over 4 end fraction

Answer 1

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 = \(\frac{-1 \pm \sqrt{33}}{4}\)**