Question

Find the zero or zeroes of the function f left-parenthesis x right-parenthesis equals log subscript 9 baseline left-parenthesis 3 x plus 2 right-parenthesis plus log subscript 9 baseline left-parenthesis x minus 3 right-parenthesis.
(1 point)
Responses

x equals Start Fraction 7 plus or minus Start Root 133 End Root over 6 End Fraction
Image with alt text: x equals Start Fraction 7 plus or minus Start Root 133 End Root over 6 End Fraction

x equals Start Fraction 7 plus Start Root 133 End Root over 6 End Fraction
Image with alt text: x equals Start Fraction 7 plus Start Root 133 End Root over 6 End Fraction

x equals Start Fraction negative 2 over 3 End Fraction and x = 3
Image with alt text: x equals Start Fraction negative 2 over 3 End Fraction and x = 3

x = 3

Answer 1

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 \( x \).

1. Set the function equal to zero:

\[
\log_9(3x + 2) + \log_9(x - 3) = 0
\]

2. Using the properties of logarithms, we can combine the logs:

\[
\log_9((3x + 2)(x - 3)) = 0
\]

3. This means:

\[
(3x + 2)(x - 3) = 9^0 = 1
\]

4. Expand the left side:

\[
3x^2 - 9x + 2x - 6 = 1
\]

\[
3x^2 - 7x - 6 = 1
\]

5. Set the equation to zero:

\[
3x^2 - 7x - 7 = 0
\]

6. 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:

\[
x = \frac{7 \pm \sqrt{133}}{6}
\]