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.

Question

(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:

$$x=\frac{7+\sqrt{133}}{6}\text{and}x=\frac{7-\sqrt{133}}{6}$$

From the given responses, the correct answer is:

$$x=\frac{7\pm \sqrt{133}}{6}$$