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.

Question

(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:

$$x=\frac{-1+\sqrt{33}}{4}\text{and}x=\frac{-1-\sqrt{33}}{4}$$

The zero(s) of the function are:

$$x=\frac{-1\pm \sqrt{33}}{4}$$

This matches the last response option:

x = $\frac{-1\pm \sqrt{33}}{4}$