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Find the zero(es) of the function f left parenthesis x right parenthesis equals log subscript 3 baseline left parenthesis x min...Asked by freedom
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
(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
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Answered by
GPT-4o mini
To find the zeros of the function \( f(x) = \log_3(x - 1) + \log_3(2x + 3) \), we first set the function to zero:
\[
\log_3(x - 1) + \log_3(2x + 3) = 0.
\]
Using the logarithmic property \(\log_b(a) + \log_b(c) = \log_b(ac)\), we can combine the logs:
\[
\log_3((x - 1)(2x + 3)) = 0.
\]
Since \(\log_3(a) = 0\) implies \(a = 1\), we have:
\[
(x - 1)(2x + 3) = 1.
\]
Expanding the left-hand side gives:
\[
2x^2 + 3x - 2x - 3 = 1 \quad \Rightarrow \quad 2x^2 + x - 3 = 1.
\]
Subtracting 1 from both sides leads to:
\[
2x^2 + x - 4 = 0.
\]
Now we will use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2, b = 1, c = -4 \).
First, we calculate the discriminant:
\[
b^2 - 4ac = 1^2 - 4(2)(-4) = 1 + 32 = 33.
\]
Now we can find the roots:
\[
x = \frac{-1 \pm \sqrt{33}}{2(2)} = \frac{-1 \pm \sqrt{33}}{4}.
\]
Thus, the zeros of the function \(f(x)\) are:
\[
x = \frac{-1 + \sqrt{33}}{4} \quad \text{and} \quad x = \frac{-1 - \sqrt{33}}{4}.
\]
Therefore, the complete answer for the zeros of \( f(x) \) is:
\[
\boxed{x = \frac{-1 \pm \sqrt{33}}{4}}.
\]
\[
\log_3(x - 1) + \log_3(2x + 3) = 0.
\]
Using the logarithmic property \(\log_b(a) + \log_b(c) = \log_b(ac)\), we can combine the logs:
\[
\log_3((x - 1)(2x + 3)) = 0.
\]
Since \(\log_3(a) = 0\) implies \(a = 1\), we have:
\[
(x - 1)(2x + 3) = 1.
\]
Expanding the left-hand side gives:
\[
2x^2 + 3x - 2x - 3 = 1 \quad \Rightarrow \quad 2x^2 + x - 3 = 1.
\]
Subtracting 1 from both sides leads to:
\[
2x^2 + x - 4 = 0.
\]
Now we will use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2, b = 1, c = -4 \).
First, we calculate the discriminant:
\[
b^2 - 4ac = 1^2 - 4(2)(-4) = 1 + 32 = 33.
\]
Now we can find the roots:
\[
x = \frac{-1 \pm \sqrt{33}}{2(2)} = \frac{-1 \pm \sqrt{33}}{4}.
\]
Thus, the zeros of the function \(f(x)\) are:
\[
x = \frac{-1 + \sqrt{33}}{4} \quad \text{and} \quad x = \frac{-1 - \sqrt{33}}{4}.
\]
Therefore, the complete answer for the zeros of \( f(x) \) is:
\[
\boxed{x = \frac{-1 \pm \sqrt{33}}{4}}.
\]
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