QuestionJuly 15, 2025

Solve the logarithic equation. log(x+6)+log(x-3)=1 a) x=-4 b) x=4 c) x=1/4 d) x=-1/4

Solve the logarithic equation. log(x+6)+log(x-3)=1 a) x=-4 b) x=4 c) x=1/4 d) x=-1/4
Solve the logarithic equation. log(x+6)+log(x-3)=1 a) x=-4 b) x=4 c) x=1/4 d) x=-1/4

Solution
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Answer

b) x=4 Explanation 1. Apply the Product Rule of Logarithms Combine the logarithms using the product rule: \log(a) + \log(b) = \log(ab). Thus, \log(x+6) + \log(x-3) = \log((x+6)(x-3)). 2. Set the Equation Equal to 1 The equation becomes \log((x+6)(x-3)) = 1. This implies (x+6)(x-3) = 10^1 = 10. 3. Expand and Simplify the Quadratic Equation Expand (x+6)(x-3) to get x^2 + 3x - 18 = 10. Simplify to x^2 + 3x - 28 = 0. 4. Solve the Quadratic Equation Use the quadratic formula x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} with a=1, b=3, c=-28. Calculate the discriminant: b^2 - 4ac = 3^2 - 4(1)(-28) = 9 + 112 = 121. Thus, x = \frac{-3 \pm \sqrt{121}}{2} = \frac{-3 \pm 11}{2}. 5. Determine the Roots The roots are x = \frac{-3 + 11}{2} = 4 and x = \frac{-3 - 11}{2} = -7. 6. Verify the Validity of the Solutions Check if the solutions satisfy the domain of the original logarithmic expressions. For x=4: x+6=10 > 0 and x-3=1 > 0. For x=-7: x+6=-1 < 0, which is invalid. Therefore, only x=4 is valid.

Explanation

1. Apply the Product Rule of Logarithms<br /> Combine the logarithms using the product rule: $\log(a) + \log(b) = \log(ab)$. Thus, $\log(x+6) + \log(x-3) = \log((x+6)(x-3))$.<br />2. Set the Equation Equal to 1<br /> The equation becomes $\log((x+6)(x-3)) = 1$. This implies $(x+6)(x-3) = 10^1 = 10$.<br />3. Expand and Simplify the Quadratic Equation<br /> Expand $(x+6)(x-3)$ to get $x^2 + 3x - 18 = 10$. Simplify to $x^2 + 3x - 28 = 0$.<br />4. Solve the Quadratic Equation<br /> Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a=1$, $b=3$, $c=-28$. Calculate the discriminant: $b^2 - 4ac = 3^2 - 4(1)(-28) = 9 + 112 = 121$. Thus, $x = \frac{-3 \pm \sqrt{121}}{2} = \frac{-3 \pm 11}{2}$.<br />5. Determine the Roots<br /> The roots are $x = \frac{-3 + 11}{2} = 4$ and $x = \frac{-3 - 11}{2} = -7$.<br />6. Verify the Validity of the Solutions<br /> Check if the solutions satisfy the domain of the original logarithmic expressions. For $x=4$: $x+6=10 > 0$ and $x-3=1 > 0$. For $x=-7$: $x+6=-1 < 0$, which is invalid. Therefore, only $x=4$ is valid.
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