Question

which is the only solution to the equation \\(\log_{3}(x^{2}+6x)=\log_{3}(2x + 12)\\)? \\(x=-6\\) \\(x=-2\\) \\(x = 0\\) \\(x=2\\) \\(x = 6\\)

Explanation:

Step1: Use logarithmic property

If $\log_a M = \log_a N$, then $M = N$ (for $a>0,a
eq1$, and $M>0,N>0$). So we set $x^2 + 6x = 2x + 12$.

Step2: Rearrange into quadratic equation

Subtract $2x + 12$ from both sides: $x^2 + 6x - 2x - 12 = 0$, which simplifies to $x^2 + 4x - 12 = 0$.

Step3: Factor the quadratic

Factor $x^2 + 4x - 12$: we need two numbers that multiply to -12 and add to 4. Those numbers are 6 and -2. So $(x + 6)(x - 2) = 0$.

Step4: Solve for x

Set each factor equal to zero: $x + 6 = 0$ gives $x = -6$; $x - 2 = 0$ gives $x = 2$.

Step5: Check domain

For $\log_3(x^2 + 6x)$ and $\log_3(2x + 12)$ to be defined, the arguments must be positive.
- For $x = -6$: $x^2 + 6x = 36 - 36 = 0$ (not positive), so $x = -6$ is invalid.
- For $x = 2$: $x^2 + 6x = 4 + 12 = 16>0$ and $2x + 12 = 4 + 12 = 16>0$, so $x = 2$ is valid? Wait, no, wait. Wait, let's re - check. Wait, maybe I made a mistake. Wait, let's go back. Wait, when we set $x^2 + 6x=2x + 12$, let's try the options directly.
Wait, let's test each option:
- Option $x=-6$: $\log_3((-6)^2+6*(-6))=\log_3(36 - 36)=\log_3(0)$ which is undefined. So $x = -6$ is out.
- Option $x=-2$: $\log_3((-2)^2+6*(-2))=\log_3(4 - 12)=\log_3(-8)$ undefined.
- Option $x = 0$: $\log_3(0 + 0)=\log_3(0)$ undefined.
- Option $x = 2$: $\log_3(4 + 12)=\log_3(16)$ and $\log_3(4 + 12)=\log_3(16)$, so it works? Wait, but earlier when we factored, we got $x=-6$ and $x = 2$, but $x=-6$ is invalid. Wait, but the options also have $x = 2$? Wait, but the original problem says "the only solution". Wait, maybe I made a mistake in factoring. Wait, wait the original equation is $\log_3(x^2 + 6x)=\log_3(2x + 12)$. Wait, let's re - do the algebra.
Wait, $x^2+6x=2x + 12$
$x^2+4x - 12 = 0$
Wait, discriminant $b^2 - 4ac=16+48 = 64$, square root of 64 is 8. So $x=\frac{-4\pm8}{2}$. So $x=\frac{-4 + 8}{2}=2$, $x=\frac{-4 - 8}{2}=-6$. But as we saw, $x=-6$ makes the argument zero, which is not allowed in logarithm (since $\log_a(0)$ is undefined). Wait, but the option $x = 2$ is there. But wait, the options also have $x = 2$? Wait, but let's check the options again. The options are $x=-6$, $x=-2$, $x = 0$, $x = 2$, $x = 6$. Wait, maybe I made a mistake in the factoring. Wait, $x^2+6x=2x + 12$
$x^2+4x-12 = 0$
Wait, maybe I factored wrong. Let's use quadratic formula: $x=\frac{-4\pm\sqrt{16 + 48}}{2}=\frac{-4\pm\sqrt{64}}{2}=\frac{-4\pm8}{2}$. So $x=\frac{-4 + 8}{2}=2$, $x=\frac{-4 - 8}{2}=-6$. But $x=-6$ is invalid. Wait, but the option $x = 2$ is present. But wait, let's check $x = 2$: $\log_3(4 + 12)=\log_3(16)$ and $\log_3(4 + 12)=\log_3(16)$, so it's valid. Wait, but the problem says "the only solution". But let's check other options again. Wait, maybe I made a mistake in the domain check for $x=-6$. Wait, $x=-6$: $x^2+6x=36-36 = 0$, so $\log_3(0)$ is undefined, so $x=-6$ is out. $x = 2$: valid. Wait, but the options have $x = 2$ as an option. But wait, let's check the original equation again. Wait, maybe I misread the equation. The equation is $\log_3(x^2 + 6x)=\log_3(2x + 12)$. So when $x = 2$, both sides are $\log_3(16)$, which is equal. When $x=-6$, left side is $\log_3(0)$ (undefined), right side is $\log_3(0)$ (undefined). Wait, no, if the argument is zero, the logarithm is undefined. So $x=-6$ is invalid. $x = 2$ is valid. Wait, but the options also have $x = 2$ as an option. But wait, maybe I made a mistake in the factoring. Wait, let's try $x = 2$: works. $x=-6$: doesn't work. What about $x=-2$: $x^2+6x=4-12=-8$, $\log_3(-8)$ is undefined. $x = 0$: $x^2+6x=0$, $\log_3(0)$ undefined. $x = 6$: $x^2+6x=36 + 36=72$, $2x + 12=12 + 12=24$, $\log_3(72)
eq\log_3(24)$. So the only valid solution is $x = 2$? Wait, but the problem says "the only solution". Wait, but let's check again. Wait, maybe I made a mistake in the algebra. Wait, $x^2+6x=2x + 12$
$x^2+4x-12 = 0$
$(x + 6)(x - 2)=0$
So solutions $x=-6$ and $x = 2$. But $x=-6$ is invalid, so $x = 2$ is the solution.

Answer:

$x = 2$ (corresponding to the option "x = 2")