Question

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

Explanation:

Step1: Set arguments equal (log property)

If $\log_b(A) = \log_b(B)$, then $A=B$ (for $A>0, B>0, b>0, b
eq1$).
So $x^2 + 6x = 2x + 12$

Step2: Rearrange to quadratic equation

$x^2 + 6x - 2x - 12 = 0$
$x^2 + 4x - 12 = 0$

Step3: Factor the quadratic

$(x+6)(x-2) = 0$
Solutions: $x=-6$ or $x=2$

Step4: Check domain validity

For $\log_3(x^2+6x)$:

• If $x=-6$: $(-6)^2 + 6(-6) = 36-36=0$, $\log_3(0)$ is undefined.
• If $x=2$: $2^2 + 6(2)=4+12=16>0$, $2(2)+12=16>0$, both logs are defined.

Answer:

$x=2$