Answer
The solutions are $x=-9$ and $x=3$.
Work Step by Step
Apply the propuct rule to have:
$\log_3{[x(x+6)]}=3$
RECALL:
$\log_a{b}=x \longrightarrow a^x=b $
where $b\gt0$
Use this rule in converting the equation to its equivalent exponential equation to have:
$3^3=x(x+6)
\\27=x^2+6x
\\0=x^2+6x-27
\\x^2+6x-27=0$
Factor the trinomial to have:
$(x+9)(x-3)=0$
Equate each factor to zero then solve each equation to have:
$x+9=0 \text{ or } x-3=0
\\x=-9 \text{ or } x=3$
Note that in $\log_a{b}$, the value of $b$ has to be positive.
This means that in $\log_3{[x(x+6)]}$, the value of $x(x+6)$ must be positive.
Check if both solutions are valid to have:
For x = -9:
$x(x+6) = -9(-9+6)=-9(-3)=27$
For x=3:
$x(x+6)=3(3+6)=3(9)=27$
Since both solutions make $x^2+x$ positive, then both are valid.
Therefore, the solutions to the given equation are -9 and 3.
