Answer
The solutions are $x=-8$ and $x=2$.
Work Step by Step
Apply the propuct rule to have:
$\log_4{[x(x+6)]}=2$
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:
$4^2=x(x+6)
\\16=x^2+6x
\\0=x^2+6x-16
\\x^2+6x-16=0$
Factor the trinomial to have:
$(x+8)(x-2)=0$
Equate each factor to zero then solve each equation to have:
$x+8=0 \text{ or } x-2=0
\\x=-8 \text{ or } x=2$
Note that in $\log_a{b}$, the value of $b$ has to be positive.
This means that in $\log_4{[x(x+6)]}$, the value of $x(x+6)$ must be positive.
Check if both solutions are valid to have:
For x = -8:
$x(x+6) = -8(-8+6)=-8(-2)=16$
For x=2:
$x(x+6)=2(2+6)=2(8)=16$
Since both solutions make $x^2+x$ positive, then both are valid.
Therefore, the solutions to the given equation are -8 and 2.
