Answer
$3$
Work Step by Step
RECALL:
(1) $\log_a{P} + \log_a{Q}=\log_a{(PQ)}$
(2) $\log_a{P}=\log_a{Q} \longrightarrow P=Q$
Use rule (1) above to obtain
$\log{[x(x+1)]}=\log{12}$
Use rule (2) above to obtain:
$x(x+1)=12
\\x(x) + x(1) = 12
\\x^2+x=12$
Subtract $12$ on both sides to obtain:
$x^2+x-12=0$
Factor the trinomial to obtain:
$(x+4)(x-3)=0$
Equate each factor to zero then solve each equation to obtain:
$\begin{array}{ccc}
&x+4=0 &\text{ or } &x-3=0
\\&x=-4 &\text{ or }&x=3
\end{array}$
$x$ cannot be $-4$ since $\log{-4}$ is undefined.
Thus, the solution is $x=3$.
