## College Algebra 7th Edition

$3$
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 $x=3$.