Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$x= -1$
Apply the logarithmic property: $\log_a M+\log_a N = \log_a (MN)$ and rearrange the given expression to obtain: $\log_6[(x+4)(x+3)] \ ...(1)$ Since, $\log_m{n} = 1$ gives: $m^{(1)}=n$, then we have: $\log (x+4)(x+3)= 6^1$ $\log (x^2+7x+12)= 6$ or, $x^2+7x+6=0$ This is a quadratic equation; thus by factoring, it will become: $(x+6)(x+1)=0$ By the zero product property, we have: $x=-6$ and $x=-1$ Since, the domain of the variable is $x \gt -3$, we cannot accept the value of $x=-6$ Thus, our answer is: $x= -1$