Respuesta :

naǫ
[tex]\hbox{the domain:} \\ 3x>0 \ \land \ x+4>0 \\ x>0 \ \land \ x>-4 \\ x \in (0,+\infty) \\ \\ \log(3x)=\log 2+\log(x+4) \\ 0=\log 2+ \log(x+4) - \log(3x) \\ 0=\log(\frac{2 \times (x+4)}{3x}) \\ 0=\log(\frac{2x+8}{3x}) \\ 10^0=\frac{2x+8}{3x} \\ 1=\frac{2x+8}{3x} \\ 3x=2x+8 \\ 3x-2x=8 \\ \boxed{x=8}[/tex]
[tex]D:\\3x > 0\ and\ x+4 > 0\\x > 0\ and\ x > -4\\therefore\ D:x\in(-4;\ \infty)\\------------------\\log(3x)=log2+log(x+4)\ \ \ |use\ loga+logb=log(a\cdot b)\\\\log(3x)=log[2(x+4)]\iff3x=2(x+4)\\\\3x=2x+8\ \ \ \ |subtract\ 2x\ from\ both\ sides\\\\\boxed{x=8\in D}[/tex]