Question

log3 (x + 2) + log3 (3x + 6) = 5

Answer 1

$\Large\boxed{\begin{array}{l}\sf C\cdot E: x+2 > 0\\\sf x > -2\\\sf \log_3(x+2)+\log_3(3x+6)=5\\\sf \log_3(x+2)+\log_33(x+2)=5\\\sf \log_33(x+2)^2=5\\\sf 3(x+2)^2=3^5\\\sf (x+2)^2=\dfrac{3^5}{3}\\\\\sf (x+2)^2=3^4\\\sf x+2=\pm\sqrt{3^4}\\\sf x+2=\pm3^2\\\sf x+2=\pm9\\\sf x+2=9\\\sf x=9-2\\\sf x=7\\\sf x+2=-9\\\sf x=-9-2\\\sf x=-11(n\tilde ao\,conv\acute em\,pois\,x > -2)\\\sf S=\{7\}\end{array}}$