Question

Log3(x+4)+Log3(x-2)=3

Answer 1

asumo que 3 es la base

 

$<var>\\\log_3(x+4)+\log_3(x-2)=3\\ D: x>2\\ \log_3(x+4)(x-2)=3\\ 3^3=(x+4)(x-2)\\ 27=x^2-2x+4x-8\\ 27=x^2+2x-8\\ x^2+2x-35=0\\ x^2+7x-5x-35=0\\ x(x+7)-5(x+7)=0\\ (x-5)(x+7)=0\\ x=5 \vee x=-7\\ -7\not \in D\\ \underline{x=5} </var>$