Question

solve for x
log(x–4)+log(x+3)=2logx

Answer 1

$D:x-4>0 \wedge x+3>0 \wedge x>0\\ D:x>4 \wedge x>-3 \wedge x>0\\ D:x>4\\\\ \log(x-4)+\log(x+3)=2\log x\\ \log(x-4)(x+3)=\log x^2\\ (x-4)(x+3)=x^2\\ x^2+3x-4x-12=x^2\\ -x=12\\ x=-12\\ -12\not\in D\\ \Downarrow\\ \boxed{x\in\emptyset}$