Question

Log base 6 (3x) + log base 6 (x-4) = 2

Answer 1

`D:3x>0 \wedge x-4>0\\ D:x>0 \wedge x>4\\ D:x>4\\\\ \log_63x+\log_6(x-4)=2\\ \log_63x(x-4)=\log_636\\ 3x(x-4)=36\\ 3x^2-12x=36\\ 3x^2-12x-36=0\\ x^2-4x-12=0\\ x^2+2x-6x-12=0\\ x(x+2)-6(x+2)=0\\ (x-6)(x+2)=0\\ x=6 \vee x=-2\\ -2\\ot \in D \Rightarrow x=6`