Answer: x = 8
(the correct answer was not provided as an option)
Step-by-step explanation:
[tex]log_2(x-6)+log_2(x-4)=log_2(x)\\\\log_2[(x-6)(x-4)]=log_2(x)\\\\(x-6)(x-4)=x\\\\x^2-10x+24=x\\\\x^2-11x+24=0\\\\(x-3)(x-8)=0\\\\x=3\qquad x=8\\\\\\\text{The term after the log symbol (inside the parenthesis) must be greater than 0!}\\\\Check:\\3-6>0\ \text{FALSE --- so 3 is not a valid solution}\\\\8-6>0\ \text{TRUE}\\8-4>0\ \text{TRUE}\\8>0\ \text{TRUE --- 8 is a valid solution because it is true for all}[/tex]
