Answer:
C
Step-by-step explanation:
Given the logarithmic equation
[tex]\log_4x+\log_4(x-3)=\log_4(-7x+21)[/tex]
First, notice that
[tex]x>0\\ \\x-3>0\Rightarrow x>3\\ \\-7x+21>0\Rightarrow 7x<21\ x<3[/tex]
So, there is no possible solutions, all possible solutions will be extraneous.
Solve the equation:
[tex]\log_4x+\log_4(x-3)=\log_4x(x-3),[/tex]
then
[tex]\log_4x(x-3)=\log_4(-7x+21)\\ \\x(x-3)=-7x+21\\ \\x^2-3x+7x-21=0\\ \\x^2+4x-21=0\\ \\D=4^2-4\cdot 1\cdot (-21)=16+84=100\\ \\x_{1,2}=\dfrac{-4\pm 10}{2}=-7,\ 3[/tex]
Hence, [tex]x=3[/tex] and [tex]x=-7[/tex] are extraneous solutions
