Respuesta :
By solving the given equation and then substituting solution in original equation we found out that x=3 and x=-7 are extraneous solution.
What is Extraneous solutions?
The extraneous solutions are the solutions that does not work in the original equation.
Original equation
[tex]$\log _{4}(x)+\log _{4}(x-3)=\log _{4}(-7x+21)$[/tex]
Now we will simplify this
[tex]\log _{4}(x(x-3))=\log _{4}(-7 x+21)$\\\\x(x-3)=-7 x+21\\\\&x^{2}-3 x=-7 x+21 \\\\\\&x^{2}-3 x+7 x-21 =0\\\\&x^{2}+4 x-21=0\\\\\\&x^{2}-3 x+7x-21=0\\\\\\(x-3)(x+7)=0\\x=3,-7[/tex]
Now, we will determine the extraneous solution, by substituting x=3 and x=-7 in the originally given equation.
Substituting x=3,
[tex]\log _{4}(3)+\log _{4}(3-3)=\log _{4}(-21+21) \\\\\log _{4}(3)+\log _{4}(0)=\log _{4}(0)[/tex]
As we know log(0) is undefined , Hence x=3 is extraneous solution
Now Substituting x=-7
[tex]\log _{4}(-7)+\log _{4}(-7-3)=\log _{4}(-7(-7)+21)\\\\\log _{4}(-7)+\log _{4}(-10)=\log _{4}(-28)[/tex]
As we know log is not defined for negative numbers hence x=-7 is extraneous solution
By solving the given equation and then substituting solution in original equation we found out that x=3 and x=-7 are extraneous solution
To know more about logarithm function visit : https://brainly.com/question/13473114
