Answer:
[tex]x_{1}=-7 \\x_{2}=3[/tex]
Step-by-step explanation:
The given equation is
[tex]log_{4} (x)+log_{4}(x-3)=log_{4}(-7x+21)[/tex]
First, we need to use the product property of logarithms
[tex]log_{4}(x)+log_{4}(x-3)=log_{4}(x(x-3))[/tex]
Replacing this in the equation, er have
[tex]log_{4}(x(x-3))=log_{4}(-7x+21)[/tex]
Now we can cancel logarithms
[tex]x(x-3)=-7x+21[/tex]
Then, we use the distributive property and solve for [tex]x[/tex]
[tex]x^{2} -3x=-7x+21\\x^{2} -3x+7x-21=0\\x^{2} +4x-21=0[/tex]
We need to find two number which product is 21 and which difference is 4. Those numbers are 7 and 3, because 7x3 = 21 and 7 - 3 = 4.
[tex]x^{2} +4x-21=(x+7)(x-3)[/tex]
There are two solutions
[tex]x_{1}=-7 \\x_{2}=3[/tex]
