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