Answer:
[tex]f(x,y) = \log_{4} (x-5-\sqrt{25-6\cdot y})+\log_{4} (x-5+\sqrt{25-6\cdot y})[/tex]
Step-by-step explanation:
Let be [tex]f(x,y) = \log_{4}(2\cdot x^{2}-20\cdot x +12\cdot y)[/tex], this expression is simplified by algebraic and trascendental means. As first step, the second order polynomial is simplified. Its roots are determined by the Quadratic Formula, that is to say:
[tex]r_{1,2} = \frac{20\pm \sqrt{(-20)^{2}-4\cdot (2)\cdot (12\cdot y)}}{2\cdot (2)}[/tex]
[tex]r_{1,2} = 5\pm \sqrt{25-6\cdot y}[/tex]
The polynomial in factorized form is:
[tex](x-5-\sqrt{25-6\cdot y})\cdot (x-5+\sqrt{25-6\cdot y})[/tex]
The function can be rewritten and simplified as follows:
[tex]f(x,y) = \log_{4} [(x-5-\sqrt{25-6\cdot y})\cdot (x-5+\sqrt{25-6\cdot y})][/tex]
[tex]f(x,y) = \log_{4} (x-5-\sqrt{25-6\cdot y})+\log_{4} (x-5+\sqrt{25-6\cdot y})[/tex]