Questions: Solve the system for all real number solutions. y = log4(x-12) - 3 y = -log4 x (x, y) = ( , )

Solution Steps

To solve the system of equations involving logarithms, we need to find the values of $x$ and $y$ that satisfy both equations simultaneously. We can start by equating the two expressions for $y$ and solving for $x$. Once $x$ is found, substitute it back into one of the original equations to find $y$.

We start with the system of equations: $$y = \log_{4}(x - 12) - 3$$ $$y = -\log_{4}(x)$$

Setting the two expressions for $y$ equal to each other gives: $$\log_{4}(x - 12) - 3 = -\log_{4}(x)$$

Rearranging the equation, we can express it as: $$\log_{4}(x - 12) + \log_{4}(x) = 3$$ Using the property of logarithms, this simplifies to: $$\log_{4}((x - 12) \cdot x) = 3$$ Exponentiating both sides results in: $$(x - 12) \cdot x = 4^3$$ $$x^2 - 12x = 64$$ Rearranging gives us the quadratic equation: $$x^2 - 12x - 64 = 0$$

Using the quadratic formula, we find: $$x = \frac{12 \pm \sqrt{(-12)^2 - 4 \cdot 1 \cdot (-64)}}{2 \cdot 1}$$ Calculating the discriminant: $$\sqrt{144 + 256} = \sqrt{400} = 20$$ Thus, the solutions for $x$ are: $$x = \frac{12 + 20}{2} = 16 \quad \text{and} \quad x = \frac{12 - 20}{2} = -4$$ Since $x$ must be greater than 12 for the logarithm to be defined, we take $x = 16$.

Substituting $x = 16$ back into either equation for $y$: $$y = -\log_{4}(16) = -2$$

Final Answer