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