Solve the Logarithmic Equation: log₄x + log₄(x+2) = 2

To solve the given logarithmic equation, let's proceed step-by-step:

• Step 1: Use the product rule of logarithms:
  Given the equation $\log_4 x + \log_4 (x+2) = 2$, apply the product rule to combine the logs:
  $\log_4 (x(x+2)) = 2$.
• Step 2: Convert the equation from logarithmic to exponential form:
  The equation becomes $x(x+2) = 4^2$, which simplifies to $x(x+2) = 16$.
• Step 3: Expand and rearrange the quadratic equation:
  We have $x^2 + 2x - 16 = 0$.
• Step 4: Solve the quadratic equation using the quadratic formula:
  The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Here, $a = 1$, $b = 2$, and $c = -16$.
  Calculate the discriminant: $b^2 - 4ac = 2^2 - 4 \cdot 1 \cdot (-16) = 4 + 64 = 68$.
  The solutions are given by:
  $x = \frac{-2 \pm \sqrt{68}}{2}$ which simplifies to $x = \frac{-2 \pm 2\sqrt{17}}{2}$.
  Thus, $x = -1 \pm \sqrt{17}$.
• Step 5: Check the solutions within the original equation's domain:
  Since $x$ must be greater than zero, $x = -1 - \sqrt{17}$ is invalid as it results in a negative value.
  Thus, the valid solution is $x = -1 + \sqrt{17}$.

Therefore, the solution to the problem is $x = -1 + \sqrt{17}$.