Question 15 Solve for \( x \) : \( \log _{4} x+\log _{4}(x+2)=6 \) \( x=\square \)

To solve the equation \( \log _{4} x + \log _{4}(x+2) = 6 \), you can use the property of logarithms that states \( \log_b(a) + \log_b(c) = \log_b(ac) \). This means you can combine the logarithms: \[ \log_{4}(x(x+2)) = 6 \] Now, you can rewrite this in exponential form. This tells you that \( x(x+2) = 4^6 \). Since \( 4^6 = 4096 \), you have: \[ x^2 + 2x - 4096 = 0 \] Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = 2, c = -4096 \): \[ x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-4096)}}{2 \cdot 1} \] \[ x = \frac{-2 \pm \sqrt{4 + 16384}}{2} \] \[ x = \frac{-2 \pm \sqrt{16388}}{2} \] \[ x = \frac{-2 \pm 128}{2} \] This gives you two possible solutions: \[ x = \frac{126}{2} = 63 \quad \text{and} \quad x = \frac{-130}{2} = -65 \] Since \( x \) must be positive in the logarithmic context, the solution is: \( x = 63 \)