Find the largest value of \( x \) that satisfies: \( \log _{4}\left(x^{2}\right)-\log _{4}(x+3)=8 \) \( x= \)

To solve the equation \( \log _{4}\left(x^{2}\right)-\log _{4}(x+3)=8 \), we can use the properties of logarithms. First, we apply the quotient rule for logarithms, which states that \( \log_b(a) - \log_b(c) = \log_b\left(\frac{a}{c}\right) \). This gives us: \[ \log_{4}\left(\frac{x^2}{x+3}\right) = 8 \] Next, we can rewrite the logarithmic equation in its exponential form. Recall that \( \log_b(a) = c \) means \( a = b^c \). Here, we have: \[ \frac{x^2}{x+3} = 4^8 \] We know that \( 4^8 = (2^2)^8 = 2^{16} \). Thus, we have: \[ \frac{x^2}{x+3} = 2^{16} \] Cross-multiplying gives us: \[ x^2 = 2^{16}(x + 3) \] Expanding this yields: \[ x^2 = 2^{16}x + 3 \cdot 2^{16} \] Rearranging the equation, we get: \[ x^2 - 2^{16}x - 3 \cdot 2^{16} = 0 \] Now, we apply the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1 \), \( b = -2^{16} \), and \( c = -3 \cdot 2^{16} \): 1. Calculate \( b^2 - 4ac \): \[ b^2 = (2^{16})^2 = 2^{32} \] \[ 4ac = 4 \cdot 1 \cdot (-3 \cdot 2^{16}) = -12 \cdot 2^{16} = -3 \cdot 2^{18} \] \[ b^2 - 4ac = 2^{32} + 3 \cdot 2^{18} = 2^{18}(2^{14} + 3) \] 2. Now substitute this value into the quadratic formula: \[ x = \frac{2^{16} \pm \sqrt{2^{18}(2^{14} + 3)}}{2} \] 3. Simplifying gives: \[ x = 2^{15} \pm 2^{9}\sqrt{2^{14} + 3} \] 4. We want the largest value of \( x \): \[ x = 2^{15} + 2^{9}\sqrt{2^{14} + 3} \] Calculating \( 2^{14} + 3 \) gives \( 16384 + 3 = 16387 \). Thus, our largest solution to \( x \) is: \[ \boxed{2^{15} + 2^{9}\sqrt{16387}} \] This is the largest value of \( x \).