Solve each equation for \( x \) (A) Solve \( 8^{(x+2)}=8 \) for \( x \) \( x=\square \) (B) Solve \( \ln x+\ln (x-10)=2 \) for \( x \) \( x=\square \)

To solve \( 8^{(x+2)} = 8 \) for \( x \), we can rewrite 8 as \( 8^1 \). Thus, we have: \[ 8^{(x+2)} = 8^1 \] Since the bases are the same, we can set the exponents equal to each other: \[ x + 2 = 1 \] Now, subtract 2 from both sides: \[ x = 1 - 2 = -1 \] So, \( x = -1 \). Now, for the equation \( \ln x + \ln (x-10) = 2 \), we can use the properties of logarithms to combine the logs: \[ \ln(x(x-10)) = 2 \] Exponentiate both sides to eliminate the logarithm: \[ x(x-10) = e^2 \] This simplifies to: \[ x^2 - 10x - e^2 = 0 \] We can use the quadratic formula here, where \( a = 1 \), \( b = -10 \), and \( c = -e^2 \): \[ x = \frac{-(-10) \pm \sqrt{(-10)^2 - 4 \cdot 1 \cdot (-e^2)}}{2 \cdot 1} \] This simplifies to: \[ x = \frac{10 \pm \sqrt{100 + 4e^2}}{2} \] Thus, \[ x = 5 \pm \sqrt{25 + e^2} \] To ensure that \( x \) is valid, we must check both solutions to see if they are greater than 10. Therefore, we consider just the positive root. So, summarizing: (A) \( x = -1 \) (B) \( x = 5 + \sqrt{25 + e^2} \) (and validate its suitability).