\( \left. \begin{array} { l l } { 9 } & { 81.9 ^ { x - 1 } = \frac { 1 } { 27 } } \\ { 10 } & { 10 ^ { 2 x ^ { 2 } } \cdot 10 ^ { x + 1 } = 10000 } \end{array} \right. \)

To solve the equations, we start with the first one: 1. The equation \( 81.9^{x - 1} = \frac{1}{27} \). We can convert both sides to the same base. Since \(81.9\) can be approximated as \(3^4\) (since \(81 = 3^4\)) and \(27 = 3^3\), we rewrite the equation as \( (3^4)^{x - 1} = 3^{-3} \). Simplifying this gives us \( 3^{4(x - 1)} = 3^{-3} \). By equating the exponents, we have \(4(x - 1) = -3\). Solving for \(x\), we find \(x = \frac{1}{4}\). 2. Now for the second equation: \(10^{2x^2} \cdot 10^{x + 1} = 10000\). Noticing that \(10000 = 10^4\), we can combine the terms on the left side: \(10^{2x^2 + x + 1} = 10^4\). By equating the exponents, we get \(2x^2 + x + 1 = 4\). Rearranging gives us \(2x^2 + x - 3 = 0\). We can solve this quadratic equation either by factoring or using the quadratic formula. The roots solved yield \(x = 1\) and \(x = -\frac{3}{2}\). Thus, the solutions to the equations are \(x = \frac{1}{4}\) and \(x = 1\) (or \(x = -\frac{3}{2}\)).