(a) Solve the simultaneous equations \[ \begin{array}{l}10^{x+2 y}=5 \\ 10^{3 x+4 y}=50\end{array} \] giving \( x \) and \( y \) in exact simplified form.

To solve the simultaneous equations: 1. Convert each equation into logarithmic form for easier manipulation. Starting with the first equation: \[ 10^{x + 2y} = 5 \] Taking log base 10 on both sides: \[ x + 2y = \log_{10}(5) \] Let's denote \( a = \log_{10}(5) \), so we have: \[ x + 2y = a \quad \text{(1)} \] Now for the second equation: \[ 10^{3x + 4y} = 50 \] Taking log base 10 on both sides: \[ 3x + 4y = \log_{10}(50) \] We can simplify \( \log_{10}(50) \) as follows: \[ \log_{10}(50) = \log_{10}(5 \times 10) = \log_{10}(5) + \log_{10}(10) = a + 1 \] So we have: \[ 3x + 4y = a + 1 \quad \text{(2)} \] Now we have the system: 1. \( x + 2y = a \) 2. \( 3x + 4y = a + 1 \) Let's solve these equations simultaneously. From equation (1): \[ x = a - 2y \] Substitute \( x \) into equation (2): \[ 3(a - 2y) + 4y = a + 1 \] Expanding this: \[ 3a - 6y + 4y = a + 1 \] Combining like terms: \[ 3a - 2y = a + 1 \] Rearranging gives: \[ 3a - a - 1 = 2y \] \[ 2a - 1 = 2y \] Dividing by 2: \[ y = a/2 - 1/2 = \frac{\log_{10}(5)}{2} - \frac{1}{2} \] Now substituting \( y \) back into equation (1) to find \( x \): \[ x + 2\left( \frac{\log_{10}(5)}{2} - \frac{1}{2} \right) = \log_{10}(5) \] Simplifying gives: \[ x + \log_{10}(5) - 1 = \log_{10}(5) \] Thus: \[ x - 1 = 0 \] \[ x = 1 \] Now we have: \[ x = 1 \] \[ y = \frac{\log_{10}(5)}{2} - \frac{1}{2} \] In conclusion, the final exact and simplified solutions are: \[ x = 1, \quad y = \frac{\log_{10}(5) - 1}{2} \]