Solve the following exponential equations, giving the final answers correct to 3 significant figures. i. \( 5^{2 x-1}=4^{300} \) \( x \approx \) ii. \( 2^{y+1}=\frac{10}{2^{y}} \). \( y \approx \)

To solve the first equation, rewrite \( 4^{300} \) as \( (2^2)^{300} = 2^{600} \). This gives us \( 5^{2x-1} = 2^{600} \). Taking the logarithm of both sides, we have: \[ (2x - 1) \log(5) = 600 \log(2) \] Solving for \( x \), we get: \[ 2x - 1 = \frac{600 \log(2)}{\log(5)} \quad \Rightarrow \quad 2x = 1 + \frac{600 \log(2)}{\log(5)} \quad \Rightarrow \quad x = \frac{1}{2} + \frac{300 \log(2)}{\log(5)} \] Calculating the values gives: \[ x \approx \frac{1}{2} + \frac{300 \times 0.301}{0.699} \approx 215.72 \quad \Rightarrow \quad x \approx 216 \] For the second equation \( 2^{y+1} = \frac{10}{2^y} \), we can rewrite it as \( 2^{y+1} \cdot 2^y = 10 \) or \( 2^{2y+1} = 10 \). Taking the logarithm gives: \[ (2y + 1) \log(2) = \log(10) \] Solving for \( y \): \[ 2y + 1 = \frac{\log(10)}{\log(2)} \quad \Rightarrow \quad 2y = \frac{\log(10)}{\log(2)} - 1 \quad \Rightarrow \quad y = \frac{\frac{\log(10)}{\log(2)} - 1}{2} \] Calculating the values gives: \[ y \approx \frac{1 - 1}{2} \approx 0 \] Thus, the final answers are: \( x \approx 216 \) and \( y \approx 0 \).