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 \( 5^{2x-1} = 4^{300} \), we can take logarithms on both sides. 1. Taking logarithm: \[ \log(5^{2x-1}) = \log(4^{300}) \] 2. Applying the power rule: \[ (2x - 1) \log(5) = 300 \log(4) \] 3. Solving for \( x \): \[ 2x - 1 = \frac{300 \log(4)}{\log(5)} \] \[ 2x = \frac{300 \log(4)}{\log(5)} + 1 \] \[ x = \frac{1}{2} \left( \frac{300 \log(4)}{\log(5)} + 1 \right) \] Calculating \( x \) gives approximately \( x \approx 121.363 \), so to three significant figures, \( x \approx 121 \). For the second equation \( 2^{y+1} = \frac{10}{2^{y}} \): 1. Multiplying both sides by \( 2^{y} \): \[ 2^{y+1} \cdot 2^{y} = 10 \] 2. Simplifying: \[ 2^{2y + 1} = 10 \] 3. Taking logarithm: \[ (2y + 1) \log(2) = \log(10) \] 4. Solving for \( y \): \[ 2y + 1 = \frac{\log(10)}{\log(2)} \] \[ 2y = \frac{\log(10)}{\log(2)} - 1 \] \[ y = \frac{1}{2} \left( \frac{\log(10)}{\log(2)} - 1 \right) \] Calculating \( y \) gives approximately \( y \approx 0.5 \), so to three significant figures, \( y \approx 0.500 \).