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 rewrite it in logarithmic form. Taking the logarithm of both sides gives us: \[ (2x - 1) \log(5) = 300 \log(4) \] From here, we can isolate \( x \): \[ 2x - 1 = \frac{300 \log(4)}{\log(5)} \] Now solving for \( x \): \[ 2x = \frac{300 \log(4)}{\log(5)} + 1 \] \[ x = \frac{1}{2} \left( \frac{300 \log(4)}{\log(5)} + 1 \right) \] After calculating this using a calculator, we find: \[ x \approx 132.75 \Rightarrow x \approx 133 \text{ (to 3 significant figures)} \] For the second equation \( 2^{y+1} = \frac{10}{2^{y}} \), we can multiply both sides by \( 2^y \): \[ 2^{y+1} \cdot 2^{y} = 10 \] \[ 2^{2y + 1} = 10 \] Now, taking the logarithm of both sides: \[ (2y + 1) \log(2) = \log(10) \] Now isolating \( y \): \[ 2y + 1 = \frac{\log(10)}{\log(2)} \] \[ 2y = \frac{\log(10)}{\log(2)} - 1 \] \[ y = \frac{1}{2} \left( \frac{\log(10)}{\log(2)} - 1 \right) \] After calculating this, we find: \[ y \approx 1.5 \Rightarrow y \approx 1.50 \text{ (to 3 significant figures)} \] So the final answers are: \( x \approx 133 \) \( y \approx 1.50 \)