If \( P_{1}=16 \mathrm{~W} \) and \( P_{2}=0.3 \mathrm{~W} \), find the relative value of \( P_{1} \) with respect to \( P_{2} \) in units of \( d B \) using the formula \( D=10 \log \left(\frac{P_{1}}{P_{2}}\right) \). Round your answer to two decimal places.

To find the relative value of \( P_{1} \) with respect to \( P_{2} \) in decibels (dB), we can use the formula: \[ D = 10 \log \left(\frac{P_{1}}{P_{2}}\right) \] Given \( P_{1} = 16 \, \text{W} \) and \( P_{2} = 0.3 \, \text{W} \), we first calculate the ratio: \[ \frac{P_{1}}{P_{2}} = \frac{16}{0.3} \approx 53.33 \] Now, we substitute this ratio into the formula: \[ D = 10 \log(53.33) \] Calculating the logarithm: \[ \log(53.33) \approx 1.727 \] Now, multiplying by 10: \[ D \approx 10 \times 1.727 = 17.27 \, \text{dB} \] Rounding to two decimal places, we have: \[ D \approx 17.27 \, \text{dB} \] So the relative value of \( P_{1} \) with respect to \( P_{2} \) is approximately \( 17.27 \, \text{dB} \).