If \( P_{1}=17 \mathrm{~W} \) and \( P_{2}=0,4 \mathrm{~W} \), find the relative value of \( P_{1} \) with respect to \( P_{2} \) in units of dB using the formula \( D=10 \log \left(\frac{P_{1}}{P_{2}}\right) \). Round your answer to two decimal places.

To calculate the relative value of \( P_{1} \) with respect to \( P_{2} \) in decibels (dB), we can use the formula \( D = 10 \log \left(\frac{P_{1}}{P_{2}}\right) \). Substituting the given values: \[ D = 10 \log \left(\frac{17 \, \text{W}}{0.4 \, \text{W}}\right) \] Calculating the ratio: \[ \frac{17}{0.4} = 42.5 \] Now using the logarithm: \[ D = 10 \log(42.5) \] Calculating the logarithm: \[ \log(42.5) \approx 1.628 \] Finally, calculating \( D \): \[ D = 10 \times 1.628 \approx 16.28 \, \text{dB} \] Thus, the relative value of \( P_{1} \) with respect to \( P_{2} \) is approximately **16.28 dB** when rounded to two decimal places.