1. The problem is to calculate the equivalent sound pressure level $L_{eq}$ given by the formula: $$L_{eq} = 10 \times \log_{10} \left( \frac{1}{3} \left(10^{\frac{72.78}{10}} + 10^{\frac{71.26}{10}} + 10^{\frac{72.77}{10}} \right) \right)$$ 2. First, convert each decibel value to its linear scale by calculating $10^{\frac{L_p}{10}}$: $$10^{\frac{72.78}{10}} = 10^{7.278} \approx 1.900 \times 10^{7}$$ $$10^{\frac{71.26}{10}} = 10^{7.126} \approx 1.337 \times 10^{7}$$ $$10^{\frac{72.77}{10}} = 10^{7.277} \approx 1.896 \times 10^{7}$$ 3. Sum these linear values: $$1.900 \times 10^{7} + 1.337 \times 10^{7} + 1.896 \times 10^{7} = 5.133 \times 10^{7}$$ 4. Divide the sum by 3: $$\frac{5.133 \times 10^{7}}{3} = 1.711 \times 10^{7}$$ 5. Take the logarithm base 10 and multiply by 10 to convert back to decibels: $$L_{eq} = 10 \times \log_{10}(1.711 \times 10^{7}) = 10 \times (\log_{10}(1.711) + \log_{10}(10^{7}))$$ $$= 10 \times (0.233 + 7) = 10 \times 7.233 = 72.33$$ 6. Therefore, the equivalent sound pressure level is approximately: $$L_{eq} \approx 72.33$$ decibels. This calculation combines the three sound levels into a single equivalent level by converting to linear scale, averaging, and converting back to decibels.