# Question #d39b2

Remember that $10 \mathrm{db}$ stands for $10 \times$ in the $\log$ scale
$\therefore 10 \log 10 = 10 \times 1 = 10$
Similarly $- 10 \mathrm{db} \implies \frac{1}{10} t h$
Required intensity of sound is $= 10 \times \left(5.81 \times {10}^{-} 9 W {m}^{-} 2\right)$
$= 5.81 \times {10}^{-} 8 W {m}^{-} 2$