Question

If a 20dB sound level (ticking clock) is replaced by 80dB (garbage disposal), by what factor has the sound intensity been multiplied?

Answer 1

The sound level is given by

$$\beta=10\lg\frac{I}{I_0}$$

Hence

$$\frac{I_2}{I_1}=10^{\Delta \beta/10}$$

$$\frac{I_2}{I_1}=10^{60/10}=10^6$$