Question

The sound level measured in a room by a person watching a movie on a home theater system varies from
60 dB
during a quiet part to
90 dB
during a loud part. Approximately how many times louder is the latter sound?

Answer 1

By definition, dB are:

$$60dB = 10\lg I_1\\ 90dB = 10\lg I_2$$

where $I_1, I_2$ are intensities of the sounds. From this obtain:

$$I_1 = 10^{6}\space (relative\space units)\\ I_2 = 10^{9}\space (relativ\space units)$$

Exact units don't matter since we are only interested in ratio $I_2/I_1$. Thus, the latter sound is

$$\dfrac{I_2}{I_1} = \dfrac{10^9}{10^6} = 1000$$

times louder.

Answer. 1000 times.