How much louder (more intense) is a $$ 42-\mathrm{dB} $$ sound than a $$ 10-\mathrm{dB} $$ sound? The $$ 42-\mathrm{dB} $$ sound is $$ \square $$ times as loud as the $$ 10-\mathrm{dB} $$ sound. (Round to the nearest integer as needed.)

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Let the sound intensities be $$ I_{42} $$ and $$ I_{10} $$ for the $$ 42\text{-dB} $$ and $$ 10\text{-dB} $$ sounds respectively. The decibel level is defined by

$$
\beta = 10 \log_{10}\left(\frac{I}{I_0}\right)
$$

where $$ I_0 $$ is the reference intensity.

For the $$ 42\text{-dB} $$ sound:

$$
42 = 10 \log_{10}\left(\frac{I_{42}}{I_0}\right) \quad \Longrightarrow \quad \frac{I_{42}}{I_0} = 10^{4.2}
$$

For the $$ 10\text{-dB} $$ sound:

$$
10 = 10 \log_{10}\left(\frac{I_{10}}{I_0}\right) \quad \Longrightarrow \quad \frac{I_{10}}{I_0} = 10^{1} = 10
$$

The ratio of the intensities is:

$$
\frac{I_{42}}{I_{10}} = \frac{10^{4.2}}{10^{1}} = 10^{4.2 - 1} = 10^{3.2}
$$

To compute $$ 10^{3.2} $$, note that:

$$
10^{3.2} = 10^3 \times 10^{0.2} \approx 1000 \times 1.585 = 1585
$$

Thus, the $$ 42\text{-dB} $$ sound is approximately $$ 1585 $$ times as intense as the $$ 10\text{-dB} $$ sound.
The $$ 42\text{-dB} $$ sound is approximately 1585 times as loud as the $$ 10\text{-dB} $$ sound.

How much louder (more intense) is a \( 42-\mathrm{dB} \) sound than a \( 10-\mathrm{dB} \) sound? The \( 42-\mathrm{dB} \) sound is \( \square \) times as loud as the \( 10-\mathrm{dB} \) sound. (Round to the nearest integer as needed.)

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