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.)

To determine how much louder a \( 42-\mathrm{dB} \) sound is compared to a \( 10-\mathrm{dB} \) sound, we can use the formula that relates decibels to intensity. The decibel level is given by: \[ L = 10 \log_{10} \left( \frac{I}{I_0} \right) \] where \( L \) is the sound level in decibels, \( I \) is the intensity of the sound, and \( I_0 \) is a reference intensity (commonly taken as \( 10^{-12} \, \mathrm{W/m^2} \)). To find the intensity ratios, we can rearrange the formula. For the \( 42-\mathrm{dB} \) sound: \[ I_{42} = I_0 \times 10^{\frac{42}{10}} = I_0 \times 10^{4.2} \] For the \( 10-\mathrm{dB} \) sound: \[ I_{10} = I_0 \times 10^{\frac{10}{10}} = I_0 \times 10^{1} \] Next, we can find the ratio of the intensities: \[ \frac{I_{42}}{I_{10}} = \frac{I_0 \times 10^{4.2}}{I_0 \times 10^{1}} = 10^{4.2 - 1} = 10^{3.2} \] Calculating \( 10^{3.2} \): \[ 10^{3.2} \approx 1585.0 \] So, rounding to the nearest integer, a \( 42-\mathrm{dB} \) sound is approximately \( 1585 \) times as loud as a \( 10-\mathrm{dB} \) sound.