When the intensity of sound is doubled, by how many decibels does...

The sound level in decibels is given by the formula:

L = 10 \log_{10} \left(\dfrac{I}{I_0} \right)

When the intensity is doubled, the new intensity is $I' = 2I$. The new sound level is:

$L' = 10 \log_{10} \left( \dfrac{2I}{I_0} \right)$

The increase in sound level is:

\Delta L = L' - L = 10 \log_{10} \left(\dfrac{2I}{I_0} \right) -10 \log_{10} \left(\dfrac{I}{I_0} \right)

\Delta L = 10 \log_{10} \left(\dfrac{2I / I_0}{I / I_0} \right) =$10 \log_{10}(2)$

Substituting the value of $\log_{10}(2) \approx 0.3010$:

$\Delta L = 10 \times 0.3010 = 3.01 \text{ dB}$

The sound level increases by approximately $3 \text{ dB}.$