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

Step 1: We have
Let the intensity of the sound be $I$.
According to the question, $I\text{'}=2I$
Step 2: Formula used
Let the intensity of the sound be $I$ and it's sound level is $\beta 1$
The sound level $\beta$ provided by, $\beta =10\log 10\left(\frac{I}{I}\right)$
Where the constant reference intensity is $I0$
Step 3: The sound level increased
The sound level $\beta 1$ provided by, $\beta 1=10\log 10\left(\frac{I}{I}\right)...\left(1\right)$
When the intensity doubles, the sound level is given by, $\beta 2=10\log 10\left(\frac{2I}{I}\right)...\left(2\right)$
By subtracting equations (2) and (1), the sound level increased is,
$\beta 2-\beta 1=\left(10\log 10\left(\frac{2I}{I}\right)\right)-\left(10\log 10\left(\frac{I}{I}\right)\right)\beta 2-\beta 1=10\left(\log 10\left(\frac{2I}{I}\right)-\log 10\left(\frac{I}{I}\right)\right)...\left(3\right)$
Since $\log a\left(x\right)-\log a\left(y\right)=\log a\left(\frac{x}{y}\right)$ equation 3 becomes,
$\beta 2-\beta 1=10\left(\log 10\left(\frac{2I}{I}\right)-\log 10\left(\frac{I}{I}\right)\right)\beta 2-\beta 1=10\log 10\left(\frac{2I}{I}\right)\beta 2-\beta 1=10\log 102\beta 2-\beta 1=10\times 0.3010\beta 2-\beta 1=3\mathrm{dB}$
Hence, the volume of the sound rises by $3\mathrm{dB}$.