# If the intensity of a sound from one source is 100 times that of another sound, how much more is the decibel level of the louder sound than the softer one?

Oct 30, 2017

$2$ times

#### Explanation:

Let the intensity of the softer sound be $x \textcolor{w h i t e}{i} \frac{W}{m} ^ 2$.
$\therefore$ the intensity of the louder sound is $100 x \textcolor{w h i t e}{i} \frac{W}{m} ^ 2$.

$\beta = 10 \textcolor{w h i t e}{i} \text{dB} \cdot \log \left(\frac{I}{I} _ o\right)$

For the softer sound:

${\beta}_{1} = 10 \textcolor{w h i t e}{i} \text{dB} \cdot \log \left(\frac{x \textcolor{w h i t e}{i} \frac{W}{m} ^ 2}{{10}^{-} 12 \textcolor{w h i t e}{i} \frac{W}{m} ^ 2}\right)$

${\beta}_{1} = 10 \cdot \log \left({10}^{14} x\right)$

For the louder sound:

${\beta}_{2} = 10 \textcolor{w h i t e}{i} \text{dB} \cdot \log \left(\frac{100 x \textcolor{w h i t e}{i} \frac{W}{m} ^ 2}{{10}^{-} 12 \textcolor{w h i t e}{i} \frac{W}{m} ^ 2}\right)$

${\beta}_{2} = 10 \cdot \log \left({10}^{12} x\right)$

How much more ?
$= {\beta}_{1} / {\beta}_{2}$

=(10*log(10^14 x))/(10*log(10^12 x)

$= 2$