Question

The formula d=10 log can be used to understand how loud a sound is in decibels, given its intensity. Let loudness of a sound be D and I the physical intensity. Suppose the decibel level is 60. The physical intensity, I, is how many times that of I0?

Answer 1

Solution:

Given that the formula

$\begin{gathered} d=10\log (\frac{I}{I_0})\text{ ----- equation 1} \\ \text{where} \\ d\Rightarrow loudness\text{ of sound, in decibels} \\ I\Rightarrow physical\text{ intensity} \end{gathered}$

is used to understand how loud a sound is in decibels, given its intensity.

Suppose the decibel level is 60, this implies that the loudness is 60 decibels.

Thus,

$d=60$

substitute the value of d into equation 1.

Thus,

$\begin{gathered} d=10\log (\frac{I}{I_0})\text{ } \\ \text{where d=60} \\ \Rightarrow60=10\log (\frac{I}{I_0}) \\ \text{divide through by 10} \\ \frac{60}{10}=\frac{10}{10}\log (\frac{I}{I_0}) \\ \Rightarrow6=\log (\frac{I}{I_0})\text{ ---- equation 2} \\ \end{gathered}$

From equation 2, take the antilogarithm of both sides.

Thus, we have

$\begin{gathered} 10^6=10^{\log (\frac{I}{I_0})} \\ \Rightarrow10^6=\frac{I}{I_0} \\ \text{Multiply through by }I_0 \\ I_0(10^6)=I_0(\frac{I}{I_0}) \\ \text{thus,} \\ I=I_0(10^6) \\ \Rightarrow I=1,000,000\times I_0 \end{gathered}$

This implies that the physical intensity is 1,000, 000 times that of I₀.

The second option is the correct answer.