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

Given that the formula

`\begin{gathered} d=10\log ((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 ((I)/(I_0))\text{ } \\ \text{where d=60} \\ \Rightarrow60=10\log ((I)/(I_0)) \\ \text{divide through by 10} \\ (60)/(10)=(10)/(10)\log ((I)/(I_0)) \\ \Rightarrow6=\log ((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 ((I)/(I_0))} \\ \Rightarrow10^6=(I)/(I_0) \\ \text{Multiply through by }I_0 \\ I_0(10^6)=I_0((I)/(I_0)) \\ \text{thus,} \\ I=I_0(10^6) \\ \Rightarrow I=1,000,000* 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.