The loudness of sound (L, measured in decibels) perceived by the human ear depends on intensity levels (I) according to , where is the standard threshold of audibility. If then using the change-of-base formula, we get . At what rate is the loudness changing with respect to x when the intensity is 1,000 times the standard threshold of audibility (that is, when x = 1,000)?

Question

The loudness of sound (L, measured in decibels) perceived by the human ear depends on intensity levels (I) according to  , where  is the standard threshold of audibility. If  then using the change-of-base formula, we get  . At what rate is the loudness changing with respect to x when the intensity is 1,000 times the standard threshold of audibility (that is, when x = 1,000)? ​

Quizplus · Accepted Answer

We have: 
$L=10\log_{10}\left(\dfrac{I}{I_0}ight)$
Taking the derivative with respect to $x$ on both sides, we get: 
$\dfrac{dL}{dx} = \dfrac{10}{\ln(10)}\dfrac{1}{I}\dfrac{dI}{dx}$
When $I=1000I_0$, we have $I=10^3I_0$, so $\dfrac{dI}{dx} = \dfrac{d}{dx}(10^3I_0)=10^3\dfrac{dI_0}{dx}=10^3(-\alpha I_0)= -\alpha I$, where we used $\dfrac{dI_0}{dx}=-\alpha I_0$, and $I_0=10^{-12} W/m^2$. Therefore, 
$\dfrac{dL}{dx} = -\dfrac{10}{\ln(10)}\dfrac{\alpha}{I}=-\dfrac{10}{\ln(10)}\dfrac{\alpha}{1000I_0}$
When $I=1000I_0$, 
$\dfrac{dL}{dx}\bigg|_{x=1000}=-\dfrac{10}{\ln(10)}\dfrac{\alpha}{1000I_0}=\boxed{	extbf{(B) }-\dfrac{10\alpha}{\ln{(10)}\cdot 1000I_0}}$

The Loudness of Sound (L, Measured in Decibels) Perceived by the Human