light intensity..question

**anitha** · August 24th 2008, 09:30 AM

on a foggy night the intensity of light from an approaching car is drastically reduced. the distance away, d, in metres, is related to the intensity, I(d), by d=166.67log I(D)/125.
a)how far from you is an approaching car if the intensity is 50 lumens?
b) what is the maximum intensity of the light from an approaching car?
c) solve the equation for I(d).

**Chris L T521** · August 24th 2008, 09:45 AM

Originally Posted by anitha

on a foggy night the intensity of light from an approaching car is drastically reduced. the distance away, d, in metres, is related to the intensity, I(d), by d=166.67log I(D)/125.
a)how far from you is an approaching car if the intensity is 50 lumens?
b) what is the maximum intensity of the light from an approaching car?
c) solve the equation for I(d).

For part (a) plug this into your calculator:

$166.67\log\left(\frac{50}{125}\right)\approx\dots$

For part (b):

The maximum intensity of the light is observed when you are 0 meters away from the car. So you would need to solve this equation:

$0=166.67\log\left(\frac{I(d)}{125}\right)$

This implies that $0=\log\left(\frac{I(d)}{125}\right)\implies 1=\frac{I(d)}{125}\implies I(d)=\dots$

For part (c):

We want to find $I(d)$ :

$d=166.67\log\left(\frac{I(d)}{125}\right)\implies \frac{d}{166.67}=\log\left(\frac{I(d)}{125}\right)$

My next step would vary, depending on your interpretation of logs. When you say log, are you referring to $\log_{10}$ or $\log_e$ ??

I will show both ways, and you pick the one that fits the context of log in this problem.

If $\log$ represents $\log_{10}$ , then

$10^{\frac{d}{166.67}}=\frac{I(d)}{125}\implies \color{red}\boxed{I(d)=125\cdot 10^{\frac{d}{166.67}}}$

If $\log$ represents $\log_e$ , then

$e^{\frac{d}{166.67}}=\frac{I(d)}{125}\implies \color{red}\boxed{I(d)=125\cdot e^{\frac{d}{166.67}}}$

I hope this makes sense!

--Chris

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