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Math Help - light intensity..question

  1. #1
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    light intensity..question

    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).
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by anitha View Post
    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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