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January 17th, 2015, 01:11 PM  #1 
Newbie Joined: Jan 2015 From: Sacramento Posts: 3 Thanks: 0  I cant figure this out! please help
In some murky harbors, the intensity of light d feet below the surface is given approximately by: I = I0e^(0.23d) What percentage of the surface light will reach a depth of: a. 10 feet? b. 20 feet? Last edited by skipjack; January 17th, 2015 at 07:15 PM. Reason: to provide the exponentiation presumably intended 
January 17th, 2015, 01:29 PM  #2 
Newbie Joined: Jan 2015 From: france Posts: 7 Thanks: 1 
I=Io*exp(0.23*d) we suppose that the unity of the term (0.23 are inverse of a foot) for d=10 ft I(10 ft)=Io * 0.10032 Percentage is 10% for d=20 ft I(20 ft)=Io * 0.01005 percentage is 1% 
January 17th, 2015, 05:13 PM  #3 
Newbie Joined: Jan 2015 From: Sacramento Posts: 3 Thanks: 0 
Thank you so much, but I still don't understand how you get the answer.

January 17th, 2015, 06:10 PM  #4 
Newbie Joined: Jan 2015 From: Washington, United States Posts: 11 Thanks: 1  I'm not sure exactly what alistrawberry did, either... I don't know much about light penetration in water, but I didn't get about 0.1 for exp(.23*10). Not sure what line 2 was supposed to mean (unity?).

January 17th, 2015, 07:10 PM  #5 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,934 Thanks: 1128 Math Focus: Elementary mathematics and beyond 
At d = 0, I = 10. We want a percent so we want "intensity at a given depth"/"intensity at d = 0". With d= 10: $\displaystyle \frac{10e^{0.23\cdot10}}{10}=e^{0.23\cdot10}\approx0.1002$ That's pretty close to 10%. With d= 20: $\displaystyle \frac{10e^{0.23\cdot20}}{10}=e^{0.23\cdot20}\approx0.0101$ That's pretty close to 1%. 
January 17th, 2015, 07:18 PM  #6 
Global Moderator Joined: Dec 2006 Posts: 20,617 Thanks: 2072  

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