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log function question

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The question states the formula D=5e^-.4h can be used to find the number of milligrams D of a certain drug that is in a patient's bloodstream h hours after the drug was administered. When the number of milligrams reaches 2, the drug is to be administered again. What is the time between injections?

I have done several problems like this before with no trouble. The function turns into 2=5e^-.4h. I know you must turn this into logarithmic form by e^y=x then y=In x. A few things that confuse me with this problem is what to do with the 5 when turning it into log form, and also how is the answer positive with the -.4h. I know the answer is h=2.29 or about 2 hours and 17 minutes between injections because I looked at the answers in the back of the book. I was just asking to see how you get this answer with showing my work. Any help is appreciated. Thank you!
 Nov 25, 2013

7+0 Answers

 #1
avatar+118653 
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I just want to clarify
is it

D=5*e^(-0.4h)
or
D=5*e^(-0.4)*h = 5*h*e(-0.4) [tecnically this is what you have written]
 Nov 26, 2013
 #2
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Sorry it is the first one you wrote. D=5*e^(-.4h)
 Nov 26, 2013
 #3
avatar+118653 
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D=5*e^(-0.4h)

Find h when D=2

2=5*e^(-0.4h)
divide both sides by 5
0.4 = e^(-0.4h)

ln 0.4 = lne^(-0.4h)

ln 0.4 = -0.4h

h = ln0.4 / -0.4 = 2.29072683hours

h = 2 hours and 17 minutes to the nearest minute.
-----------------------------
now looking at your specific concerns
1) what to do with the 5 when turning it into log form (hopefully you can see what happened to the 5)

2) and also how is the answer positive with the -.4h.

-0.4h is the power of e. Any positive number raised to any number (positive or negative) will be positive. e is a positive number.

I don't know how good you can manipulate indices but maybe if you think about this it might help

16^(-0.5) = 1 / 16^(0.5) = 1 / the sqrt of 16 = 1/4
 Nov 26, 2013
 #4
avatar+9 
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Wow thank you so much! I kept trying to turn it into log form before dividing by 5 and just getting ridiculous answers. Thank you for explaining to me, I understand how to find the inverse function with a negative exponent now and it actually makes sense! Thanks again for your help, much appreciated.
 Nov 26, 2013
 #5
avatar+118653 
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I am really glad that i could help you. I always like getting feedback and it is great when the feedback is so positive.
Thank you.
 Nov 26, 2013
 #6
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I really appreciate that you took all this time to try and help us.I truly astounded by your publish regards. I really appreciate that you took all this time to try and help us.Your work is carry the information to the other people.
 Nov 26, 2013
 #7
avatar+118653 
0
Thank you. I love mathematics and I want to share my enthusiasm with others.
If I help others to understand and to enjoy the experience that is all I want.
It is always really nice when you are appreciated even it you are doing what you want to do most.
so again, thank you.
 Nov 26, 2013

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