anonymous
  • anonymous
What does the Exponential Decay Variables Mean? For Half Lifes Y=Ae^-bX
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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campbell_st
  • campbell_st
ok so your formula is A is the initial population b is the decay coefficient X is normally the time. \[Y = Ae^{-bx}\] then then is the population half of the initial population of A for half life you will have \[\frac{Y}{A} = \frac{1}{2} \] so the equation becomes \[\frac{1}{2} = e^{-bX}\] you can find - bX by taking the log of both sides of the equation
anonymous
  • anonymous
Another thing is if i know the half life in years = 5715 and the amount after 1000 years is 2 grams then what is the initial amount?
anonymous
  • anonymous
so normal time is the amount of time passed?

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campbell_st
  • campbell_st
so then X = 5715 this will help to find the decay constant - b \[\frac{1}{2} = e^{-b \times 5715}\] you'll need to solve for b by using logs... the when you have b you will be able to find A given Y = 2 and X = 1000
campbell_st
  • campbell_st
thats correct.
anonymous
  • anonymous
ok so is 1/2 the grams?
anonymous
  • anonymous
also if it is then why is it divided by 1
campbell_st
  • campbell_st
well half life means you end up with 1/2 of what you start with... and because you are given X = 5715 you can calculate b
anonymous
  • anonymous
give me a moment...
campbell_st
  • campbell_st
the calculation is \[\ln(\frac{1}{2}) = -b \times 5715\]
anonymous
  • anonymous
i see that but what i don't understand is the 1/2
campbell_st
  • campbell_st
half life if you start with 100 gm then 50gm is the half life, 20gm then 10gm is the half life. 3 gm then 1.5 gm is the half life. so its when the initial quantity gets to half its size.
campbell_st
  • campbell_st
so if you really don't need to know the initial quantity if you know how long it takes to get 1/2 life...
anonymous
  • anonymous
oh so is 1/2 just to solve to get b then plug in to the equation for the 1000 years =x to get the answer...
campbell_st
  • campbell_st
thats it... b will be a positive decimal...
anonymous
  • anonymous
ok let me see if this checks out for a second
anonymous
  • anonymous
one thing were did A go?
anonymous
  • anonymous
oh wait i see
anonymous
  • anonymous
y/a is 1/2
campbell_st
  • campbell_st
the ratio of Y/A = 1/2

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