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andreadesirepen

How do I write a half life function of this info? F(x)=45e^-.89x y=45e^-.89x y/45 = e^(-.89x) in y/45 = In e^(-.89x) in y - in 45 = in e^-.89x in y = in e^-.89x+3.81 in y = -.89x+3.81

  • one year ago
  • one year ago

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  1. andreadesirepen
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    HELP! It says it should look like this: F(x)=A(1/2)^x. Where A is the initial amount of the substance and x represents the time for decay. I have A, it´s 25. How do i find x?

    • one year ago
  2. asnaseer
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    if:\[F(x)=45e^{-.89x}\]then the value of F(0) will tell you how much the initial amount is equal to. In this case it would be:\[F(0)=45*e^0=45\]

    • one year ago
  3. asnaseer
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    what you need to find is the value of x to get half this value

    • one year ago
  4. asnaseer
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    i.e. what value of x will give you F(x) = 22.5

    • one year ago
  5. andreadesirepen
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    So it would be, 45(1/2)^22.5 ?

    • one year ago
  6. asnaseer
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    no, you need to solve the equation:\[22.5=45*e^{-0.89x}\]

    • one year ago
  7. asnaseer
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    this will give you a value for x for which F(x) is equal to half its initial value

    • one year ago
  8. andreadesirepen
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    i plugged, 45*e^-.89 into my calculator and got 18.5. is that right?

    • one year ago
  9. asnaseer
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    how does that solve the equation I listed above?

    • one year ago
  10. asnaseer
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    you need to rearrange that equation to get an expression for x

    • one year ago
  11. asnaseer
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    I can show you the first few steps...

    • one year ago
  12. asnaseer
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    so, starting with:\[22.5=45*e^{-0.89x}\]we divide both sides by 45 to get:\[0.5=e^{-0.89x}\]then take logs of both sides to get:\[\ln(0.5)=-0.89x\]can you do the rest?

    • one year ago
  13. andreadesirepen
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    I´m lost sorry, no.

    • one year ago
  14. asnaseer
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    which part are you stuck on?

    • one year ago
  15. andreadesirepen
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    what to do next

    • one year ago
  16. asnaseer
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    do you understand that you need to find x?

    • one year ago
  17. asnaseer
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    did you follow the reasoning above?

    • one year ago
  18. andreadesirepen
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    i understand i need to find x, but i dont understand how

    • one year ago
  19. asnaseer
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    If I said:\[12=4x\]then would you be able to find the value of x here?

    • one year ago
  20. andreadesirepen
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    yes

    • one year ago
  21. andreadesirepen
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    3

    • one year ago
  22. asnaseer
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    good, so now use the same principals to solve this for x:\[\ln(0.5)=-0.89x\]

    • one year ago
  23. andreadesirepen
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    i got -.56

    • one year ago
  24. asnaseer
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    tat does not look right to me - please show me your steps so that I can check where you may have made a mistake

    • one year ago
  25. asnaseer
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    *that

    • one year ago
  26. andreadesirepen
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    i divided .5 by -.89 or is it the other way around?

    • one year ago
  27. asnaseer
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    you cannot do that - the left-hand-side of the equals sign has \(\ln(0.5)\) not just 0.5

    • one year ago
  28. asnaseer
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    so first find the value for \(\ln(0.5)\), then divide that by -0.89

    • one year ago
  29. andreadesirepen
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    .78

    • one year ago
  30. asnaseer
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    that looks more like it.

    • one year ago
  31. andreadesirepen
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    is that x?

    • one year ago
  32. asnaseer
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    yes - you can double check that you have the right value for x by substituting it into your original equation:\[F(x)=45e^{-.89x}\]and see if you get roughly 22.5 as the answer.

    • one year ago
  33. andreadesirepen
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    why would my A be 45?

    • one year ago
  34. andreadesirepen
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    45(1/2)^.78

    • one year ago
  35. asnaseer
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    In the function given to you, the value of the function when x=0 is 45.

    • one year ago
  36. andreadesirepen
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    is that the half life function?

    • one year ago
  37. asnaseer
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    no

    • one year ago
  38. andreadesirepen
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    oh mah gaaahh. this is tuff

    • one year ago
  39. asnaseer
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    In general you ay have an exponentially decaying function defined as:\[F(x)=Ae^{-bt}\]

    • one year ago
  40. asnaseer
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    in this function t represents the time

    • one year ago
  41. asnaseer
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    and you can see (I hope) that when t=0 F(0)=A

    • one year ago
  42. asnaseer
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    so A represents the initial amount - i.e. the amount at the start

    • one year ago
  43. asnaseer
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    as time passes, the mount decays and gets less and less

    • one year ago
  44. asnaseer
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    *amount

    • one year ago
  45. asnaseer
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    at some point, it decays to half its original value

    • one year ago
  46. asnaseer
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    so, at that point F(t) = A/2

    • one year ago
  47. asnaseer
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    now, since we know that:\[F(t)=Ae^{-bt}\]then we can replace F(t) by A/2 to get:\[\frac{A}{2}=Ae^{-bt}\]divide both sides by A to get:\[\frac{1}{2}=e^{-bt}\]take logs of both sides to get:\[\ln(0.5)=-bt\]divide both sides by -b to get:\[t=-\frac{\ln(0.5)}{b}\]this is what the half life is

    • one year ago
  48. asnaseer
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    it represents the amount of time that has to elapse before the amount of material decays to half its original value

    • one year ago
  49. andreadesirepen
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    I understand the process but it tells me that it needs to look like this: A(1/2)^x

    • one year ago
  50. asnaseer
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    I don't know where you are getting that from - it makes no sense to me.

    • one year ago
  51. andreadesirepen
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    The half-life of a substance is the time it takes for half of the substance to decay. The exponential function representing half-life is f(x) = A(1/2) where A is the initial amount of the substance and x represents the time for decay. The half-life of the substance will depend on the initial amount of substance you have. In this activity, you will experience this formula in action.

    • one year ago
  52. andreadesirepen
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    I had to use pennies in the beggining if that helps, so i thought that was A, i used 25 pennies.

    • one year ago
  53. asnaseer
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    where you wrote: f(x) = A(1/2) that corresponds to what I said above when I said that at its half life: so, at that point F(t) = A/2

    • one year ago
  54. asnaseer
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    what this is saying is that when the material (whatever it may be) reaches its half life, then the value of the function f(x) will be equal to HALF the value of f(0). and we know that f(0) = A so, at its half life, f(x) = A/2

    • one year ago
  55. andreadesirepen
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    that´s the function? why doesnt it look like f(x) = A(1/2)^x ?

    • one year ago
  56. asnaseer
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    it is saying that the VALUE of the function will equal A/2

    • one year ago
  57. asnaseer
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    e.g. if I had a function defined as:\[f(x)=x^2\]and I asked you to find the value of x for which f(x) = 16, then you would do:\[16=x^2\]therefore:\[x=\sqrt{16}=4\]

    • one year ago
  58. asnaseer
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    so note here that I said you need to find the value of x for which: f(x) = 16

    • one year ago
  59. asnaseer
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    so here we are saying that the VALUE of f(x) should be equal to 16

    • one year ago
  60. asnaseer
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    regardless of how the function itself is defined

    • one year ago
  61. andreadesirepen
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    So how will the half life equation of my probelm look? like this f(x) = A/2 ?

    • one year ago
  62. asnaseer
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    the HALF LIFE is defined as the point at which the VALUE of the function is equal to half its initial value - i.e. the point at which f(x) = A/2

    • one year ago
  63. asnaseer
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    this is not stating that the FUNCTION definition is f(x) = A/2 instead it is stating that the VALUE of f(x) = A/2 and you need to find a suitable x for this to be true

    • one year ago
  64. asnaseer
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    I need to go now - it is very late here and I need some sleep - hope you understand this concept better now.

    • one year ago
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