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Exponential growth/decay question

  1. Mar 8, 2005 #1
    hi all! i hope ur all doing great! :rolleyes:

    ok my question is kind of simple but hard for me. the thing is that im trying to get a head start on a concept that we will soon solve aldebraically in my class. the reason why i want to get a head start is because if i get a pretty good idea of a concept prior to the actual going over it in class, i'll get it down much quicker and better, because im not that good of a math whiz! :cry: i just want to get this so that when it comes time for the teacher to show us, i will understand it much more better! thanks!

    ok so heres a general example:

    The estimated population of planet Kee2000 was 5,900,000 in year 2000, and 6,100,000 in year 2003.

    Assume that population is growing EXPONENTIALLY. SEt up variables and write a forumula for the populatioon as a function of time.

    i already did this part and got this:

    (read as "y of x" here ) y(x)= 5,900,000 (1.034)^(x/3)

    i worked it out to get an approximate value for "m" and that's what i got, 1.034^(1/3)


    now here's what i dont get:

    In how many years will the population equal 7,000,000 people?

    How the heck would you find that one out??

    i tried this:

    7,000,000=5,900,000(1.034)^(x/3)

    here do u divide both sides by 5,900,000 to get:

    about 1.67=1.034^x

    and now im stuck...help! :yuck:
     
    Last edited: Mar 8, 2005
  2. jcsd
  3. Mar 8, 2005 #2

    dextercioby

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    It's a transcendental equation.It can be solved approximately by logarithmation.

    Daniel.
     
  4. Mar 9, 2005 #3
    You're trying to solve:

    [tex]1.67 = 1.034^x[/tex]

    Take the natural log of both sides and you get.

    [tex]ln(1.67) = ln(1.034^x)[/tex]

    Using one of the log rules you can rewrite this as:

    [tex]ln(1.67) = x*ln(1.034)[/tex]

    I think you can solve it from there.


    Jameson
     
  5. Mar 9, 2005 #4

    ZapperZ

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    And this is correct?

    When something is said to grow "exponentially", isn't it usually meant as

    [tex] y(x) = A e^{kx}[/tex]

    where k is the growth rate? This is certainly what we mean when we talk about the exponential decay in radioactivity, etc.

    Zz.
     
  6. Mar 9, 2005 #5
    Yes I agree. I didn't really look at the work when I answered my first post.

    So you use the equation listed above. Remember that A is the initial population.
     
  7. Mar 9, 2005 #6

    dextercioby

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    Zapper,i'm sure this is not a physics question,of course to physicists,exponential decay means ~e^{-x},but the exponential function in mathematics is much more general...The base is arbitrary.Except 1...

    Daniel.
     
  8. Mar 9, 2005 #7
    So are you saying the way to find the solution is not by the forumla

    [tex]y(x) = P e^{kx}[/tex]

    ?

    It makes sense to use "e" because of its very definition which you know...

    [tex]\lim_{x\rightarrow\infty}(1 + \frac{1}{x})^x = e[/tex]

    I don't see why Zapper's way is incorrect.

    Jameson
     
  9. Mar 9, 2005 #8

    ZapperZ

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    If "exponential function" is a general, generic description, then what is the justification to choose the first one over this one?

    .. and maybe my interpretation of it is based on physics, but considering that I haven't seen it (in physics) interpreted any other way..... If I had to choose one, I'd do my version of what I think the question is asking.

    Zz.
     
  10. Mar 9, 2005 #9

    dextercioby

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    Yes,i agree with both.It's up to one's interpretation,because both are correct.

    Daniel.
     
  11. Mar 14, 2005 #10
    yes, thank you! thats the way that seems most relevant right now!


    i'm not sure about the other ones because to put it bluntly, they're WAY OVER MY HEAD right now, although i must admit, i am really intrigued by those two other methods, is there any place where i can read up on them a bit?
     
  12. Mar 14, 2005 #11
    Well, his solution really is in the form [tex]Ae^{kx}[/tex], he just took part of the [tex]k[/tex] into the base, so he's written it instead as [tex]A(e^{3k})^{\frac{x}{3}}[/tex].
     
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