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Math Help - Help with the inverse of a exponential decay (i think?) graph

  1. #1
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    Help with the inverse of a exponential decay (i think?) graph

    Hi, I'm new here, I also don't know if this is in the right section.
    I'm not looking for anyone to fully answer this, I just am wondering what do to with it basically. I know the core ideas about exponential graphs, y = yi * (1 + r)^x where yi is initial value and x is time. but its clear that the graph below has to be an inverse. something around the lines of y = (log x)/(log .99) or something...

    yea, i'm really lost. my ultimate question is questiong #3 in the picture below,
    how would i find an equation or model to represent the two curves below?
    Thanks and sorry again if this is in the wrong section.


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  2. #2
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    Re: Help with the inverse of a exponential decay (i think?) graph

    Quote Originally Posted by tbracket View Post
    Hi, I'm new here, I also don't know if this is in the right section.
    I'm not looking for anyone to fully answer this, I just am wondering what do to with it basically. I know the core ideas about exponential graphs, y = yi * (1 + r)^x where yi is initial value and x is time. but its clear that the graph below has to be an inverse. something around the lines of y = (log x)/(log .99) or something...

    yea, i'm really lost. my ultimate question is questiong #3 in the picture below,
    how would i find an equation or model to represent the two curves below?
    Thanks and sorry again if this is in the wrong section.


    In item one, I found that the coal would decrease linearly at a rate of y = -5x + 2500. I don't know if that helps
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  3. #3
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    Re: Help with the inverse of a exponential decay (i think?) graph

    You'ld better take a harder look at those charts. That is the amount of coal remaining in the resevoir. They don't much look linear.

    Example:

    Start with Coal = 100 (some units)
    1st year usage = 10
    Available for year 2= 100 - 10 = 90
    2nd year usage = 10 * 1.03 = 10.3 -- A 3% increase
    Available for year 3 = 90 - 10.3 = 79.7
    3rd year usage = 10 * 1.03^2 = 10.609
    Available for year 3 = 79.7 - 10.609 = 69.091
    4th year usage = 10 * 1.03^3 = 10.927
    Available for year 4 = 69.091 - 10.927 = 58.164

    There is nothing linear about it.

    The USAGE is growing exponentially. There should not be any straight lines associated with this model.
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