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Math Help - Exponential problem

  1. #1
    Junior Member
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    Post Exponential problem

    Hi,

    I'm stuck in this problem. Can somebody help me?

    After discontinuing all advertising for a certain product in 1994 ,the manufacturer noted that sales began to drop according to the model

    s = 500,000/(1+.6e^(kt))

    where s represents the number of units sold and t = 0 represents 1994. In 1996, the company sold 300,000 units

    a) complete the model by solving for k
    b) estimates sales in 1999

    Thank you ...(really!)

    Isabel
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by cuteisa89 View Post
    After discontinuing all advertising for a certain product in 1994 ,the manufacturer noted that sales began to drop according to the model

    s = 500,000/(1+.6e^(kt))

    where s represents the number of units sold and t = 0 represents 1994. In 1996, the company sold 300,000 units

    a) complete the model by solving for k
    i suppose t is the time in years.

    in 1996, the company sold 300,000 units, that means when t = 2, s = 300,000. now we can use this to solve for k

    => 300000 = 500000/(1 + 0.6e^(2k))
    => 300000(1 + 0.6e^(2k)) = 500000
    => 1 + 0.6e^(2k) = 500000/300000 = 5/3
    => 0.6e^(2k) = 5/3 - 1 = 2/3
    => e^(2k) = 2/3*10/6 = 10/9 ...................0.6 = 6/10 so i multiplied both sides by 10/6

    now we take the log of both sides

    => lne^(2k) = ln(10/9)
    => 2k*ln(e) = ln(10/9)
    => 2k = ln(10/9) ........................lne = 1
    => k = ln(10/9)/2 = 0.0527


    so the complete model is s = 500,000/(1 + 0.6e^(0.0527t))


    b) estimates sales in 1999
    in 1999, t = 5

    => s = 500,000/(1 + 0.6e^(0.0527(5)))
    => s = 500000/(1 + 0.7809)
    => s = 280756.92 units
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