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Math Help - Right track?

  1. #1
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    Lightbulb Right track?

    980=75(1-75e^(-.07t) find t
    Do I divide 75 from each side to give me
    980/75=1-75e^(.07t)

    How do I bring the 1-75 to the other side now. The book I have on explaining this sucks big time.

    Any clue?

    Ok I came up with something?
    now I divide both sides by 1-75 which gives me
    (980/75)/1-75 = e^ -.07t
    Then I multiply both sides by Ln which gives me
    ln(980/75)/(1-75) = -.07t
    then just divide again by -0.07 to isolate t.

    Now if it makes sense not sure.

    I got an answer of 0.9252???
    Last edited by bradycat; December 4th 2010 at 01:26 PM.
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Detroit, MI
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    Quote Originally Posted by bradycat View Post
    980=75(1-75e^(-.07t) find t
    Do I divide 75 from each side to give me
    980/75=1-75e^(.07t)

    How do I bring the 1-75 to the other side now. The book I have on explaining this sucks big time.

    Any clue?
    980=75\left(1-75e^{-.07t}\right)

    \frac{980}{75}=1-75e^{-.07t}

    \frac{980}{75}-1=-75e^{-.07t}

    \frac{\frac{980}{75}-1}{-75}=e^{-.07t}

    Now since \frac{\frac{980}{75}-1}{-75}<0, and e^{-.07t}>0 for all values of t. We conclude that there is no solution.
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  3. #3
    Member
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    Awesome pic of the cat.
    Ok so on line 4, this is where you apply Ln to each side, and being the left side is negative, you can get a number for it. Ln or Log can never be negative numbers before using the function then???
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