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Math Help - Am I allowed to do this?

  1. #1
    Junior Member ImaCowOK's Avatar
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    Am I allowed to do this?

    Put into Cartesian form with x>0

    x=5e^t y=3e^{-t}

    I solved for t:

    t=ln_e(x/5)

    Found the Cartesian:

    y=3e^{-ln_e(x/5)}

    This is how I got the final answer of 15/x and I'd like to know if, mathematically, I'm allowed to do it this way(I might leave out some steps, but I think you will understand. Let me know if you want me to add those steps):

    y=3/e^{ln(x/5)}

    Multiply both sides by e^{ln(x/5)}

    Change to the form:  yln_e(x/5) = ln_e3

    Divide both sides by ln_e

    (yx)/5 = 3

    Multiply both sides by 5, then divide both sides by x.

    Final answer 15/x

    The correct answer is 15/x

    Sorry about not using MATH tags all the way through. I'm under time constraints and am not very familiar/savvy with MATH tags.
    Last edited by ImaCowOK; October 20th 2010 at 02:57 PM.
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  2. #2
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    If you have more than one character in an exponent, set off the whole exponent in braces.
    [tex]e^{-t}[/tex] gives  e^{-t} instead of  e^- t .
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  3. #3
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    x = 5 e^t

    \frac{x}{5} = e^t

    \frac{5}{x} = e^{-t}


    3e^{-t} = 3 \cdot \frac{5}{x} = \frac{15}{x}
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  4. #4
    Junior Member ImaCowOK's Avatar
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    I don't understand what you did there. Just posting that without an explanation doesn't help. Was the way I solved for the final answer wrong? That's what I'm asking. Are you showing a simpler and quicker way to do it? Those are welcome too, but I need to understand what you did to get there.
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  5. #5
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    Quote Originally Posted by ImaCowOK View Post
    Was the way I solved for the final answer wrong?
    What you did is correct.
    It is a matter of choice the way to do it.
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  6. #6
    Junior Member ImaCowOK's Avatar
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    Okai, thanks cupcake.
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  7. #7
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    Quote Originally Posted by ImaCowOK View Post
    Okai, thanks cupcake.
    Your solution is actually partially correct - you don't actually "divide by ln_e" because it is not a number.

    Also, ln denotes the logarithm to the base of e, so writing ln_e is redundant.

    Generally, if you have lnx = lny then you can raise e to both sides of the equation to get e^{lnx} = e^{lny} and since e^{lnx} = ln(e^x) = x, you get that x=y.

    Note that you could've actually finished in the third line using logarithm \ exponent rules:

    y=3e^{-ln(x/5)} = 3e^{ln(\frac{5}{x})} = 3 \cdot \frac{5}{x} = \frac{15}{x}
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