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Is this right? e multipled out oddly

  1. Nov 18, 2005 #1

    Pengwuino

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    [tex]
    \frac{{{\rm e}^{{\rm 2t}} }}{{\sqrt {e^{4t} + t} }}*2e^{2t} = \frac{{{\rm 2e}^{{\rm 4t}} }}{{\sqrt {e^{4t} + t} }}
    [/tex]
    Is that true???? What am i not realizing here...
     
    Last edited: Nov 18, 2005
  2. jcsd
  3. Nov 18, 2005 #2
    you left out the factor of 2 in the numerator
     
  4. Nov 18, 2005 #3

    Pengwuino

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    oops yah i did... ill fix that

    but is that true? I can't seem to grasp it... and its midnight so i dont think i will be able to easily realize whats going on

    I was thinking that it should just be 3e^2t.... and I don't understand how it just got turned into e^4t
     
  5. Nov 18, 2005 #4

    uart

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    No, [tex]e^{2t} + 2 e^{2t}[/tex] would be equal to [tex]3 e^{2t}[/tex].
    [tex]e^{2t} * 2 e^{2t} = 2 (e^{2t})^2 = 2 e^{4t}[/tex]

    Or for an even more straight forward example,
    x + 2x = 3x
    x * 2x = 2x^2
     
    Last edited: Nov 18, 2005
  6. Nov 18, 2005 #5
    youre lucky though, since the [tex]e^x[/tex] function was proven to be exponential, thus it follows the rules of exponents, just like anything else. the rest is just algebra.
    [tex]2^2 * 2^2=4 * 4=16=2^4[/tex]
    or more generally
    [tex]x^a * x^b=x^{ab}[/tex] for all bases that correspond to exponential functions. since the "e" function is exponential, you can use these laws.
    have fun
     
    Last edited: Nov 18, 2005
  7. Nov 18, 2005 #6
    there is a square missing:
    [tex]e^{2t} * 2 e^{2t} = 2 (e^{2t})^{2t} = 2 e^{4t^2}[/tex]
     
  8. Nov 18, 2005 #7

    shmoe

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    [tex]e^{2t}* e^{2t}=e^{2t+2t}[/tex], not [tex](e^{2t})^{2t}[/tex]
     
  9. Nov 18, 2005 #8
    lol i thought that's what the problem was, where did the 2 go? :tongue2:
     
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