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E^(2lnt) = t^2 Whyyy?

  1. Sep 13, 2005 #1

    CinderBlockFist

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    ok guys, i dont see how e^2ln|t| = t^2 can someone explain it to me please? Seems so easy but i dont see it.
     
    CinderBlockFist, Sep 13, 2005
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  3. Sep 13, 2005 #2

    CinderBlockFist

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    ok, so far i tried this. I know e^(lnx) = x


    so, i broke the e^(2ln|t|) into to parts:


    e^2 times e^(ln|t|) which equals t (from the top identity)


    so im left with e^2 times t. which is te^2

    but the book says it equals t^2..so what happened to the e? (exponential function)
     
    CinderBlockFist, Sep 13, 2005
  4. Sep 13, 2005 #3

    Brad Barker

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    you have

    e^2 e^(ln t) = e^(2+ln t),

    which is incorrect.


    you want to use the properties:


    a ln b = ln b^a

    and

    e^ln a = a.


    the rest should be straightfoward.
     
    Last edited: Sep 13, 2005
    Brad Barker, Sep 13, 2005
  5. Sep 13, 2005 #4

    Tom Mattson

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    Staff Emeritus
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    First look at this rule for logarithms:

    [itex]\log_b(a^x)=x\log_b(a)[/itex].

    Now apply that to your exponent:[itex]2\ln |t|[/itex]. What do you get?

    Then note that [itex]f(x)=\ln (x)[/itex] and [itex]g(x)=e^x[/itex] are inverse functions, which means that [itex]f(g(x))=g(f(x))=x[/itex].

    Those two rules together will give you the answer.
     
    Tom Mattson, Sep 13, 2005
  6. Sep 13, 2005 #5

    CinderBlockFist

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    SWEEET! THANK YOU GUYS. I got my laws of exponents mixed up. :yuck:
     
    CinderBlockFist, Sep 13, 2005
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