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Tricky exponential question

  1. Sep 17, 2009 #1
    1. The problem statement, all variables and given/known data
    for x and y satisfying x-y=2 and 2[tex]^{x}[/tex]-2[tex]^{y}[/tex]=6
    find 2[tex]^{x}[/tex]+2[tex]^{y}[/tex].






    3. The attempt at a solution
    x-y=2 ; x=2+y
    log[tex]_{2}[/tex] 2[tex]^{x}[/tex]-log[tex]_{2}[/tex]2[tex]^{y}[/tex]=log[tex]_{2}[/tex]6

    log 6/log2=2.585

    x-y=2.585 but this violates the intitial condition x-y=2

    where am i going wrong?
     
  2. jcsd
  3. Sep 17, 2009 #2
    I don't think you need to use logs.

    x-y=2
    x=y+2

    2^(y+2) - 2^y = 6

    I think you should be able to figure it out from that.
     
  4. Sep 17, 2009 #3
    Yes, use substitution. Also
    log2(a + b) [itex]\neq[/itex] log2a + log2b
     
  5. Sep 17, 2009 #4
    I'm not seeing what to do. I see the substitution but then I don't know what to do.
     
  6. Sep 18, 2009 #5
    Let's take
    x - y = 2
    x = y + 2

    substitute into your equation

    2y+2 - 2y = 6
    (2y)(22) - 2y = 6
    2y(4-1) = 6
    ...

    You don't even have to solve for x and y explicitly.
     
  7. Sep 18, 2009 #6
    thank you very much.
     
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