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Complicated logarithm problem!or at least i think it is

  1. Sep 22, 2006 #1
    I was doing some assignment i have to give in, for math, and came upon this exponential equation: (2^x+1) + (2^x+2) = (2^1-x) + (2^3-x)

    I thought, pfft, thats easy...so i did it, wrong answer, tried something else, wrong answer..tried another tactic, and i think you can guess what happened..Each answer i get is wrong and, to my judment, im using a method of development that should work. but it doesnt. looked easy at first glance but isnt..lol, can anyone start me off on the right track?
  2. jcsd
  3. Sep 22, 2006 #2

    Hi my name is Mace, and i was woundering how do u post something like wat u did?:confused:
  4. Sep 22, 2006 #3
    haha, what do you mean?
  5. Sep 22, 2006 #4

    put up a post?:confused:
    can u tell me?
    i am kinda new.
  6. Sep 22, 2006 #5
    On the main forum page, press "New Topic" in the upper left. TO get to the main forum page from this page, scroll up and press "Introductory Physics". Welcome to PF

    ps - I will work on the main question and get back to you in a minute.
    Last edited: Sep 22, 2006
  7. Sep 22, 2006 #6
    Can you rewrite the equation as
    [tex]2^{x+1}+3=10-2x [/tex] ?

    Does that help?
  8. Sep 22, 2006 #7
    ummm, how did you get that??
  9. Sep 22, 2006 #8
    I should also explain the trick... it's kind of subtle.

    If you have a common base and a common exponent and you want to add them, you MUST have the same number of terms as the base! If you wanted to add [tex]3^{x} + 3^{x} + 3^{x} [/tex] it would be [tex]3^{x+1} [/tex] but of course 3^x + 3^x can't be added. And if you wanted to add 4^x you would need four of them, etc. Kind of a useless trick really, but it helps here!

    Perhaps someone else can explain why it works, I think it's probably simple but I just remember the trick, not the reason. lol
    Last edited: Sep 22, 2006
  10. Sep 22, 2006 #9
    but i dont see where your trick applies here, lol..i have two terms with the base=2..but different exponents.
  11. Sep 22, 2006 #10
    lol! You have to use brackets! I thought it was 2^x + 1 not 2^(x+1) Sorry. lol
  12. Sep 22, 2006 #11


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    If your equation is
    2^(x+1) + 2^(x+2) = 2^(1-x) + 2^(3-x)
    then you have
    2^(x+1) + 2*2^(x+1) = 2^(1-x) + 4*2^(1-x)
    Do you see what to do next?
  13. Sep 23, 2006 #12
    kewl kewl, got it thanks
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