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Homework Help: Simplify and express with positive indices

  1. Jun 13, 2010 #1
    Ok, so I have this Expression, (6x+2 x 42x-4 x 35-x x 2x-6)/(124x+3 x 92x-3)

    But it needs to be simplifyed, and expressed with positive indices. Now as far as my knowledge takes me, this can't be simplifyed, but then I am probably wrong. If anybody could help to solve, that would be great.
  2. jcsd
  3. Jun 13, 2010 #2


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    maybe you could start by trying breaking things down into prime components eg.
    [tex] 6^{x+2} = 3^{x+2}2^{x+2}[/tex]
  4. Jun 13, 2010 #3
    I'm sorry but that example made no sense to me at all? I need more explanation.
  5. Jun 13, 2010 #4


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    well say you have, separate powers of 2 multiplied toegther, you can simplify by
    [tex] 2^{a}2^{b} = 2^{a+b}[/tex]

    so break down each number into its prime components, then multiply similar numbers together
    6 = 3.2
    4 = 2.2 etc.
  6. Jun 13, 2010 #5


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    The rules of exponents,

    [itex]a^xa^y= a^{x+y}[/itex] (which lanedance gave) and [itex](a^x)^y= a^{xy}[/itex], are usually learned long before taking Calculus!

    You will also need to know that [itex]4= 2^2[/itex], [itex]6= 3(2)[/itex], [itex]9= 3^2[/itex], and [itex]12= 2^2 (3)[/itex]. Break everything down into powers of 2 and 3 and use the laws of exponents.
  7. Jun 13, 2010 #6
    Could somebody please do this as an example, this is just the first one in the list I have to do.

  8. Jun 13, 2010 #7


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    you've got all the info, why not try it out?
  9. Jun 13, 2010 #8
    All this made no sence to me either. Can someone show an example?
  10. Jun 13, 2010 #9


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    do for the first terms
    [tex] 6^{x+2}.4^{2x-4} = (3.2)^{x+2}.(2.2)^{2x-4}
    = (3^{x+2}.2^{x+2})(2^{2x-4}. 2^{2x-4})
    = (3^{x+2})(2^{x+2}.2^{2x-4}. 2^{2x-4})
    = (3^{x+2})(2^{(x+2) + (2x-4) + (2x-4)})
    = (3^{x+2})(2^{5x+6}) [/tex]

    then for terms that are divided, the power will be negative
  11. Jun 13, 2010 #10
    sorry to be a bother but, that really just confused me even more.
    If u can could you please show the whole example, and explain everything?
    That would help heeps.
  12. Jun 13, 2010 #11


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    i think you've got way more than enough to attempt something... even though its not your post, have a try & i'll guide you the through

    basically all you need to do is the same for the last 4 terms
    [itex] 3^{5-x} [/itex] - this is in lowest form
    [itex] 2^{x-6} [/itex] - this is in lowest form
    [itex] \frac{1}{12^{4x+3}} = 12^{-4x-3} = (2.3.3)^{-4x-3}[/itex]

    and then i've leave the last for you... then multiply all the terms together & collect exponents
  13. Jun 13, 2010 #12
    There may be heeps on info, but it makes no sence to me, thats y i wanted an example so i could attempt questions just like that.
  14. Jun 14, 2010 #13


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    ok so what doesn't make sense?

    post #9 is in effect a simple example
  15. Jun 14, 2010 #14
    the whole thing makes sence, can u tell me what answer u got, cus i got 1.5x^3.
    Is this correct???
  16. Jun 14, 2010 #15
    wait thats not it, never mind.
    can u please show me how u did ur working out cus im completly stuck?
  17. Jun 14, 2010 #16


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    I have in post 9..

    part of the confusion may be how its written:
  18. Jun 14, 2010 #17
    sorry but thats even more confusing. can u show me how u worked it out?
  19. Jun 14, 2010 #18
    [tex] \frac{6^{(x + 2)} \ \cdot \ 4^{(2x - 4)} \ \cdot \ 3^{(5 - x)} \ \cdot \ 2^{(x - 6)}}{12^{(4x + 3)} \ \cdot \ 9^{(2x - 3)}} [/tex]

    Is that the equation??? (the dot means multiplication).

    Start off by recognizing that you can do things like;

    [tex]6^{(x + 2)} = 6^x \cdot 6^2 [/tex]

    Play with this one first, compare the values of the Left Hand Side & The Right Hand side by giving different x values, you'll see something surprising, then generalise this idea.
  20. Jun 14, 2010 #19
    We can continue to try to make sense of what doesn't make sense to you. Which concept is confusing: prime factorization or the laws of exponents alluded to by HallsofIvy and lanedance?
  21. Jun 14, 2010 #20
    This will sound dumb but i dont even really know wat that means, thats y i wanted the answer, or the working so i can use it withother similar questions.
  22. Jun 14, 2010 #21
  23. Jun 14, 2010 #22
  24. Jun 14, 2010 #23
    I think you should watch this video

    http://khanexercises.appspot.com/video?v=zM_p7tfWvLU [Broken]

    It's number 53 in the list underneath it, if you understand it then watch numbers 54 & 55 after it.

    I think it will help you out here a bit.

    Come back after you watch these and read people's comments again, I bet everything will be clearer.

    Then you should watch the videos on logarithms, ( I think I saw you in another thread with logarithm questions)
    Last edited by a moderator: May 4, 2017
  25. Jun 14, 2010 #24
    the video helped a little, but not with this question. it just makes me more fustrated. i cant do it no matter how many time u interpret it.
  26. Jun 14, 2010 #25
    If you watched 53,54 & 55 and then look at my first post here that should explain everything.

    You've also got about 3 other ways to do this problem just given to you from people on this thread.

    I can only guess that you haven't studied this stuff before or haven't looked at a good book explaining it.

    There are these notes here = http://tutorial.math.lamar.edu/Classes/Alg/Alg.aspx that explicitly taught me how to do these problems ages ago, just read the 'Preliminaries' chapter, but really all you need to do is study this page here http://tutorial.math.lamar.edu/Classes/Alg/IntegerExponents.aspx that is from that link, I've just checked & the method to do this problem is taught explicitly.

    If you give studying this page a shot & still can't do it then I'll show you step by step but you have to give this a shot first to learn for yourself.

    I thought the video would just remind you of something you'd learned but you must not have ever seen this before, (or just understood it).

    The above chapter I gave you should be 100% all you need to master this problem.
    Last edited by a moderator: Apr 25, 2017
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