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Non Calc 3 x √27 = 3^N

  1. Mar 31, 2007 #1
    Hi how do I go about finding out what N is to
    3 x √27 = 3^N

    this can be written as

    3 x (√3 x √3 x √3)
    which makes
    9√3 = 3^N

    now where though

    i presume n must be a fraction somthing like x/2 ? x is a number? Is this right. How do i go about finding the answer?
     
  2. jcsd
  3. Mar 31, 2007 #2

    Hurkyl

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    Of only you could rewrite the equation as 3^{something} = 3^N, it would be obvious...
     
  4. Mar 31, 2007 #3
    huh? Sorry I dont see...was that a suggestion/hint or comment

    as ive got a root it does the n have a 2 as it's denominator?
     
    Last edited: Mar 31, 2007
  5. Mar 31, 2007 #4
    An eager student would take the logs of both sides. A clever student would notice something special about this question enabling him to skip taking logs and immediately solve an equation which would give him his answer.

    If you're not that clever student, take logs. Solve from there. Then see if you can see what the clever student would do.
     
  6. Mar 31, 2007 #5
    GCSE students dont learn logs. This is on a non-calculator paper. I cant remeber what to do. Ive looked in my textbook, revision book, bitesize, s-cool.co.uk and cant see somthing to help... so I turn here. Can someone show me the steps or continue my work so far? I'm doing a pratice paper and their designed to show you what you dont know.... I searched books and web and had no luck :(
     
  7. Mar 31, 2007 #6
    Since you can have LHS in terms of [itex]\ \sqrt{3}[/itex]why not try expressing RHS as [itex] \ ( \sqrt{3} )^{something} [/itex]
     
  8. Mar 31, 2007 #7

    cristo

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    Well, you're forced to be the clever student then, aren't you! Hurkyl's post was a hint. To put it more blatantly-- try writing the left hand side as 3 to the power of something. Can you write sqrt(3) as 3^x, for some x? Can 9 be expressed as 3 to some power? Finally, can you express a product of two numbers with the same base as one number?
     
  9. Mar 31, 2007 #8
    Interestingly enough you could write it 9√3 OR you could write it as √81 x √3... Try going on from that approach. Oh, and remember √x = x^(1/2).
     
  10. Mar 31, 2007 #9

    Gib Z

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    I have no idea how Feldoh's first comment helps, but yes his second does. BIG Hint. Have you learned your Index Laws?
     
  11. Mar 31, 2007 #10
    If you haven't got it yet, I'll give you a more explicit hint. x->3^x is a well defined map, ie, if x=y then 3^x=3^y.
     
  12. Apr 1, 2007 #11

    HallsofIvy

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    Finally, if you still don't understand,
    [tex]\sqrt{a}= a^{\frac{1}{2}}[/tex].
     
  13. Apr 1, 2007 #12

    Gib Z

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    Even More Finally if you still don't get it, give up.

    N=2.5
     
  14. Apr 1, 2007 #13
    I still dont see...

    I dont see what that means...

    I know that [tex]\sqrt{a}= a^{\frac{1}{2}}[/tex] but because I have a whole number and a surd im not sure what to do. Can someone just WRITE OUT the steps they did one by one using latex?

    Thx
     
  15. Apr 1, 2007 #14

    Gib Z

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    [tex]3\sqrt{27} = 3^n[/tex]
    [tex]\sqrt{27}=3\sqrt{3}[/tex]
    Therefore:
    [tex]9\sqrt{3}=3^n[/tex]
    [tex]3^2 \cdot 3^{\frac{1}{2}} = 3^n[/tex]
    Since [tex]a^m \cdot a^n = a^{m+n}[/tex], then
    [tex]3^2 \cdot 3^{\frac{1}{2}} = 3^{2.5}[/tex]
    [tex]3^{2.5} = 3^n[/tex]
    [tex]n=2.5[/tex]
     
  16. Apr 1, 2007 #15
    NICE! Yeh I see now...ingenius.
    Thanks for all the help :redface:
     
  17. Apr 1, 2007 #16

    HallsofIvy

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    No, not ingenious- just basic formulas. Long before you attempt a problem like this you should have learned
    [tex]a^x a^y= a^{x+ y}[/tex]
    and
    [tex](a^x)^y= a^{xy}[/tex]

    (I am not saying that Gib Z isn't ingenious- just that it wasn't needed here!)
     
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