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Simple alg. problem

  1. Aug 26, 2013 #1

    462chevelle

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    1. The problem statement, all variables and given/known data

    (k^(1/5))/k^4/5 * k^-1

    2. Relevant equations



    3. The attempt at a solution step 1(k^(1/5))/k^4/5 * k^-1
    step 2: (k^1/5)/(k^4/5) * (k^5/5) changed from neg exp. 1/k^1 is implied there
    step 3: (k^1/5)/(k^9/5)
    answer: 1/(k^8/5)
    I have to be making a stupid mistake. but this is such a simple problem.. can anyone help me figure out where im messing this up?
    thanks
     
  2. jcsd
  3. Aug 26, 2013 #2

    Mark44

    Staff: Mentor

    Looks OK to me if I'm reading your problem correctly, which I assume is this:
    $$ \frac{k^{1/5}}{k^{4/5}} k^{-1}$$

    This simplifies to k-8/5, which is equal to what you have.
     
  4. Aug 26, 2013 #3

    462chevelle

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    I may have typed that wrong but off the top of my head I don't see where it would matter. but the k^-1 is in the denominator after k^4/5. theyre multiplied.
     
  5. Aug 26, 2013 #4

    HallsofIvy

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    I have no idea what this is supposed to mean. a/b/c is NOT standard mathematics notation. If I try to interpret the "/" as division it could still be interpreted as (a/b)/c or as a/(b/c) which are NOT the same thing.

    If you mean (k^(1/5)/k^4)/5k^-1 then k^(1/5)/k^4= k^(1/5- 4)= k^(1/5- 20/5)= k^(-19/5). Then (k^(1/5)/k^4)/5k^-1= (k^(-19/5)/5k^-1= (1/5)k^(-19/5+1)= (1/5)k^(-14/5).

    But k^4/5k^-1= (1/5)k^(4+1)= (1/5)k^5. So k^1/5/(k^4/5k^-1)= (k^1/5)/((1/5)k^5= 5 k^(1/5- 5)= 5k^(1/5- 25/5)= 5k^(-24/5)

     
  6. Aug 26, 2013 #5

    462chevelle

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    disregard this. I donno what i was thinking>>>>>>>wait. I just redid it like that and it is 1/k^4/5 I think
    second step would be
    (k^1/5)/(k^4/5)(1/k)
    third would be (k^1/5)/((k^4/5)/(k^5/5))
    is that correct for that format?
     
    Last edited: Aug 26, 2013
  7. Aug 26, 2013 #6

    462chevelle

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    those are fraction exponents. I don't know how to use latex
     
  8. Aug 26, 2013 #7

    462chevelle

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    here this is crude but it is my first attempt
     

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  9. Aug 26, 2013 #8

    Mark44

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    Apparently this is the problem:
    $$\frac{k^{1/5}}{k^{4/5}k^{-1}} $$

    Simplify the denominator first. What do you get? Note that k-1 ≠ k5/5.
     
  10. Aug 26, 2013 #9

    462chevelle

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    here is another attempt. though I don't think its correct. the first attempt seems correct to me.
     

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  11. Aug 26, 2013 #10

    Mark44

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    That's correct, but you're taking the long way in getting to it.

    Instead of writing k-1 as 1/k in the denominator, simply add the exponents to get k-1/5 in the denominator. The idea is that ar * as = ar + s.
     
  12. Aug 26, 2013 #11

    462chevelle

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    does it not equal 1/k^5/5 though? because it would equal 1/k^1?? I was just using 5/5 for the lcd of the denominator but ill try again.
     
  13. Aug 26, 2013 #12

    462chevelle

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    so this is how youre saying I should do it?
     

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  14. Aug 26, 2013 #13

    462chevelle

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    put it in and that was correct. thanks! knew it was a stupid mistake.
     
  15. Aug 26, 2013 #14

    haruspex

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    You appear to have turned ##k^{-1}## into ##k^{\frac 55}##.
    LaTex is pretty easy for something like this. Type a double hash symbol (two times '#') at each end of the expression. To raise to a power use ^, as you did in the OP, but if the exponent consists of more than one item, as here, you need to wrap it in curly braces, {}. Fractions take the form \frac{ }{ }, with the numerator inside the first curly braces and the denominator inside the second pair. Again, you can omit a pair of braces if there's only one item inside. Click the Quote button to see how I wrote it above. Always click Preview Post before submitting a Post with LaTex.
    One gotcha - if you notice a mistake after posting and click Edit, when you submit ("save") the edited form it will not interpret the LaTex on your screen. The post is probably ok, but it will look wrong to you. Just do a screen refresh.
     
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