MATH TIME!

Discussion in 'User Topics' started by Mr.Lame, Jan 9, 2018.

  1. Mr.Lame

    Mr.Lame Veteran

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    BUT I do get it, so seperate for scientific notation.
     
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  2. MarkFL

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    lame008.png
     
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  3. Mr.Lame

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  4. Mr.Lame

    Mr.Lame Veteran

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    @MarkFL

    (1/8)^-4/3

    SO This time I didn't make a mistake on the question hahaha

    do I make it a cube root ?
     
  5. MarkFL

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    I would first use:

    1/8 = 2^(-3)

    like so:

    (1/8)^(-4/3) = (2^(-3))^(-4/3)

    And now multiply the exponents:

    (1/8)^(-4/3) = (2^(-3))^(-4/3) = 2^4 = 16
     
  6. Mr.Lame

    Mr.Lame Veteran

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    A bit lost on the (2^(-3))^(-4/3)
     
  7. MarkFL

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    1/8 = 1/(2^3) = 2^(-3)

    Is that the part that needed clarification?
     
  8. Mr.Lame

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    1/8 = 2^(-3) < I don't understand how you get this any chance you could break that down.

    From a previous I'd root with the denominator [(ex. 3/2) Square root (4)= 2^3)] and use the numerator as the exponent outside the root
     
  9. Mr.Lame

    Mr.Lame Veteran

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    So if I did this.

    Cuberoot of 1/8 = 1/2^-4 = 16
     
  10. MarkFL

    MarkFL Guest

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    We have:

    1/8

    Now, if we recognize that 8 is the cube of 2, that is 8 = 2^3, then we can write:

    1/8 = 1/(2^3)

    Now, when an exponent is down in the denominator, we can bring that term up into the numerator by negating the exponent

    1/8 = 1/(2^3) = 2^(-3)

    We can think of this as:

    1/(2^3) = (2^0)/(2^3) = 2^(0 - 3) = 2^(-3)
     
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  11. MarkFL

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    Yes, there's lots of ways to go about it...I try to show the method that I think will be the most useful in general. :)
     
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  12. Mr.Lame

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    I like that because there may be a situation where this method may not work!
     
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  13. Mr.Lame

    Mr.Lame Veteran

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    Hey @MarkFL,

    When there are exponents inside a root, you divided by the root?

    example 4root (x^8y^12) would be x^2y^3 right? Or am I doing it wrong?
     
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  14. MarkFL

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    Yes, exactly:

    lame009.png
     
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  15. Mr.Lame

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  16. MarkFL

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    Just a note here...the simplification we stated only works for a non-negative value of y. :)
     
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  17. Mr.Lame

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    O: Good to know, I'm sure I'm going to be seeing a negative one soon, so be prepared hahaha!
     
  18. MarkFL

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    Yes, technically, we should write:

    lame010.png
     
  19. Mr.Lame

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    Is it in ABS because it's neg?
     
  20. MarkFL

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    The short story here is that we started out with an even exponent on y and ended up with an odd exponent, and when a negative number has an odd exponent, the result is a negative value. But, when a negative number has an even exponent, the result is positive So, if y is negative, the original expression is positive, and we need to insure the end result is too. :)
     
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