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    learning C1 atm and the text book doesnt really go into details about writing equation using indices -_-

    why does 6/x² become 6x⁻²
    and why does 3/2x^-1/2 become 3/2√4

    (so sorry for the second one, im not familiar with making the -1/2 on computer )
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    x^(-n) = 1/(x^n). The - in a power makes it the denominator. The number in front of the x values isn't affected because you are multiplying the function of x^-n by that value.
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    Here's the picture to make it more clear

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    (Original post by ihatePE)
    learning C1 atm and the text book doesnt really go into details about writing equation using indices -_-

    why does 6/x² become 6x⁻²
    and why does 3/2x^-1/2 become 3/2√4

    (so sorry for the second one, im not familiar with making the -1/2 on computer )
    Well by definition x^{-n} = \frac{1}{x^n}, that's something you just have to learn. So \frac{6}{x^2}=6 \times \frac{1}{x^2} = 6x^{-2}. Using the exact same reasoning, \frac{3}{2}x^{-\frac{1}{2}}=\frac{3}{2}\times \frac{1}{x^{\frac{1}{2}}} = \frac{3}{2} \times \frac{1}{\sqrt{x}} = \frac{3}{2\sqrt{x}}
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    (Original post by Plagioclase)
    Well by definition x^{-n} = \frac{1}{x^n}, that's something you just have to learn. So \frac{6}{x^2}=6 \times \frac{1}{x^2} = 6x^{-2}. Using the exact same reasoning, \frac{3}{2}x^{-\frac{1}{2}}=\frac{3}{2}\times \frac{1}{x^{\frac{1}{2}}} = \frac{3}{2} \times \frac{1}{\sqrt{x}} = \frac{3}{2\sqrt{x}}
     x*\frac{1}{x}=1 and similarly,  x*x^{-1}=x^1x^{-1}=x^{(1-1)}=x^0=1 therefore x*\frac{1}{x}=x*x^{-1} ; divide both sides by x, \frac{1}{x}=x^{-1}
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    (Original post by Vikingninja)
    x^(-n) = 1/(x^n). The - in a power makes it the denominator. The number in front of the x values isn't affected because you are multiplying the function of x^-n by that value.
    (Original post by Plagioclase)
    Well by definition x^{-n} = \frac{1}{x^n}, that's something you just have to learn. So \frac{6}{x^2}=6 \times \frac{1}{x^2} = 6x^{-2}. Using the exact same reasoning, \frac{3}{2}x^{-\frac{1}{2}}</b>=\frac{3}{2}\times \frac{1}{x^{\frac{1}{2}}} = \frac{3}{2} \times \frac{1}{\sqrt{x}} = \frac{3}{2\sqrt{x}}
    thanks, i stared at this for 5 minutes and finally got it

    would u say 3/2x ^ -1/2 is the hardest i'll get?
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    (Original post by ihatePE)
    thanks, i stared at this for 5 minutes and finally got it

    would u say 3/2x ^ -1/2 is the hardest i'll get?
    It'll be harder in exams but you will have a while to practice it. You can get stuff like -4/3.
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    (Original post by Vikingninja)
    It'll be harder in exams but you will have a while to practice it. You can get stuff like -4/3.
    to the power of -4/3?

    so i would cube root it and raise it to power of 4 but only for x, not the number before it as it's unaffected by the power? am i right?
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    (Original post by ihatePE)
    thanks, i stared at this for 5 minutes and finally got it

    would u say 3/2x ^ -1/2 is the hardest i'll get?
    No, but these things will get a lot easier over time as you get used to them
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    (Original post by Plagioclase)
    No, but these things will get a lot easier over time as you get used to them
    how do you remember it? do you write notes about little things like this or do you just do questions on it to let it sink in? i feel like i shouldnt study maths like i would with history with tons of notes and stuff
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    (Original post by ihatePE)
    to the power of -4/3?

    so i would cube root it and raise it to power of 4 but only for x, not the number before it as it's unaffected by the power? am i right?
    Well first you would flip it to be 1 over that number raised to the 4/3 (we call this to be the reciprocal). Then you cube root that number while raising it to the 4th power.

    Alternatively, after you take the reciprocal, you can split the 4/3 into 1+1/3 and think of that power as a sum, hence you can separate the variable using laws of indices.
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    (Original post by ihatePE)
    how do you remember it? do you write notes about little things like this or do you just do questions on it to let it sink in? i feel like i shouldnt study maths like i would with history with tons of notes and stuff
    You will do loads of questions. Practice makes perfect. After GCSE maths and 2 A-Level maths subjects, I can safely say I never looked back on my books in terms of notes unless it is a subject I'm really not confident on. You just practice loads of questions and it sinks in.
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    (Original post by RDKGames)
    Well first you would flip it to be 1 over that number raised to the 4/3 (we call this to be the reciprocal). Then you cube root that number while raising it to the 4th power.

    Alternatively, after you take the reciprocal, you can split the 4/3 into 1+1/3 and think of that power as a sum, hence you can separate this using laws of indices.
    that made more sense to me :hoppy:
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    (Original post by ihatePE)
    to the power of -4/3?

    so i would cube root it and raise it to power of 4 but only for x, not the number before it as it's unaffected by the power? am i right?
    You would put it as the denominator first so then its to the power of 4/3, then get the value to the power of 4 and then cube root that. You won't necessarily have to always find a value of x to a power with this stuff.
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    (Original post by ihatePE)
    how do you remember it? do you write notes about little things like this or do you just do questions on it to let it sink in? i feel like i shouldnt study maths like i would with history with tons of notes and stuff
    Lots, and lots, and lots of practice.
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    (Original post by Vikingninja)
    You would put it as the denominator first so then its to the power of 4/3, then get the value to the power of 4 and then cube root that. You won't necessarily have to always find a value of x to a power with this stuff.
    (Original post by Plagioclase)
    Lots, and lots, and lots of practice.
    thanks, i think i can finally move on from indices now
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    (Original post by ihatePE)
    that made more sense to me :hoppy:
    Also you should note that the order in which you perform the raise of power and the root is irrelevant. This is simply due to laws of indices.

     (5^4)^\frac{1}{3} = (5^\frac{1}{3})^4 because they just multiply anyway by the laws of indices.
 
 
 
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