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    As you can see I got this question wrong and indices laws always confuse me. In the context of this question, how do you know when to add or multiply the indices together?


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    (Original post by jessyjellytot14)
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    As you can see I got this question wrong and indices laws always confuse me. In the context of this question, how do you know when to add or multiply the indices together?


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    You add indices when you multiply two things together. e.g. (x^2)(x^3) = x^5

    You multiply indices when you are raising something to a power, e.g. (x^3)^4 = x^{12}
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    If the power is on the outside then you times the thing inside the bracket by itself the same number as the power, so for example (2x^3)^2 would become 4x^6 due to the 2 being squared and the 3 being multiplied by 2

    or (2x^3)^3 would become 8x^9

    However, in the case of something like 125 x 5^(0.5) it would become 5^3 x 5^(0/5), so that would become 5^(3+0.5) to become 5^(3.5)

    I'm not great at explaining, but try these

    12X^(4)Y^(2) x 2X^(2)Y^(3)

    (12X^(4)Y^(2))^2



    edit:
    (Original post by 16Characters....)
    You add indices when you multiply two things together. e.g. (x^2)(x^3) = x^5You multiply indices when you are raising something to a power, e.g. (x^3)^4 = x^{12}

    nvrmind, the other guys is way simpler to understand
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    (Original post by jessyjellytot14)

    As you can see I got this question wrong and indices laws always confuse me. In the context of this question, how do you know when to add or multiply the indices together?


    Posted from TSR Mobile
    Here's a quick intuitive feel for a^x \times a^y = a^{x+y}.

    I'll do it with a=3 here to make life simpler, but it works for any a:

    \displaystyle 3^x = \underbrace{3 \times 3 \times \cdots \times 3}_{x \, \text{times}} and 3^y = \underbrace{3 \times 3 \times \cdots \times 3}_{y \, \text{times}}

    So:

    \displaystyle

\begin{equation*} 3^x \times 3^y = \underbrace{3 \times 3 \times \cdots \times 3}_{x \, \text{times}} \times \underbrace{3 \times 3 \times \cdots \times 3}_{y \, \text{times}} = \underbrace{3 \times 3 \times \cdots \times 3}_{x+y \, \text{times}}\end{equation*}

    You now have a product of x 3's and y 3's and there are x+y threes being multiplied in total.

    Hence 3^x \times 3^y = \underbrace{3 \times 3 \times \cdots \times 3}_{x+y\,\text{times}}\end{equat  ion*} = 3^{x+y} as expected.

    -----------------------------------------------------------------------------------------------------

    And here's a quick intuitive argument as to why (a^x)^y is a^{xy}, it'll use the other theorem we 'proved' above.

    You're comfortable that a^x \times a^y = a^{x+y} now, I hope.

    So, since: \displaystyle (a^x)^y = \underbrace{a^x \times a^x \times \cdots \times a^x}_{y \, \text{times}}

    then we can simply add their powers:

    \displaystyle

\begin{equation*} (a^x)^y = a^{\overbrace{x + x + \cdots + x}^{y \, \text{times}}} = a^{xy}\end{equation*}

    Since \underbrace{x + x + \cdots + x}_{y \, \text{times}} (i.e: x added to itself y times) is just xy.

    --------------------------------------------------------------------------------------

    Basically, when you multiply two things, you add their powers. (if they are the same base). If you have a number to a power and raise it to another power then you multiply their powers.

    But really, if you just sit down and read and understand the above, it should all make sense!
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    16Characters.... C-rated Zacken Oh okay thanks, I understand it now in terms of bases
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    (Original post by jessyjellytot14)
    16Characters.... C-rated Zacken Oh okay thanks, I understand it now in terms of bases
    No problem!
 
 
 
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