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Fractions

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    Hi,

    I have some queries.

    There are some questions on the attachment below of which I am not too sure about.

    Click image for larger version. 

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    For i)0
    ii)9
    iii) I'm not too sure

    Help would be appreciated.
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    i) That's incorrect. Any number to the power of 0 is equal to 1.
    ii) Correct.
    iii) Use the index law:

    (a^b)^c = a^{b\times c}
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    (Original post by notnek)
    i) That's incorrect. Any number to the power of 0 is equal to 1.
    ii) Correct.
    iii) Use the index law:

    (a^b)^c = a^{b\times c}
    So would it be -6/4
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    (Original post by zed963)
    So would it be -6/4
    Two negatives make a plus (it should be +6/4 in the exponent).
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    (Original post by f1mad)
    Two negatives make a plus (it should be +6/4 in the exponent).
    So whats 16^6/4
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    (Original post by zed963)
    So whats 16^6/4
    Remember,  16=2^4
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    (Original post by zed963)
    So whats 16^6/4
    16^(6/4)= 16^(3/2)
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    (Original post by raheem94)
    Remember,  16=2^4
    I don't understand what u mean here.
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    (Original post by zed963)
    I don't understand what u mean here.
    Raise both sides to the power of 6/4.
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    (Original post by zed963)
    I don't understand what u mean here.
     16^{\frac64} = (2^4)^{\frac64}
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    (Original post by raheem94)
     16^{\frac64} = (2^4)^{\frac64}
    I would like you to explain why you are doing the things.
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    would the answer be 18 by any chance?

    I meant 2 to the power 6
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    (Original post by zed963)
    I would like you to explain why you are doing the things.
    We have got  16^{\frac64}

    Now we need to simplify it further, so we will notice that,  16=2^4

    Hence,  16^{\frac64} = (2^4)^{\frac64}

    Now use the rule,  (a^x)^{\frac{b}{d}} = a^{x \times \frac{b}{d}}

     16^{\frac64} = (2^4)^{\frac64} = 2^{4 \times \frac64} = 2^6

    Do you get it?
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    (Original post by zed963)
    would the answer be 18 by any chance?
    The answer is 64.
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    (Original post by zed963)
    I would like you to explain why you are doing the things.
    You seem to be struggling so I'll show you an alternative method that you may be more used to:

    \displaystyle 16^{\frac{6}{4}} = 16^{\frac{3}{2}}

    Raising a number to the power of \frac{3}{2} is the same as square-rooting it (the 2 on the denominator tells you this) and then cubing it (from the numerator). So you get:

    \displaystyle 16^{\frac{3}{2}} = (\sqrt{16})^3 = 4^3 = 64
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    (Original post by notnek)
    You seem to be struggling so I'll show you an alternative method that you may be more used to:

    \displaystyle 16^{\frac{6}{4}} = 16^{\frac{3}{2}}

    Raising a number to the power of \frac{3}{2} is the same as squaring it (the 2 on the denominator tells you this) and then cubing it (from the numerator). So you get:

    \displaystyle 16^{\frac{3}{2}} = (\sqrt{16})^3 = 4^3 = 64
    Its just that we've been taught using a different method.
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    (Original post by zed963)
    Its just that we've been taught using a different method.
    Do you understand the method I gave? Which method have you been taught?
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    (Original post by notnek)
    Do you understand the method I gave? Which method have you been taught?
    I've been taught that you have 4 boxes and all those 4 boxes multiplied by something that has to = 16 and now since its asking for 6 we would just multiply the number inside the box to the power of 6.

    4 boxes come from the denominator of 6/4
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    (Original post by zed963)
    I've been taught that you have 4 boxes and all those 4 boxes multiplied by something that has to = 16 and now since its asking for 6 we would just multiply the number inside the box to the power of 6.

    4 boxes come from the denominator of 6/4
    This seems very odd. Show your method and perhaps we can make sense of it.
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    (Original post by steve2005)
    This seems very odd. Show your method and perhaps we can make sense of it.
    I also can't understand what he is saying. But i also don't see any point for explaining things in this way, a GCSE student shouldn't have any problem doing these in the way it is done in this thread.

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