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    Name:  lololo.png
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Size:  3.3 KBHi there I have provided a picture which an arrow highlighting my problem..

    If you would help me that would be really good! thanks
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    This comes down to the fact that 4=2^2 and cancelling.
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    (Original post by madfish)
    Name:  lololo.png
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Size:  3.3 KBHi there I have provided a picture which an arrow highlighting my problem..

    If you would help me that would be really good! thanks
    \displaystyle 2^{n-1} \times  2^{-2} = 2^{n-1-2} = 2^{n-3}

    2 to the power of negative two is a quarter, so it's the same as dividing by 4
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    Divide top and bottom by 4

    2^2=4, 2^{-2}=\frac{1}{4}
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    (Original post by Indeterminate)
    \displaystyle 2^{n-1} \times  2^{-2} = 2^{n-1-2} = 2^{n-3}

    2 to the power of negative two is a quarter, so it's the same as dividing by 4
    why do we have to divide by 4?:P
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    (Original post by madfish)
    why do we have to divide by 4?:P
    We divide by 4 to reduce the fraction to its simplest form
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    (Original post by Indeterminate)
    We divide by 4 to reduce the fraction to its simplest form
    Okay I am going to sound really dumb for an alevel student but I just CANNOT see how we have went from n-1 to n-3.. i think I missing something really obvious or else I am thick... I think it's probably the latter :rolleyes: haha
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    (Original post by madfish)
    Okay I am going to sound really dumb for an alevel student but I just CANNOT see how we have went from n-1 to n-3.. i think I missing something really obvious or else I am thick... I think it's probably the latter :rolleyes: haha
    It's ok.

    \displaystyle \frac{4\times 25}{2^{n-1}} = \frac{25}{\frac{1}{4}2^{n-1}}

    2^{-2} = \frac{1}{2^2} = \frac{1}{4}

    using the fact that

    \frac{1}{x^n}=x^{-n}

    for all non-zero x.

    Now can you see it?
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    (Original post by SubAtomic)
    Divide top and bottom by 4

    2^2=4, 2^{-2}=\frac{1}{4}
    thank you but I still am not sure about how it becomes n-3.. i think I am missing something really obvious

    Would you mind explaining further? really sorry to sound silly thanks
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    (Original post by madfish)
    thank you but I still am not sure about how it becomes n-3.. i think I am missing something really obvious

    Would you mind explaining further? really sorry to sound silly thanks
    I forgot to mention that

    \displaystyle a^b \times a^c = a^{b+c}
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    (Original post by Indeterminate)
    It's ok.

    \displaystyle \frac{4\times 25}{2^{n-1}} = \frac{25}{\frac{1}{4}2^{n-1}}

    2^{-2} = \frac{1}{2^2} = 2^{-2}

    using the fact that

    \frac{1}{x^n}=x^{-n}

    for all non-zero x.

    Now can you see it?
    ahhh yes okay thank you so much for that! + 1 rep
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    (Original post by Indeterminate)
    I forgot to mention that

    \displaystyle a^b \times a^c = a^{b+c}
    ohh yea I know that rule, cheers (i repped you on this post btw
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    (Original post by madfish)
    ...
    I think you have it now right?

    \displaystyle\frac{25}{2^{-2} \cdot 2^{n-1}}=\frac{25}{2^{n-3}}
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    (Original post by SubAtomic)
    I think you have it now right?

    \displaystyle\frac{25}{2^{-2} \cdot 2^{n-1}}=\frac{25}{2^{n-3}}
    yea got it now thanks dude thanks for your replies anyway.. +1 rep
 
 
 
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