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    Does 5y^-2 equal
    \frac{1}{5y^2} or \frac{5}{y^2}

    We've been taught all the basic indices rules but I'm a bit confused about this one...

    +rep to any help
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    The 5/y^2 one
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    The second one.
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    It's the 2nd option. Since you have 5 lots of y^-2. If you put 5 on the bottom you change the value.
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    Note that ab^n = a \times (b^n), so it wouldn't make sense for 5 to be on the denominator.
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    Okay, so using that logic;
    (3y^-2)*(4y^-3) = \frac{3}{y^2}*\frac{4}{y^3} = \frac{12}{y^5}?
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    (Original post by jbeach09)
    Okay, so using that logic;
    (3y^-2)*(4y^-3) = \frac{3}{y^2}*\frac{4}{y^3} = \frac{12}{y^5}?
    Correct.
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    (Original post by zxh800)
    It's the 2nd option. Since you have 5 lots of y^-2. If you put 5 on the bottom you change the value.
    Thankyou, I'll rep now

    One more thing, I get that rule now, but how would you get \frac{1}{2y^3} in the form y^n or ky^n as everything I try just results in the two being on the numerator instead
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    (Original post by jbeach09)
    Thankyou, I'll rep now

    One more thing, I get that rule now, but how would you get \frac{1}{2y^3} in the form y^n or ky^n as everything I try just results in the two being on the numerator instead
    You need to use the rule \dfrac{1}{ab} = \dfrac{1}{a} \times \dfrac{1}{b}. When you put it in the form ky^n, k is allowed to be a fraction!
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    (Original post by jbeach09)
    Thankyou, I'll rep now

    One more thing, I get that rule now, but how would you get \frac{1}{2y^3} in the form y^n or ky^n as everything I try just results in the two being on the numerator instead
    Note that \frac{1}{2y^3} = \frac{1}{2} \times \frac{1}{y^3}.
    You know how to write \frac{1}{y^3} as a power of y.
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    (Original post by Farhan.Hanif93)
    Note that \frac{1}{2y^3} = \frac{1}{2} \times \frac{1}{y^3}.
    You know how to write \frac{1}{y^3} as a power of y.
    So it would be

    \frac{1}{2}y^{-3}?
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    (Original post by jbeach09)
    So it would be

    \frac{1}{2}y^{-3}?
    :yep:
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    (Original post by jbeach09)
    So it would be

    \frac{1}{2}y^{-3}?
    Thanks

    Would y^-\frac{5}{4} equal \frac{1}{(\sqrt[4]y)^5}

    if it doesnt show, the power of y is negative 5/4
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    (Original post by jbeach09)
    Thanks

    Would y^-\frac{5}{4} equal \frac{1}{(\sqrt[4]y)^5}

    if it doesnt show, the power of y is negative 5/4
    :yep:
 
 
 
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