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Math Help - Exponents question

  1. #1
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    Exponents question

    Hi I'm not too sure how this works so would appreciate if someone could explain how to tackle these questions and what indices rules are used to solve this problem.

    Question: Use the laws of exponents to simplify;

    1. a^3 / a(a^2)^-1 (a to the power of 3, divided by a times, brackets a to the power of 2, close brackets, to the power of minus 1).

    2. Square root of z^4 (z to the power of 4)

    Sorry if I have'nt writtern them out in a proper manner.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by komodo View Post
    Hi I'm not too sure how this works so would appreciate if someone could explain how to tackle these questions and what indices rules are used to solve this problem.

    Question: Use the laws of exponents to simplify;

    1. a^3 / a(a^2)^-1 (a to the power of 3, divided by a times, brackets a to the power of 2, close brackets, to the power of minus 1).

    2. Square root of z^4 (z to the power of 4)

    Sorry if I have'nt writtern them out in a proper manner.
    The basic rules are
    a^na^m = a^{m + n} <-- This holds when the exponents are negative as well.

    a^{-n} = \frac{1}{a^n} \text{  and  } \frac{1}{a^{-n}} = a^n

    (a^n)^m = a^{nm}

    So let's look at these problems with these rules in mind.

    1.
     \frac{a^3}{(a^2)^{-1}}

    We know that (a^2)^{-1} = a^{2 \cdot -1} = a^{-2}

    So the expression is
     \frac{a^3}{(a^2)^{-1}} = \frac{a^3}{a^{-2}}

    What does the \frac{1}{a^{-n}} rule tell you what to do with this?

    2. One further rule: Square roots represent a 1/2 power. So \sqrt{a} = a^{1/2}. Note that
    \sqrt{a^n} = (a^n)^{1/2} = a^{n/2}

    (Last step by the "power to the power" rule.) What can you do using this?

    -Dan
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  3. #3
    Newbie Auri's Avatar
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    Edit: Sorry, just over all ignore this post, made a mistake reading the question.
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  4. #4
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    Thanks for your help guys

    Dan going by your given rule which was realy helpful for the second question I'm assuming that I have to multiply the 1/2 by the 4 which would leave me with Z^2 as the answer?

    The 1st question is still not clear to me as there was an (a) outside the brackets which either you have not taken into account or I have missed a step...would be greatful if you could shed some light on the matter.
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by komodo View Post
    Thanks for your help guys

    Dan going by your given rule which was realy helpful for the second question I'm assuming that I have to multiply the 1/2 by the 4 which would leave me with Z^2 as the answer?

    The 1st question is still not clear to me as there was an (a) outside the brackets which either you have not taken into account or I have missed a step...would be greatful if you could shed some light on the matter.
    Sorry, there was, wasn't there? That's okay, it only adds one more step.

    Like last time let me work with the denominator.
    a(a^2)^{-1} = a \cdot a^{-2}

    Notice that a = a^1, so we have
    a(a^2)^{-1} = a \cdot a^{-2} = a^1 a^{-2}

    What does the a^na^m = a^{n + m} rule have to say about this?

    -Dan

    Edit: And yes, you have the second one correct.
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  6. #6
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    Quote Originally Posted by topsquark View Post
    Sorry, there was, wasn't there? That's okay, it only adds one more step.

    Like last time let me work with the denominator.
    a(a^2)^{-1} = a \cdot a^{-2}

    Notice that a = a^1, so we have
    a(a^2)^{-1} = a \cdot a^{-2} = a^1 a^{-2}

    What does the a^na^m = a^{n + m} rule have to say about this?

    -Dan

    Edit: And yes, you have the second one correct.
    Thanks for your help Dan its clear as day now and just to think that I was contemplating suicide an hour ago

    Just to make sure the final answer for 1 should be a^3/a^-1 ? Or should I take it a step further by using the division rule to end up with a^4 as the final answer?

    Thanks again
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  7. #7
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    e^(i*pi)'s Avatar
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    You use the division rule to end up with a^4
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