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Thread: exponent division question

  1. #1
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    exponent division question

    Can someone please show me (or direct me to) a method for solving these two.
    I cannot figure out how you would go about getting the answer.

    Thanks for any help.
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  2. #2
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    First try to reduce each term to the same base. $\displaystyle 9= 3^2$ and $\displaystyle 27= 3^3$ so $\displaystyle \frac{(9*3^2)^3}{(3*27)^2}= \frac{(3^2*3^2)^3}{(3*3^3)^2}$$\displaystyle = \frac{(3^{2+2})^3}{(3^{1+3})}= \frac{(3^4)^3}{(3^4)^2}= \frac{3^{12}}{3^8}$.

    Can you finish that?

    Similarly, $\displaystyle 16= 2^4$, $\displaystyle 4= 2^2$, and $\displaystyle 8= 2^3$ so $\displaystyle \frac{(16*4)^2}{(2*8)^3}= \frac{(2^4*2^2)^2}{(2^1*2^3)^3}$$\displaystyle = \frac{(2^6)^2}{(2^4)^3}= \frac{2^{12}}{2^{12}}$.

    Can you finish that?
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  3. #3
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    Here are some hints to start you off:

    Get a common base. For example, $\displaystyle 9=3^2$ and $\displaystyle 27=3^3$.

    Use the following laws of exponents:

    $\displaystyle x^ax^b=x^{a+b}$

    $\displaystyle \frac{x^a}{x^b}=x^{a-b}$

    $\displaystyle (x^a)^b=x^{ab}$
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