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Math Help - Exponents, check.

  1. #1
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    Exponents, check.

    Hey, could someone please check if I simplified this problem correctly?

    \frac {(2^{(a+b)}x^{2b})^{(a-b)}}{(\frac {2^ax}{x^{-b}})^a*(\frac {x^{(a-b)}}{2^b})^b}

    First I got rid of the exponents by multiplying.

    = \frac {2^{(a^2-b^2)}x^{(2ab-2b^2)}}{(\frac {x^{(ab-b^2)}}{2^{b^2}})*(\frac {2^{a^2}x^a}{x^{-ab}})}

    Then I simplified and combined the bottom fractions.

    = \frac {2^{(a^2-b^2)}x^{(2ab-2b^2)}}{x^{(2ab-b^2+a)}2^{(a^2-b^2)}}

    And then repeated that step and got this answer.

    = \frac {1}{x^{b^2+a}}

    Thanks in advance.

    Mi986
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  2. #2
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    yes that is correct.
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  3. #3
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    Hello, mi986!

    I can't follow your second line . . .


    \displaystyle{\frac{\left(2^{a+b}\cdot x^{2b}\right)^{a-b}}{\left(\frac {2^ax}{x^{-b}}\right)^a*(\frac {x^{a-b}}{2^b})^b}}

    We have: . \frac{2^{a^2-b^2}\cdot x^{2ab -b^2}}{\frac{2^{a^2}\cdot x^a}{x^{-ab}} \cdot \frac{x^{ab-b^2}}{2^{b^2}}} \;= \;\frac{2^{a^2-b^2}\cdot x^{2ab-b^2}}{2^{a^2-b^2}\cdot x^{a + 2ab -b^2}} \;=\;\frac{1}{x^a}

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  4. #4
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    Did you mean this part?
    <br />
{x^{2b}}^{(a-b)}<br />
    Don't I multiply the 2b by the a and then to the -b? So it becomes {x^{2ab-2b^2}}
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