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Math Help - Solving for an exponent within a multiplication of radical expressions

  1. #1
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    Solving for an exponent within a multiplication of radical expressions

    Thumbnail below, problem #99. Solve for k.

    I don't really know how to write as far as I got. :[ But basically I started by combining like terms, making it the 5th root of 32a^2k+8 = 2a^4 (after the ^ those are part of the exponent). Then from there I took the 5th root of 32, 2, and then divided both sides by two making it:

    5th root of a^2k+8 = a^4

    From there I don't know what to do, or if the previous steps of combining the exponents were correct.


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  2. #2
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    Quote Originally Posted by Kyrie View Post
    Thumbnail below, problem #99. Solve for k.

    I don't really know how to write as far as I got. :[ But basically I started by combining like terms, making it the 5th root of 32a^2k+8 = 2a^4 (after the ^ those are part of the exponent). Then from there I took the 5th root of 32, 2, and then divided both sides by two making it:

    5th root of a^2k+8 = a^4

    From there I don't know what to do, or if the previous steps of combining the exponents were correct.


    Even after clicking on the thumbnail, it's still too small for me to read. Please type the equation (click on the relevant link in my signature to learn how to format equations).
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  3. #3
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    This image should work. http://a.yfrog.com/img843/4405/scan0001c.jpg

    I couldn't figure out how to get the addition of exponents after looking at the tutorial images. :[ Sorry. But that image, if you click on it, should blow up quite big.
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  4. #4
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    \displaystyle \sqrt[5]{4a^{3k+2}} \sqrt[5]{8a^{6-k}}=2a^4

    Taking both sides to the power of 5 gives

    \displaystyle 4a^{3k+2}\times  8a^{6-k}=(2a^4)^5

    \displaystyle 32a^{(3k+2)+(6-k)}=32a^{20}

    \displaystyle a^{(3k+2)+(6-k)}=a^{20}

    \displaystyle (3k+2)+(6-k)=20

    Can you take it from here?
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