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  1. #1
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    solve for a

    look.............................................. ..................solve for a
    a a-2 a+1
    3 4 25/ a-1 a+1 a-2
    9 2 5

    common.......................
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  2. #2
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    Quote Originally Posted by pickbrain View Post
    look.............................................. ..................solve for a
    a a-2 a+1
    3 4 25/ a-1 a+1 a-2
    9 2 5

    common.......................
    Is this supposed to mean anything? Please write this clearly if you expect anyone to help you. Is this a matrix multiplication? Where is the equal sign?

    -Dan
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  3. #3
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    Hello, pickbrain!

    Did you see what you wrote?


    solve for a
    a a-2 a+1
    3 4 25/ a-1 a+1 a-2
    9 2 5

    common.......................
    I think it's supposed to look like this: . \frac{3^a\cdot4^{a-2}\cdot25^{a+1}}{9^{a-1}\cdot2^{a+1}\cdot5^{a-2}}


    The numerator is: . 3^2\cdot(2^2)^{a-2}\cdot(5^2)^{a+1} \;=\;3^2\cdot2^{2a-4}\cdot5^{2a+2}

    The denominator is: . (3^2)^{a-1}\cdot2^{a+1}\cdot5^{a-2} \;=\;3^{2a-2}\cdot2^{a+1}\cdot5^{a-2}


    The problem becomes: . \frac{3^a\cdot2^{2a-4}\cdot5^{2a+2}}{3^{2a-2}\cdot2^{a+1}\cdot5^{a-2}} \;=\;\frac{3^a}{3^{2a-2}}\cdot\frac{2^{2a-4}}{2^{a+1}}\cdot\frac{5^{2a+2}}{5^{a-2}}


    Divide ("subtract exponents"): . 3^{a-(2a-2)}\cdot 2^{(2a-4)-(a+1)}\cdot 5^{(2a+2)-(a-2)}

    . . . . . . . . . . . . . . . . . . . . . = \;3^{a-2a+2}\cdot 2^{2a-4-a-1}\cdot 5^{2a+2-a+2}

    . . . . . . . . . . . . . . . . . . . . . = \;3^{-a+2}\cdot2^{a-5}\cdot5^{a+4}

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