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Thread: algebra refresher problems

  1. #1
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    algebra refresher problems

    1. $\displaystyle ^3\sqrt{x}=4$
    2. $\displaystyle 2x^2-17=-8$
    3.$\displaystyle -7*2^x=-224$
    4. $\displaystyle (\frac{2}{3})^x=\frac{3}{2}$
    5. $\displaystyle 3^x=\frac{1}{9}$

    I've picked out these problems in relation to one question. What do I do with all these indexes and powers? I totally forgot.

    Thanks.
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  2. #2
    Super Member 11rdc11's Avatar
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    problem 2

    $\displaystyle 2x^2 -17 = -8$

    $\displaystyle 2x^2 = 9$

    $\displaystyle x^2 = \frac{9}{2}$

    $\displaystyle x =^+_- \sqrt{\frac{9}{2}}$
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  3. #3
    Super Member 11rdc11's Avatar
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    problem 5

    $\displaystyle 3^x = \frac{1}{9}$

    $\displaystyle 3^x = \frac{1}{3^2}$

    $\displaystyle 3^x = 3^{-2}$

    $\displaystyle x = -2$
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  4. #4
    Super Member 11rdc11's Avatar
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    problem 1

    $\displaystyle ^3\sqrt{x}=4$

    $\displaystyle x= 4^3$

    $\displaystyle x = 64$
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  5. #5
    Super Member 11rdc11's Avatar
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    Problem 3

    $\displaystyle (-7)(2^x) = -224$

    $\displaystyle 2^x = 32$

    $\displaystyle 2^x = 2^5$

    $\displaystyle x =5$

    problem 4

    $\displaystyle \bigg(\frac{2}{3}\bigg)^x = \bigg(\frac{3}{2}\bigg)$

    $\displaystyle \bigg(\frac{2}{3}\bigg)^x = \bigg(\frac{2}{3}\bigg)^{-1}$

    $\displaystyle x = -1$
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