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Math Help - Mid-Terms Review

  1. #16
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Rocher View Post
    1) 32^{\frac{3}{5}}a^{-\frac{2}{3}}b^2 \div (216a^4b^2)^\frac{1}{3}
    I will do the first, you do the second.
    32^{3/5} \cdot a^{-2/3} \cdot b^2 \div (216a^4b^2)^{1/3}

    The first step is to write this thing out as a single fraction:
    = \frac{32^{3/5} \cdot a^{-2/3} \cdot b^2}{(216a^4b^2)^{1/3}}

    The next step is to work out the parenthesis. There is only one set of parenthesis, in the denominator:
    = \frac{32^{3/5} \cdot a^{-2/3} \cdot b^2}{216^{1/3} \cdot a^{4/3} \cdot b^{2/3}}

    Now write the denominator as negative powers of the numerator:
    = 32^{3/5} \cdot a^{-2/3} \cdot b^2 \cdot 216^{-1/3} \cdot a^{-4/3} \cdot b^{-2/3}

    Now combine powers of like bases:
    = 32^{3/5} \cdot 216^{-1/3} \cdot a^{-2/3 - 4/3} \cdot b^{2 - 2/3}

    = 32^{3/5} \cdot 216^{-1/3} \cdot a^{-2} \cdot b^{4/3}

    Now go to work on the constants. You are expected to know (or somehow figure out) that 2^5 = 32 and 6^3 = 216:
    = (2^5)^{3/5} \cdot (6^3)^{-1/3} \cdot a^{-2} \cdot b^{4/3}

    = 2^3 \cdot 6^{-1} \cdot a^{-2} \cdot b^{4/3}

    = 8 \cdot 6^{-1} \cdot a^{-2} \cdot b^{4/3}

    Now put negative exponents in a denominator:
    = \frac{8b^{4/3}}{6a^2}

    And simplify:
    = \frac{4b^{4/3}}{3a^2}

    You do the second one and post your solution. We'll check it.

    -Dan
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  2. #17
    Forum Admin topsquark's Avatar
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    Here is a list, in no particular order, of the exponent rules. In this a and b are real numbers greater than 0, and n and m are any real number.

    a^m \cdot a^n = a^{m + n}

    a^{-m} = \frac{1}{a^m} likewise \frac{1}{a^{-m}} = a^m

    \left ( a^m \right ) ^n = a^{mn}

    \frac{a^m}{a^n} = a^m \cdot a^{-n} = a^{m - n}

    a^mb^m = (ab)^m

    a^0 = 1 for all a

    And finally:
    a^{m/n} = \sqrt[n]{a^m} <-- for n \neq 0

    -Dan
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  3. #18
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    Is the answer for two \frac{27x^\frac{10}{3}}{125y^\frac{4}{3}}??
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  4. #19
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Rocher View Post
    Is the answer for two \frac{27x^\frac{10}{3}}{125y^\frac{4}{3}}??
    You neglected to

    1) Apply the overall exponent to both numbers

    2) Keep the negative sign when you did the y variable.

    But the exponent on the x factor is correct.

    -Dan
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  5. #20
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    What do you mean by applying overall exponent to the numbers? I did that I think... You mean like -5 * -2/3 right? Multiply them together? That's how I got my final exponents.

    For the y, it would be correct if I just added the negative sign to the exponent right?
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  6. #21
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Rocher View Post
    What do you mean by applying overall exponent to the numbers? I did that I think... You mean like -5 * -2/3 right? Multiply them together? That's how I got my final exponents.
    I mean if you have something like
    (125x^3y^2)^{2/3}
    you can't just apply the 2/3 power to just the x and y. You have to apply it to the 125 as well!

    Quote Originally Posted by Rocher View Post
    For the y, it would be correct if I just added the negative sign to the exponent right?
    Yes, because a negative times a positive is a negative.

    -Dan
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  7. #22
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    Quote Originally Posted by topsquark View Post
    I mean if you have something like
    (125x^3y^2)^{2/3}
    you can't just apply the 2/3 power to just the x and y. You have to apply it to the 125 as well!
    -Dan
    125^{\frac{2}{3}}=5208\frac{1}{3}

    o.O No way that can be right... can it?
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  8. #23
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Rocher View Post
    125^{\frac{2}{3}}=5208\frac{1}{3}

    o.O No way that can be right... can it?
    Hint: 125 = 5^3

    What's 3 \cdot \frac{2}{3}?

    -Dan
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  9. #24
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    Is the answer \frac{3x^6}{5y^4}??
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  10. #25
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Rocher View Post
    Is the answer \frac{3x^6}{5y^4}??
    I told you before that your exponent for the x was correct! Why did you change it?

    \left ( \frac{27x^{-5}}{125y^2} \right )^{-2/3}

    27^{-2/3} = \left ( 3^3 \right ) ^{-2/3} = 3^{3 \cdot -2/3} = 3^{-2}

    A similar comment goes for the 125. In this updated answer you don't have any of the exponents, nor the constants correct.

    Please note the rule I posted for you below:
    \left ( a^m \right ) ^n = a^{mn}

    -Dan
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  11. #26
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    Omg, this is really frustrating . Once again, is the answer \frac{3^{-2}x^\frac{10}{3}}{5^{-2}y^-\frac{4}{3}}
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  12. #27
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Rocher View Post
    Omg, this is really frustrating . Once again, is the answer \frac{3^{-2}}x^{\frac{10}{3}}{5^{-2}y^{-\frac{4}{3}}
    Yes!! Now all you need to do is get rid of the negative exponents.

    Again:
    a^{-m} = \frac{1}{a^m}
    and
    \frac{1}{a^{-m}} = a^m

    -Dan
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  13. #28
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    So do I put all the currently negative exponents at the bottom and all the positive ones at the top or something?
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  14. #29
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Rocher View Post
    So do I put all the currently negative exponents at the bottom and all the positive ones at the top or something?
    In a word, "yup."

    -Dan
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  15. #30
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    \frac{5^2y^{\frac{4}{3}}x{\frac{10}{3}}}{3^2}
    Hopefully yes!

    [EDIT]Oooh wait...

    \frac{3^25^2y^{\frac{4}{3}}}{x^{\frac{10}{3}}}

    ?? One of those two? XD
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