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Thread: Too easy?

  1. #1
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    Too easy?

    Please help me on these ones guys. I'm not as smart as you! :P

    Simplify the following expressions and express the result with positive indices. All letters given represent non-zero numbers.

    1. $\displaystyle b^ {-\frac{2}{3}}$
    2. $\displaystyle c^{-1\frac{1}{4}}$
    3. $\displaystyle \sqrt[b^4]{b^3}$
    4. $\displaystyle (36a^2)^\frac{3}{2}$
    5. $\displaystyle (1000b^3)^\frac{2}{3}$
    6. $\displaystyle (64c^2)^{-1.5}$
    7. $\displaystyle (\frac{8b^3}{a^3})^{-\frac{1}{3}}$
    8. $\displaystyle (\frac{-a^3}{8b^3})^{-\frac{1}{3}}$
    9. $\displaystyle (\frac{27a^3}{8b^3})^{-\frac{2}{3}}$

    Find the values of the following:

    1. $\displaystyle 32^{-\frac{3}{5}}$
    2. $\displaystyle (0.01)^{-\frac{3}{2}}$
    3. $\displaystyle (0.4)^2\times(0.125){^\frac{1}{3}}\div(2.5)^{-3}$
    4. $\displaystyle \frac{8^{n+2}-16\times8^{n-1}}{31\times8^n}$
    5. $\displaystyle \frac{3\times2^n-4\times2^{n-2}}{2^n-2^{n-1}}$

    Simplify:

    1. $\displaystyle 2a\sqrt{a^3b^2}+3a^2b\sqrt{36a}-a^2\sqrt{25ab^2}$

    Express the following in their simplest form:

    1. $\displaystyle \sqrt{192}$
    2. $\displaystyle \sqrt[3]{40}$
    3. $\displaystyle \sqrt[4]{567}$
    4. $\displaystyle \sqrt[5]{-64a^6b^7}$
    5. [tex]\sqrt[4]{243c^8d^3}

    Arrange the following in descending order:

    1. $\displaystyle \sqrt{2}$,$\displaystyle \sqrt[3]{3}$,$\displaystyle \sqrt[6]{5}$

    Simplify and express with positive indices:

    1. $\displaystyle a\sqrt{a\sqrt[3]{a}\sqrt[4]{a}}$
    2. $\displaystyle (\sqrt2^3b^6)^4$
    3. $\displaystyle \sqrt[6]{a^8b^6}\times(a^\frac{2}{3}b^{-1})^{-2}$

    Thanks for your help guys!!
    Last edited by Rocher; Nov 2nd 2007 at 02:36 PM.
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  2. #2
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    I can only do so much but here goes:
    sorry, no LateX here

    1. b^(-2/3) = b^0 / b^(2/3)
    = 1 / cube rt. (b^2)

    2. c^(-1 1/4) = c^(-5/4)
    = c^0 / c^(5/4)
    = 1 / 4 root (c^5)

    3. b^4 root (b^3) = b^(3/b^4)

    4. (36a^2)^(3/2) = (36^3/2)(a^6/2)
    = (sq.root 36^3)(a^3)
    = (6^3)(a^3)
    = 216a^3

    Well, I hope you get the picture, it's better for you to work it out yourself and I am feeling quite tired
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  3. #3
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    Hello, Rocher!

    Here's the last set . . .


    Simplify and express with positive indices:

    $\displaystyle 1.\;\;a\sqrt{a\sqrt[3]{a}\sqrt[4]{a}}$

    We have: .$\displaystyle a\cdot\sqrt{a^1\cdot a^{\frac{1}{3}}\cdot a^{\frac{1}{4}}} \;=\;a\cdot\sqrt{a^{\frac{19}{12}}} \;=\;a\cdot\left(a^{\frac{19}{12}}\right)^{\frac{1 }{2}}\;=\;a^1\cdot a^{\frac{19}{24}} \;=\;a^{\frac{43}{24}} $



    $\displaystyle 2.\;\;(\sqrt{2^3}b^6)^4$

    We have: .$\displaystyle \left[\left(2^3\right)^{\frac{1}{2}}b^6\right]^4 \;=\;\left(2^{\frac{3}{2}}b^6\right)^4\;=\;\left(2 ^{\frac{3}{2}}\right)^4\left(b^6\right)^4 \;=\;2^6b^{24} \;=\;64b^6$



    $\displaystyle 3.\;\;\sqrt[6]{a^8b^6}\times(a^\frac{2}{3}b^{-1})^{-2}$

    We have: .$\displaystyle \left(a^8b^6\right)^{\frac{1}{6}} \times \left(a^{\frac{2}{3}}\right)^{-2}\left(b^{-1}\right)^{-2}\;=\;\left(a^8\right)^{\frac{1}{6}}\left(b^6\rig ht)^{\frac{1}{6}} \times a^{-\frac{4}{3}} b^2 \;=\;a^{\frac{4}{3}}b^1 \times a^{-\frac{4}{3}}b^2
    $

    . . . $\displaystyle =\;\left(a^{\frac{4}{3}}\cdot a^{-\frac{4}{3}}\right)\left(b^1\cdot b^2\right) \;=\;a^0\cdot b^3 \;=\;1\cdot b^3 \;=\;b^3$

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  4. #4
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    Quote Originally Posted by Rocher View Post
    Find the values of the following:

    1. $\displaystyle 32^{-\frac{3}{5}}$
    2. $\displaystyle (0.01)^{-\frac{3}{2}}$
    3. $\displaystyle (0.4)^2\times(0.125){^\frac{1}{3}}\div(2.5)^{-3}$
    4. $\displaystyle \frac{8^{n+2}-16\times8^{n-1}}{31\times8^n}$
    5. $\displaystyle \frac{3\times2^n-4\times2^{n-2}}{2^n-2^{n-1}}$
    Not asking for much, are you?

    All of these depend on the laws of exponents:
    $\displaystyle a^na^m = a^{n + m}$

    $\displaystyle a^nb^n = (ab)^n$

    $\displaystyle a^{-n} = \frac{1}{a^n}$

    $\displaystyle \sqrt[n]{a} = a^{1/n}$

    $\displaystyle (a^n)^m = a^{nm}$

    So let's look at these.
    1)
    $\displaystyle 32^{-\frac{3}{5}}$

    $\displaystyle = \frac{1}{32^{3/5}}$ <-- It's always a good idea to get rid of the negative powers as soon as possible.

    $\displaystyle = \frac{1}{(32^{1/5})^3}$

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

    $\displaystyle = \frac{1}{8}$

    2) This is the same thing as 1.

    3)
    $\displaystyle \frac{(0.4)^2 \cdot 0.125^{1/3}}{2.5^{-3}}$

    $\displaystyle = (0.4)^2 \cdot 0.125^{1/3} \cdot 2.5^3$

    $\displaystyle = 0.16 \cdot 0.5 \cdot 15.625$

    You finish from here.

    4) I've got to do this one, simply because it looks cool.
    $\displaystyle \frac{8^{n+2}-16 \cdot 8^{n-1}}{31 \cdot 8^n}$

    $\displaystyle = \frac{8^{n+2} - 2 \cdot 8 \cdot 8^{n-1}}{31 \cdot 8^n}$

    $\displaystyle = \frac{8^2 \cdot 8^n - 2 \cdot 8^n}{31 \cdot 8^n}$

    $\displaystyle = \frac{8^n}{8^n} \cdot \frac{8^2 - 2}{31}$

    $\displaystyle = \frac{62}{31}$

    $\displaystyle = 2$

    5) This works on a similar basis as 4). Get as many terms as you can into a form with powers of 2. Then factor the numerator and denominator, getting rid of as many powers of 2 that you can. Then simplify.

    -Dan
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  5. #5
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    Thank you all, but class is starting soon and damnit still ain't done the homework :P Oh well, thanks!
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