# Too easy?

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• Nov 2nd 2007, 01:43 PM
Rocher
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. $b^ {-\frac{2}{3}}$
2. $c^{-1\frac{1}{4}}$
3. $\sqrt[b^4]{b^3}$
4. $(36a^2)^\frac{3}{2}$
5. $(1000b^3)^\frac{2}{3}$
6. $(64c^2)^{-1.5}$
7. $(\frac{8b^3}{a^3})^{-\frac{1}{3}}$
8. $(\frac{-a^3}{8b^3})^{-\frac{1}{3}}$
9. $(\frac{27a^3}{8b^3})^{-\frac{2}{3}}$

Find the values of the following:

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

Simplify:

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

Express the following in their simplest form:

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

Arrange the following in descending order:

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

Simplify and express with positive indices:

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

Thanks for your help guys!!
• Nov 2nd 2007, 02:56 PM
Geometor
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 :D
• Nov 2nd 2007, 07:06 PM
Soroban
Hello, Rocher!

Here's the last set . . .

Quote:

Simplify and express with positive indices:

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

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

Quote:

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

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

Quote:

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

We have: . $\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
$

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

• Nov 2nd 2007, 07:41 PM
topsquark
Quote:

Originally Posted by Rocher
Find the values of the following:

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

Not asking for much, are you?

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

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

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

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

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

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

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

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

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

$= \frac{1}{8}$

2) This is the same thing as 1.

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

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

$= 0.16 \cdot 0.5 \cdot 15.625$

You finish from here.

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

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

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

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

$= \frac{62}{31}$

$= 2$

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
• Nov 3rd 2007, 07:17 AM
Rocher
Thank you all, but class is starting soon and damnit still ain't done the homework :P Oh well, thanks!