# Mid-Terms Review

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• Jan 18th 2008, 04:52 PM
topsquark
Quote:

Originally Posted by Rocher
1) $\displaystyle 32^{\frac{3}{5}}a^{-\frac{2}{3}}b^2 \div (216a^4b^2)^\frac{1}{3}$

I will do the first, you do the second.
$\displaystyle 32^{3/5} \cdot a^{-2/3} \cdot b^2 \div (216a^4b^2)^{1/3}$

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

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

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

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

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

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

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

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

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

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

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

-Dan
• Jan 18th 2008, 04:58 PM
topsquark
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.

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

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

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

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

$\displaystyle a^mb^m = (ab)^m$

$\displaystyle a^0 = 1$ for all a

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

-Dan
• Jan 18th 2008, 05:10 PM
Rocher
Is the answer for two $\displaystyle \frac{27x^\frac{10}{3}}{125y^\frac{4}{3}}$??
• Jan 18th 2008, 05:12 PM
topsquark
Quote:

Originally Posted by Rocher
Is the answer for two $\displaystyle \frac{27x^\frac{10}{3}}{125y^\frac{4}{3}}$??

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
• Jan 18th 2008, 05:18 PM
Rocher
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?
• Jan 18th 2008, 06:48 PM
topsquark
Quote:

Originally Posted by Rocher
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
$\displaystyle (125x^3y^2)^{2/3}$
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
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
• Jan 18th 2008, 06:52 PM
Rocher
Quote:

Originally Posted by topsquark
I mean if you have something like
$\displaystyle (125x^3y^2)^{2/3}$
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

$\displaystyle 125^{\frac{2}{3}}=5208\frac{1}{3}$

o.O No way that can be right... can it?
• Jan 18th 2008, 07:09 PM
topsquark
Quote:

Originally Posted by Rocher
$\displaystyle 125^{\frac{2}{3}}=5208\frac{1}{3}$

o.O No way that can be right... can it?

Hint: $\displaystyle 125 = 5^3$

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

-Dan
• Jan 18th 2008, 07:13 PM
Rocher
Is the answer $\displaystyle \frac{3x^6}{5y^4}$??
• Jan 18th 2008, 07:20 PM
topsquark
Quote:

Originally Posted by Rocher
Is the answer $\displaystyle \frac{3x^6}{5y^4}$??

I told you before that your exponent for the x was correct! Why did you change it?

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

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

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

-Dan
• Jan 18th 2008, 07:29 PM
Rocher
Omg, this is really frustrating :(. Once again, is the answer $\displaystyle \frac{3^{-2}x^\frac{10}{3}}{5^{-2}y^-\frac{4}{3}}$
• Jan 18th 2008, 07:31 PM
topsquark
Quote:

Originally Posted by Rocher
Omg, this is really frustrating :(. Once again, is the answer $\displaystyle \frac{3^{-2}}x^{\frac{10}{3}}{5^{-2}y^{-\frac{4}{3}}$

Yes!! Now all you need to do is get rid of the negative exponents.

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

-Dan
• Jan 18th 2008, 07:38 PM
Rocher
So do I put all the currently negative exponents at the bottom and all the positive ones at the top or something?
• Jan 18th 2008, 07:40 PM
topsquark
Quote:

Originally Posted by Rocher
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." (Nod)

-Dan
• Jan 18th 2008, 07:45 PM
Rocher
$\displaystyle \frac{5^2y^{\frac{4}{3}}x{\frac{10}{3}}}{3^2}$
Hopefully yes!

[EDIT]Oooh wait...

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

?? One of those two? XD
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