# Need Help Exponents Due Tomorow

• April 23rd 2008, 02:21 PM
TH1
Need Help Exponents Due Tomorow

(-3x^2y)^3 (-3x^-3y)^2

(5a^-1b^2)^-2 / 125a^5b^-3

(8x^6y^-3)^1/3 / (2xy)^3

Any help would be greatly appreciated
• April 23rd 2008, 02:23 PM
Mathstud28
Quote:

Originally Posted by TH1

(-3x^2y)^3 (-3x^-3y)^2

(5a^-1b^2)^-2 / 125a^5b^-3

(8x^6y^-3)^1/3 / (2xy)^3

Any help would be greatly appreciated

Do you mean this
$(-3x^{2y})^3\cdot(-3x^{-3y})^2=-3x^0=-3$?
• April 23rd 2008, 02:29 PM
TH1
No since the variable y isn't suppose to be an exponent but I meant everything else
• April 23rd 2008, 02:35 PM
Mathstud28
Quote:

Originally Posted by TH1
No since the variable y isn't suppose to be an exponent but I meant everything else

I will show you all the tools then

YOU apply them

$(ab)^{n}=a^{n}\cdot{b^{n}}$

$a^b\cdot{a^c}=a^{b+c}$

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

using this last definition in conjunction with the second one we can see that

$\frac{a^c}{a^b}=a^c\cdot{a^{-b}}=a^{c-b}$

$\sqrt[c]{a}=a^{\frac{1}{c}}$

Apply those..not all are used...it is up to you to figure it out...if you have any specific problems post them and we will help

Good luck
• April 23rd 2008, 02:37 PM
Moo
Hello,

Applying the formulas : $(a^b)^c=a^{bc}$ and $a^b a^c=a^{b+c}$

$(-3x^2 y)^3 (-3x^{-3} y)^2=(-3)^3 x^6 y^3 (-3)^2 x^{-6} y^2$

$=(-3)^5 x^{6-6} y^5=(-3)^5 y^5=(-3y)^5$

Can you calculate the followings ?
• April 23rd 2008, 02:40 PM
TH1
Would the final answer be 243y^5 and what about the rest of the other questions
• April 23rd 2008, 02:42 PM
Moo
There's a - sign missing...

It would be exactly the same method for the followings..
• April 23rd 2008, 03:14 PM
TH1
For the last one how would you do that one since the exponent is a fraction and for the second one would the answer be 25a^3b
• April 23rd 2008, 03:27 PM
Mathstud28
Quote:

Originally Posted by TH1
For the last one how would you do that one since the exponent is a fraction and for the second one would the answer be 25a^3b

$\frac{1}{3}+3=\frac{1}{3}+\frac{9}{3}=\frac{10}{3}$
• April 23rd 2008, 04:08 PM
TH1
I dont understand what you wrote since for the last one the exponent is 1/3 and I dont know what to do for that part
• April 23rd 2008, 04:10 PM
Mathstud28
Quote:

Originally Posted by TH1
I dont understand what you wrote since for the last one the exponent is 1/3 and I dont know what to do for that part

What do you think you do?
• April 23rd 2008, 04:18 PM
Mathstud28
• April 23rd 2008, 04:30 PM
TH1
For the exponent 1/3 i know you have to evaluate it but when you evaluate it on 8 would it be 8/3 or would you simplify 8 as a power of 2^3 and the answer be 2 since the 3 cancels out
2^3 1/3
• April 23rd 2008, 04:33 PM
Mathstud28
Quote:

Originally Posted by TH1
For the exponent 1/3 i know you have to evaluate it but when you evaluate it on 8 would it be 8/3 or would you simplify 8 as a power of 2^3 and the answer be 2 since the 3 cancels out
2^3 1/3

if you are talking about $8^{\frac{1}{3}}=(2^3)^{\frac{1}{3}}=2^{\frac{3}{3} }=2^1=2$ then yes
• April 23rd 2008, 04:40 PM
TH1
Thanks I have another question
if x=1/2 evaluate the expression (-2x^-2)^3(6x)^2/ 2(-3x^-1)^3

I have evaluated part of it without substituting x=1/2 and I have
(-8x^-6)(36x^2)/ 2(-27x^-3) so what do I do from there