# Thread: Need Help Exponents Due Tomorow

1. ## 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

2. 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
$(-3x^{2y})^3\cdot(-3x^{-3y})^2=-3x^0=-3$?

3. No since the variable y isn't suppose to be an exponent but I meant everything else

4. 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

5. 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 ?

6. Would the final answer be 243y^5 and what about the rest of the other questions

7. There's a - sign missing...

It would be exactly the same method for the followings..

8. 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

9. 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
$\frac{1}{3}+3=\frac{1}{3}+\frac{9}{3}=\frac{10}{3}$

10. 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

11. 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?

12. 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

13. 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

14. 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