## anonymous one year ago (x^-2y^-4x^3)^-2 First remove the parenthesis now we use the ab = 1/ab property x^-2 = 1/x^2 now we have a positive exponent x^2. Next y^-4 = 1/y^4 now we have three positive exponents now we need to make the negative 2 positive. -2= 1/-2 now we have x^2y^4x^3 add the like terms x^2+ x^3 = x^5 (x^5y^4)^2

1. anonymous

Can you help me with this @misssunshinexxoxo

2. anonymous

@SolomonZelman?

3. SolomonZelman

$$\LARGE\color{black}{ \displaystyle \large \left(x^{-2}~y^{-4}~x^3\right)^{-2} }$$ like this?

4. anonymous

Yeah. Brb five minutes or less sorry making lunch.

5. SolomonZelman

:)

6. SolomonZelman

Ok, I will tell you another property..... $$\displaystyle \Large {\rm w}^\color{red}{\rm \LARGE a}\times{\rm w}^\color{blue}{\rm \LARGE b}={\rm w}^{\color{red}{\rm \LARGE a}+\color{blue}{\rm \LARGE b}}$$

7. SolomonZelman

this property, can be applied to the terms with an x. lets do a little replacement, just to visualize it better: $$\LARGE\color{black}{ \displaystyle \large \left(x^{-2}~\cdot~y^{-4}~\cdot~x^3\right)^{-2} }$$ $$\LARGE\color{black}{ \displaystyle \large \left(x^{-2}~\cdot~x^3~\cdot~y^{-4}\right)^{-2} }$$

8. SolomonZelman

so, you can apply the property I told you just now to the first 2 terms in this expression.

9. anonymous

Back Okay.

10. SolomonZelman

ok, take your time to read, and apply the property i have provided...

11. anonymous

So we don't change the terms into positive exponents?

12. SolomonZelman

we will do that later, but for now we just want to simplify inside the parenthesis (using this new property i told you).....

13. SolomonZelman

$$\LARGE\color{black}{ \displaystyle \large \left(x^{-2}~\cdot~x^3~\cdot~y^{-4}\right)^{-2} }$$ (this is your expression - the order of how you multiply doesn't matter) ---------------------------------------------------- Knowing that: $$\displaystyle \Large {\rm w}^\color{red}{\rm \LARGE a}\times{\rm w}^\color{blue}{\rm \LARGE b}={\rm w}^{\color{red}{\rm \LARGE a}+\color{blue}{\rm \LARGE b}}$$ then, $$\displaystyle \Large {\rm x}^\color{red}{\rm \LARGE -2}\times{\rm w}^\color{blue}{\rm \LARGE 3}={\rm x}^{?}$$

14. anonymous

By the bases being different i can't add them together right. so i guess x would be still x^-2

15. SolomonZelman

you can apply it to the " x$$^{-2}$$ • x$$^3$$ " part (where the base is same, the base is x)

16. anonymous

we would just have x^1 which we don't count so we just have x.

17. anonymous

Sorry thought i already enter it forgot i had to click on post. >~< @SolomonZelman

18. SolomonZelman

yes, x^1 is correct.

19. SolomonZelman

And x^1 is the same exact thing as x (besides the fact that you waste extra space and time when writing ^1 there)

20. SolomonZelman

we had: $$\LARGE\color{black}{ \displaystyle \large \left(x^{-2}~\cdot~x^3~\cdot~y^{-4}\right)^{-2} }$$ our new expression (that is equivalent to the previous one, of course) is: $$\LARGE\color{black}{ \displaystyle \large \left(x~\cdot~y^{-4}\right)^{-2} }$$

21. SolomonZelman

now, apply the property of $$a^{-b}$$=$$\dfrac{1}{a^b}$$ to the " y$$^{-4}$$ " part

22. anonymous

Okay y^-4 =1/y^4 @SolomonZelman

23. SolomonZelman

yes

24. anonymous

So how do i solve for -2 then?

25. SolomonZelman

our expression: $$\LARGE\color{black}{ \displaystyle \large \left(x~\cdot~y^{-4}\right)^{-2} }$$ is going to become: $$\LARGE\color{black}{ \displaystyle \large \left(x~\cdot~\dfrac{1}{y^{4}}\right)^{-2} }$$ $$\LARGE\color{black}{ \displaystyle \large \left(\dfrac{x}{y^{4}}\right)^{-2} }$$

26. anonymous

Okay, so we don't change two?

27. SolomonZelman

I will give you another propery (which can is derived from $$a^{-b}$$=1 /$$a^b$$) the property is: $$\color{black}{ \displaystyle \large \left(\dfrac{\color{red}{\rm a}}{\color{blue}{\rm b}}\right)^{\Large-w} = \left(\dfrac{\color{blue}{\rm b}}{\color{red}{\rm a}}\right)^{\Large w} }$$

28. SolomonZelman

Text - correct: which can *be* derived...

29. SolomonZelman

Knowing that: $$\color{black}{ \displaystyle \large \left(\dfrac{\color{red}{\rm a}}{\color{blue}{\rm b}}\right)^{\Large-w} = \left(\dfrac{\color{blue}{\rm b}}{\color{red}{\rm a}}\right)^{\Large w} }$$ then, therefore: $$\color{black}{ \displaystyle \large \left(\dfrac{\color{red}{ x}}{\color{blue}{ y^4}}\right)^{\Large-2} = \left(\dfrac{\color{blue}{y^4}}{\color{red}{x}}\right)^{\Large 2} }$$

30. SolomonZelman

so far so good?

31. anonymous

Yeah just make the reciprocal and it turn the negative into a positive.

32. anonymous

Thanks for your help again. I appreciate it. @SolomonZelman

33. SolomonZelman

this is not it

34. anonymous

There's more DX

35. SolomonZelman

$$\color{black}{ \displaystyle \large \left(\dfrac{\color{red}{ x}}{\color{blue}{ y^4}}\right)^{\Large-2} = \left(\dfrac{\color{blue}{y^4}}{\color{red}{x}}\right)^{\Large 2}=\dfrac{\left(\color{blue}{y^4}\right)^2}{\left(\color{red}{x}\right)^2} }$$

36. SolomonZelman

yes, this one more step...

37. anonymous

Alright.

38. SolomonZelman

$$($$y$$^4$$$$)$$$$\large^{^2}$$ = y$$^{4~\times~ 4}$$ = y$$^?$$

39. anonymous

y^16

40. SolomonZelman

yes, y$$^{16}$$/x$$^2$$

41. SolomonZelman

everything makes sense?

42. anonymous

Yeah so far do i divide by x or i'm just writing it as y^16 over x^2

43. SolomonZelman

you had $$\color{black}{ \displaystyle \large \left(\dfrac{\color{blue}{y^4}}{\color{red}{x}}\right)^{\Large 2}=\dfrac{\left(\color{blue}{y^4}\right)^2}{\left(\color{red}{x}\right)^2} }$$

44. anonymous

so we multiply the exponent y^4 with 2 and we get 16/x^2 Yeah it makes sense :)

45. SolomonZelman

applying the rule: $$\color{black}{ \displaystyle \large \left(\dfrac{\color{blue}{\rm A}}{\color{red}{\rm B}}\right)^{\Large w}=\dfrac{\left(\color{blue}{\rm A}\right)^w}{\left(\color{red}{\rm B}\right)^w} }$$

46. SolomonZelman

and then after we applied this rule above, we simplify on the top.

47. SolomonZelman

Which gives the result of: y$$^{16}$$/x$$^2$$

48. SolomonZelman

and that is it. can't do anything with this....

49. anonymous

Alright thanks @SolomonZelman

50. SolomonZelman

yw, gtg.