## calculusxy one year ago Help with exponents! Question attached below...

1. calculusxy

$\large (\frac{ 3n }{ 5m^{-3} })^2 (\frac{ 3m^4 }{ 2n^2 })^{-3}$

2. calculusxy

@jim_thompson5910 @Nnesha @satellite73

3. anonymous

you want to work inside first or outside first?

4. calculusxy

I don't know... anything is fine as long as i get a correct solution

5. anonymous

ok lets take care of the outside exponents first, not that it matter

6. anonymous

$\large (\frac{ 3n }{ 5m^{-3} })^2 (\frac{ 3m^4 }{ 2n^2 })^{-3}$multiply each exponent in the first parentheses by 2, in the second by $$-3$$

7. calculusxy

$(\frac{ a^2b }{ b^{-3}c^4 })^2 -> \frac{ a^6b^3 }{ b^{-9}c^{12} }$

8. calculusxy

i meant to say (a^3)^2

9. anonymous

you lost me there

10. anonymous

your question has m and n in it, not a and b

11. calculusxy

oh god.. i wrote the wrong question.. wait a minute . (sorry)

12. calculusxy

$\huge (\frac{ a^2b^3 }{ b^{-3}c^4 })(a^{-3}b)^{-2}$

13. anonymous

oooh

14. calculusxy

The first parenthesis is raised to the third power.

15. anonymous

second one is $\huge a^6b^{-2}=\frac{a^6}{b^2}$

16. calculusxy

and the b doesn't have the third power...

17. anonymous

is that part ok? multiplied each exponent by $$-2$$

18. calculusxy

the whole fraction is raised to the power of three.

19. anonymous

can you post a screen shot?

20. calculusxy

$\huge (\frac{ a^2b }{ b^{-3}c^4 })^3(a^{-3}b)^{-2}$

21. calculusxy

@satellite73

22. calculusxy

@jim_thompson5910

23. anonymous

ooh k

24. anonymous

multiply each exponent in the first parentheses by 3

25. calculusxy

26. anonymous

correct, minus the typo

27. anonymous

$(\frac{ a^2b }{ b^{-3}c^4 })^3 \to \frac{ a^6b^3 }{ b^{-9}c^{12} }$

28. anonymous

now before we mess with the second part, lets clean this up first

29. anonymous

the exponent of $$-9$$ in the denominator brings the term up to the numerator, adding the exponetns makes this $\large \frac{a^6b^{12}}{c^{12}}$

30. calculusxy

okay

31. anonymous

so far so good?

32. calculusxy

yes

33. anonymous

next term is $(a^{-3}b)^{-2}$ multiply each exponent by $$-2$$ get $\large a^6b^{-2}$

34. calculusxy

$\frac{ 1 }{ a^{-6}b^2 }$

35. anonymous

on no don't introduce a negative exponent, leave it be

36. anonymous

we got $\large \frac{a^6b^{12}}{c^{12}}\times a^6b^{-2}$ add up the exponents to finish it

37. calculusxy

but since the exponent is negative doesn't it let the base become it's reciprocal?

38. anonymous

ok lets back a second and look at $a^6b^{-2}$

39. anonymous

the exponent that is negative is attached to the $$b$$ only

40. calculusxy

no it's the whole expression including the a^-3 and b

41. anonymous

if for some reason you want to write this with positive exponents only, it would be $\large a^6b^{-2}=\frac{a^6}{b^2}$

42. calculusxy

since the whole thing has a parenthesis around it and there is a -2 as the exponent.

43. anonymous

yes it was $\large (a^{-3}b)^{-2}$ right?

44. calculusxy

yes

45. anonymous

ok so you COULD write this as $\frac{1}{(a^{-3}b)^2}$ but you do not need to or want to

46. calculusxy

but my teacher preferred for us to go this way, and that's why i feel comfortable using it.

47. anonymous

what you want to do is multiply each exponent by $$-2$$ to get $\large a^6b^{-2}$

48. anonymous

than multiplying that by the first part is easy add the exponents

49. anonymous

if you do it the way your teacher likes you will get the same thing because $\frac{1}{(a^{-3}b)^2}=\frac{1}{a^{-6}b^2}=\frac{a^6}{b^2}$

50. calculusxy

i just want to know if my final answer is correct or not: a^12c^12b^10

51. anonymous

the c is in the denominator right?

52. calculusxy

oh yes

53. anonymous

if you want to write it in one line it should be $a^{12}b^{10}c^{-12}$

54. calculusxy

i prefer the fraction way... but thanks for the "one line" thing

55. calculusxy

and i just want to make sure my answer for another question as well.

56. anonymous

yw

57. anonymous

kk

58. calculusxy

$\large (5x^3y^{-3})^{-2}(2x^5y^{-4})^{-3} = \frac{ y^18 }{ 200x^{21} }$

59. calculusxy

that was a y^{18}

60. anonymous

i knew that ~

61. anonymous

looks good to me!

62. anonymous

not to make light of this but it is really just bookkeping right? $(-3)(-2)+(-4)(-3)=6+12=18$ for example

63. calculusxy