## some.random.cool.kid one year ago What is the simplified form of left parenthesis 27 x to the power of negative six end power y to the power of 12 end power right parenthesis to the power of start fraction two over three end fraction end power

1. some.random.cool.kid

@zepdrix

2. some.random.cool.kid

@triciaal

3. some.random.cool.kid

4. some.random.cool.kid

@nincompoop @whpalmer4 @pooja195 @paki @ganeshie8

5. some.random.cool.kid

mind helping?

6. zepdrix

$\large\rm \left[abc\right]^{n}=a^n b^n c^n$When we have a bunch of stuff being raised to a power, we have to apply the exponent to each of them.

7. zepdrix

$\large\rm \left[27x^{-6}y^{12}\right]^{2/3}=27^{2/3}(x^{-6})^{2/3}(y^{12})^{2/3}$Ok with that step? :)

8. some.random.cool.kid

mm got it am i supposed to add across? for the powers part?

9. some.random.cool.kid

or exponents i should say?

10. zepdrix

Have to apply one of your exponent rules from that point:$\large\rm \color{orangered}{(x^a)^b=x^{a\cdot b}}$When you have a power and another power, you multiply

11. some.random.cool.kid

mmm ok ...

12. zepdrix

So for the exponent on x, we'll multiply -6 and 2/3.

13. some.random.cool.kid

-6 x 12? I belive. what each of those two thirds?

14. some.random.cool.kid

oh

15. some.random.cool.kid

I belive that is -4? correct...

16. zepdrix

$\large\rm \left[27x^{-6}y^{12}\right]^{2/3}=27^{2/3}x^{-4}(y^{12})^{2/3}$k looks good :)

17. some.random.cool.kid

mmm next step is ? -4 x 12 or -4 x the other 2/3rds?

18. some.random.cool.kid

or is it -4 x 12 and 2/3 x 2/3?

19. some.random.cool.kid

or is it just multiply across lol? i believe its what i said first though

20. zepdrix

Notice that these two things have different bases. One has a base of x, the other a base of y, that's letting us know that they won't interact with one another. $$\large\rm x^{a}\cdot x^{b}=x^{a+b}$$ This rule that you're familiar with only works when they have the same base.

21. zepdrix

So the -4 on the x isn't going to interact in any way with the powers on y.

22. some.random.cool.kid

right so find the lcm?

23. zepdrix

You have $$\large\rm (y^{12})^{2/3}$$. Our exponent rule tells us to multiply these powers, ya? 12 * 2/3?

24. some.random.cool.kid

8

25. zepdrix

$\large\rm \left[27x^{-6}y^{12}\right]^{2/3}=\color{orangered}{27^{2/3}}x^{-4}y^{8}$k good :) almost done

26. some.random.cool.kid

is this going to be 8 over 4?

27. some.random.cool.kid

no

28. zepdrix

Try to pay attention to what I've been saying. They have different bases, so they won't interact with one another in any way.

29. some.random.cool.kid

its not nvm i answered my own thing i belive you leave those alon now right so now i have 9 x^4y^8?

30. zepdrix

yay good job \c:/

31. some.random.cool.kid

:D mind helping me with a few more?

32. zepdrix

stay cool kid (⌐▨_▨) oh sure

33. some.random.cool.kid

Ok great I have another one this one is a bit different then the last a little bit ...

34. some.random.cool.kid

A student uses the quadratic formula to solve a quadratic equation and determines that one of the solutions is x equals negative five plus square root of negative 147 end square root. What are the values of a and b if this solution is written in the form x = a + bi, where a and b are real numbers?

35. some.random.cool.kid

hold on i need to attach the numbers part this is too messy

36. zepdrix

$\large\rm x=-5+\sqrt{-147}$

37. some.random.cool.kid

38. some.random.cool.kid

there we go :D

39. zepdrix

Well, recall that we define the imaginary unit i, to be $$\large\rm \sqrt{-1}$$

40. zepdrix

So we can do something clever with the square root in our problem.

41. zepdrix

$\large\rm \sqrt{-147}=\quad\sqrt{-1\cdot147}=\quad\sqrt{-1}\cdot\sqrt{147}$I broke it up in a clever way to pull the sqrt(-1) outside like that. Do you see the i?

42. some.random.cool.kid

yep I have used this method before...

43. zepdrix

$\large\rm x=-5+\sqrt{-147}$$\large\rm x=-5+\color{orangered}{\sqrt{-1}}\cdot\sqrt{147}$

44. some.random.cool.kid

first before you continue

45. some.random.cool.kid

shouldnt I be trying to find the sqrt?

46. some.random.cool.kid

and trying to figure out whetther its irrational or not?

47. some.random.cool.kid

because I believe that is 12.124355653

48. zepdrix

Yes, try to simplify the square root if you're able to. In doesn't matter which order you do these steps in though :) No no, we don't want an ugly decimal approximation.

49. some.random.cool.kid

which is irrational so now you may preceed

50. some.random.cool.kid

eh 12.12

51. zepdrix

no...

52. some.random.cool.kid

...

53. zepdrix

You don't use your calculator for this step.

54. some.random.cool.kid

no i rounded lol

55. zepdrix

But you first used a calculator.. that's not what we're doing here.

56. zepdrix

You try to simplify the root by pulling out a perfect square. Example: $$\large\rm \sqrt{72}=\sqrt{2\cdot36}=\sqrt{2}\cdot\sqrt{36}=\sqrt{2}\cdot 6$$

57. some.random.cool.kid

oooo nvm its not 12.12

58. zepdrix

So in my example here, I noticed that 72 was a multiple of 36, 36 can be taken out of the root since it's a perfect square. it comes out as a 6

59. some.random.cool.kid

its either 21 or 7sqrtof 3 I believe correct? not sure which one yet tough

60. some.random.cool.kid

though*

61. some.random.cool.kid

im not sure 6 is simplest of forms though?

62. zepdrix

6sqrt2 is the simplified form of sqrt(72). That was just an example though.

63. zepdrix

Break down 147, what are some factors that we can use? One little math trick is to add up the digits, 1 + 4 + 7 = 12. 12 is a multiple of 3, so this number is divisible by 3. So 147 = 3 * _

64. some.random.cool.kid

/uh

65. some.random.cool.kid

not sure

66. zepdrix

Just use a calculator then :U 147 / 3

67. some.random.cool.kid

49

68. zepdrix

ok$\large\rm x=-5+\color{orangered}{\sqrt{-1}}\cdot\sqrt{147}$$\large\rm x=-5+\color{orangered}{\sqrt{-1}}\cdot\sqrt{3\cdot49}$$\large\rm x=-5+\color{orangered}{\sqrt{-1}}\cdot\sqrt{3}\cdot\sqrt{49}$

69. zepdrix

$\large\rm x=-5+\color{orangered}{\mathcal i}\cdot\sqrt{3}\cdot\sqrt{49}$What does sqrt(49) simplify to?

70. some.random.cool.kid

ok

71. some.random.cool.kid

7

72. zepdrix

Good good good. Let's write the i in the back, so it looks a lil bit cleaner.$\large\rm x=-5+7\sqrt3~\mathcal i$

73. some.random.cool.kid

ok