Question

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

Answer 1

@zepdrix

Answer 2

@triciaal

Answer 3

@nincompoop @whpalmer4 @pooja195 @paki @ganeshie8

Answer 4

mind helping?

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

or exponents i should say?

Answer 9

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`

Answer 10

mmm ok ...

Answer 11

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

Answer 12

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

Answer 13

oh

Answer 14

I belive that is -4? correct...

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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`.

Answer 20

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

Answer 21

right so find the lcm?

Answer 22

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

Answer 23

8

Answer 24

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

Answer 25

is this going to be 8 over 4?

Answer 26

no

Answer 27

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.

Answer 28

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

Answer 29

yay good job \c:/

Answer 30

:D mind helping me with a few more?

Answer 31

stay cool kid (⌐▨_▨)
oh sure

Answer 32

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

Answer 33

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?

Answer 34

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

Answer 35

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

Answer 36

there we go :D

Answer 37

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

Answer 38

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

Answer 39

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

Answer 40

yep I have used this method before...

Answer 41

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

Answer 42

first before you continue

Answer 43

shouldnt I be trying to find the sqrt?

Answer 44

and trying to figure out whetther its irrational or not?

Answer 45

because I believe that is 12.124355653

Answer 46

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.

Answer 47

which is irrational so now you may preceed

Answer 48

eh 12.12

Answer 49

no...

Answer 50

...

Answer 51

You don't use your calculator for this step.

Answer 52

no i rounded lol

Answer 53

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

Answer 54

You try to simplify the root by pulling out a `perfect square`.

Example:
\(\large\rm \sqrt{72}=\sqrt{2\cdot36}=\sqrt{2}\cdot\sqrt{36}=\sqrt{2}\cdot 6\)

Answer 55

oooo nvm its not 12.12

Answer 56

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

Answer 57

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

Answer 58

though*

Answer 59

im not sure 6 is simplest of forms though?

Answer 60

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

Answer 61

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 * _

Answer 62

/uh

Answer 63

not sure

Answer 64

Just use a calculator then :U

147 / 3

Answer 65

49

Answer 66

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

Answer 67

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

Answer 68

ok

Answer 69

7

Answer 70

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

Answer 71

ok