Question

how do I simplify (27x^(1/2)y^(6))^(-1/3)

Answer 1

http://www.math-play.com/image-exponents-rules.jpg

Answer 2

any part you need clearing out?

Answer 3

my answer is 1/3x^(1/6) y^2

Answer 4

but I need to show my steps and i don't know how

Answer 5

hello?

Answer 6

$$
\pmatrix{27x^{\frac{1}{2}}y^6}^{-\frac{1}{3}}
$$

Answer 7

start by the outtermost exponent, what would that give you?

Answer 8

yup thats right

Answer 9

-1/3?

Answer 10

(27x^(1/2)y^(6))^(-1/3) can you rewrite it like shown on our paper use the equation botton

Answer 11

yes, but what would say... \(a^{-1} = ? \)

Answer 12

-a?

Answer 13

ahh I am so confused

Answer 14

ahemm, ok
well
have you checked http://www.math-play.com/image-exponents-rules.jpg yet?

Answer 15

yeah

Answer 16

because if you dunno the "exponent rules", then there's no use for this test

Answer 17

ok so do I just apply each individual part to the rules?

Answer 18

yes, starting from the outtermost exponent

Answer 19

o

Answer 20

so, from the sheet there in the link, look at the \(a^{-m} \)

Answer 21

it'd be the -1/3

Answer 22

in your case, yes

Answer 23

so, what would that leave you with?

Answer 24

3

Answer 25

$$
a^{-\color{red}{2}} = \frac{1}{a^\color{red}{2}}\\
so\\
\pmatrix{27x^{\frac{1}{2}}y^6}^{-\color{red}{\frac{1}{3}}} \implies ?
$$

Answer 26

\[27x ^{1/2}y ^{6}/ (1/3)\]

Answer 27

right?

Answer 28

sorry I got a bit lagged

Answer 29

well, I'll put this way

$$
a^{-\color{red}{2}} = \frac{1}{a^\color{red}{2}}\\
so\\
\pmatrix{a}^{-\color{red}{\frac{1}{3}}} \implies ?
$$

Answer 30

\[1/(27x ^{1/2}y^6)^{-1/3}\]

Answer 31

right?

Answer 32

well, very close, yes, it becomes a fraction, but the "negative exponent" once at the denominator becomes "positive"

Answer 33

oh whoops I forgot

Answer 34

$$
a^{-\color{red}{2}} = \frac{1}{a^\color{red}{2}}\\
so\\
\pmatrix{27x^{\frac{1}{2}}y^6}^{-\color{red}{\frac{1}{3}}} \implies
\cfrac{1}{\pmatrix{27x^{\frac{1}{2}}y^6}^{\color{red}{\frac{1}{3}}}}
$$

Answer 35

\[1/(27^{1/2}y ^{6})^{1/3}\]

Answer 36

as far as having an exponent outside a parentheses, see the rules sheet that shows

\(\large (a^m)^n = a^{mn}\)

Answer 37

so it comes into parenthesis

Answer 38

I would multiply it in but by which one?

Answer 39

yes

Answer 40

well -> \(\large (a^2b^3c)^4 = (a^{2\times4} b^{3\times 4} c^{1\times 4}) \)

Answer 41

oh so all of them

Answer 42

6*1/3 is 2 and 6* 1/2 is 3

Answer 43

so my new equation would look like \[1/27x^3 y^2\]

Answer 44

is that right?

Answer 45

oh wait I found an error

Answer 46

ahem, you're multiplying by a fraction, \(\frac{1}{3}\) not by "3"

Answer 47

it'd be 1/3 * 1/2 which =1/6 and 1/3*6=2

Answer 48

yes

Answer 49

yeah so my new equation would be 1/27x^(1/6) y^2

Answer 50

is that right?

Answer 51

well, yes, just that, any number by itself, usually would have an exponent of 1, so
25 = 25^1
27 = 27^1

\(\large 27^{1\times \frac{1}{3}}\)

Answer 52

ok now what?

Answer 53

$$\huge {
\pmatrix{27x^{\frac{1}{2}}y^6}^{-\color{red}{\frac{1}{3}}} \implies
\cfrac{1}{\pmatrix{27x^{\frac{1}{2}}y^6}^{\color{red}{\frac{1}{3}}}}\\
\implies \cfrac{1}{27^{\color{red}{\frac{1}{3}}}x^{\color{red}{\frac{1}{6}}}y^\color{red}{2}}
\implies \cfrac{1}{\sqrt[\color{red}{3}]{27}\sqrt[\color{red}{6}]{x}y^\color{red}{2}}
}
$$

Answer 54

ok and then I get

Answer 55

1/3x^(1/6)y^2

Answer 56

right?

Answer 57

yes

Answer 58

yay

Answer 59

thank you!

Answer 60

yw

Answer 61

can you help me with another problem?

Answer 62

if I can

Answer 63

ok it looks like \[-5/\sqrt{5}-\sqrt{3}\]

Answer 64

right, a fractional exponent, just means a root, so

$$\large {
a^{\frac{n}{m}} = \sqrt[m]{a^n}
}
$$

Answer 65

is that => $$\large {
\cfrac{-5}{\sqrt{5}-\sqrt{3}}
}?
$$