Question

Help with exponents!

Question attached below...

Answer 1

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

Answer 2

@jim_thompson5910 @Nnesha @satellite73

Answer 3

you want to work inside first or outside first?

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

i meant to say (a^3)^2

Answer 9

you lost me there

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

oooh

Answer 14

The first parenthesis is raised to the third power.

Answer 15

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

Answer 16

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

Answer 17

is that part ok? multiplied each exponent by \(-2\)

Answer 18

the whole fraction is raised to the power of three.

Answer 19

can you post a screen shot?

Answer 20

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

Answer 21

@satellite73

Answer 22

@jim_thompson5910

Answer 23

ooh k

Answer 24

multiply each exponent in the first parentheses by 3

Answer 25

correct, minus the typo

Answer 26

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

Answer 27

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

Answer 28

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

Answer 29

okay

Answer 30

so far so good?

Answer 31

yes

Answer 32

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

Answer 33

\[\frac{ 1 }{ a^{-6}b^2 }\]

Answer 34

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

Answer 35

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

Answer 36

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

Answer 37

ok lets back a second and look at \[a^6b^{-2}\]

Answer 38

the exponent that is negative is attached to the \(b\) only

Answer 39

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

Answer 40

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

Answer 41

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

Answer 42

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

Answer 43

yes

Answer 44

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

Answer 45

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

Answer 46

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

Answer 47

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

Answer 48

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

Answer 49

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

Answer 50

the c is in the denominator right?

Answer 51

oh yes

Answer 52

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

Answer 53

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

Answer 54

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

Answer 55

yw

Answer 56

kk

Answer 57

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

Answer 58

that was a y^{18}

Answer 59

i knew that ~

Answer 60

looks good to me!

Answer 61

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

Answer 62

THANK YOU FOR YOUR HELP!

Answer 63

yw