- calculusxy

Help with exponents!
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- chestercat

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- calculusxy

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

- calculusxy

- anonymous

you want to work inside first or outside first?

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## More answers

- calculusxy

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

- anonymous

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

- 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 \(-3\)

- calculusxy

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

- calculusxy

i meant to say (a^3)^2

- anonymous

you lost me there

- anonymous

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

- calculusxy

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

- calculusxy

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

- anonymous

oooh

- calculusxy

The first parenthesis is raised to the third power.

- anonymous

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

- calculusxy

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

- anonymous

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

- calculusxy

the whole fraction is raised to the power of three.

- anonymous

can you post a screen shot?

- calculusxy

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

- calculusxy

- calculusxy

- anonymous

ooh k

- anonymous

multiply each exponent in the first parentheses by 3

- calculusxy

##### 1 Attachment

- anonymous

correct, minus the typo

- anonymous

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

- anonymous

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

- anonymous

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

- calculusxy

okay

- anonymous

so far so good?

- calculusxy

yes

- anonymous

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

- calculusxy

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

- anonymous

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

- anonymous

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

- calculusxy

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

- anonymous

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

- anonymous

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

- calculusxy

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

- anonymous

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

- calculusxy

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

- anonymous

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

- calculusxy

yes

- anonymous

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

- calculusxy

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

- anonymous

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

- anonymous

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

- anonymous

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

- calculusxy

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

- anonymous

the c is in the denominator right?

- calculusxy

oh yes

- anonymous

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

- calculusxy

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

- calculusxy

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

- anonymous

yw

- anonymous

kk

- calculusxy

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

- calculusxy

that was a y^{18}

- anonymous

i knew that ~

- anonymous

looks good to me!

- anonymous

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

- calculusxy

THANK YOU FOR YOUR HELP!

- anonymous

yw

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