calculusxy
  • calculusxy
Help with exponents! Question attached below...
Mathematics
chestercat
  • chestercat
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calculusxy
  • calculusxy
\[\large (\frac{ 3n }{ 5m^{-3} })^2 (\frac{ 3m^4 }{ 2n^2 })^{-3}\]
calculusxy
  • calculusxy
anonymous
  • anonymous
you want to work inside first or outside first?

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calculusxy
  • calculusxy
I don't know... anything is fine as long as i get a correct solution
anonymous
  • anonymous
ok lets take care of the outside exponents first, not that it matter
anonymous
  • 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
  • calculusxy
\[(\frac{ a^2b }{ b^{-3}c^4 })^2 -> \frac{ a^6b^3 }{ b^{-9}c^{12} }\]
calculusxy
  • calculusxy
i meant to say (a^3)^2
anonymous
  • anonymous
you lost me there
anonymous
  • anonymous
your question has m and n in it, not a and b
calculusxy
  • calculusxy
oh god.. i wrote the wrong question.. wait a minute . (sorry)
calculusxy
  • calculusxy
\[\huge (\frac{ a^2b^3 }{ b^{-3}c^4 })(a^{-3}b)^{-2}\]
anonymous
  • anonymous
oooh
calculusxy
  • calculusxy
The first parenthesis is raised to the third power.
anonymous
  • anonymous
second one is \[\huge a^6b^{-2}=\frac{a^6}{b^2}\]
calculusxy
  • calculusxy
and the b doesn't have the third power...
anonymous
  • anonymous
is that part ok? multiplied each exponent by \(-2\)
calculusxy
  • calculusxy
the whole fraction is raised to the power of three.
anonymous
  • anonymous
can you post a screen shot?
calculusxy
  • calculusxy
\[\huge (\frac{ a^2b }{ b^{-3}c^4 })^3(a^{-3}b)^{-2}\]
calculusxy
  • calculusxy
calculusxy
  • calculusxy
anonymous
  • anonymous
ooh k
anonymous
  • anonymous
multiply each exponent in the first parentheses by 3
calculusxy
  • calculusxy
anonymous
  • anonymous
correct, minus the typo
anonymous
  • anonymous
\[(\frac{ a^2b }{ b^{-3}c^4 })^3 \to \frac{ a^6b^3 }{ b^{-9}c^{12} }\]
anonymous
  • anonymous
now before we mess with the second part, lets clean this up first
anonymous
  • 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
  • calculusxy
okay
anonymous
  • anonymous
so far so good?
calculusxy
  • calculusxy
yes
anonymous
  • anonymous
next term is \[(a^{-3}b)^{-2}\] multiply each exponent by \(-2\) get \[\large a^6b^{-2}\]
calculusxy
  • calculusxy
\[\frac{ 1 }{ a^{-6}b^2 }\]
anonymous
  • anonymous
on no don't introduce a negative exponent, leave it be
anonymous
  • anonymous
we got \[\large \frac{a^6b^{12}}{c^{12}}\times a^6b^{-2}\] add up the exponents to finish it
calculusxy
  • calculusxy
but since the exponent is negative doesn't it let the base become it's reciprocal?
anonymous
  • anonymous
ok lets back a second and look at \[a^6b^{-2}\]
anonymous
  • anonymous
the exponent that is negative is attached to the \(b\) only
calculusxy
  • calculusxy
no it's the whole expression including the a^-3 and b
anonymous
  • 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
  • calculusxy
since the whole thing has a parenthesis around it and there is a -2 as the exponent.
anonymous
  • anonymous
yes it was \[\large (a^{-3}b)^{-2}\] right?
calculusxy
  • calculusxy
yes
anonymous
  • anonymous
ok so you COULD write this as \[\frac{1}{(a^{-3}b)^2}\] but you do not need to or want to
calculusxy
  • calculusxy
but my teacher preferred for us to go this way, and that's why i feel comfortable using it.
anonymous
  • anonymous
what you want to do is multiply each exponent by \(-2\) to get \[\large a^6b^{-2}\]
anonymous
  • anonymous
than multiplying that by the first part is easy add the exponents
anonymous
  • 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
  • calculusxy
i just want to know if my final answer is correct or not: a^12c^12b^10
anonymous
  • anonymous
the c is in the denominator right?
calculusxy
  • calculusxy
oh yes
anonymous
  • anonymous
if you want to write it in one line it should be \[a^{12}b^{10}c^{-12}\]
calculusxy
  • calculusxy
i prefer the fraction way... but thanks for the "one line" thing
calculusxy
  • calculusxy
and i just want to make sure my answer for another question as well.
anonymous
  • anonymous
yw
anonymous
  • anonymous
kk
calculusxy
  • calculusxy
\[\large (5x^3y^{-3})^{-2}(2x^5y^{-4})^{-3} = \frac{ y^18 }{ 200x^{21} }\]
calculusxy
  • calculusxy
that was a y^{18}
anonymous
  • anonymous
i knew that ~
anonymous
  • anonymous
looks good to me!
anonymous
  • anonymous
not to make light of this but it is really just bookkeping right? \[(-3)(-2)+(-4)(-3)=6+12=18\] for example
calculusxy
  • calculusxy
THANK YOU FOR YOUR HELP!
anonymous
  • anonymous
yw

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