anonymous
  • anonymous
simplify the expression (a^2)^-3(a^3b)^2(b^3)^4
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
Solve for b over the real numbers: b^14 = 0 Eliminate the exponent. Take the 14^th root of both sides: Answer: | | b = 0
anonymous
  • anonymous
how do i do that
DebbieG
  • DebbieG
Is this the expression? \(\Large (a^2)^{-3}(a^3b)^2(b^3)^4\) There is nothing to "solve for b", it isn't an equation... it's just an expression to be simplified. But is what I have above, correct? It's hard to tell from plain text sometimes. :)

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anonymous
  • anonymous
yes
DebbieG
  • DebbieG
You'll just need a few of the rules for exponents: \(\Large (x^m)^{n}=x^{mn}\) \(\Large (xy)^m=x^{m}y^m\) \(\Large x^{-1}=\dfrac{1}{x} \) \(\Large x^{-m}=\dfrac{1}{x^m} \)
anonymous
  • anonymous
ok
DebbieG
  • DebbieG
So start with the exponents outside each set of ( ) and simplify that.... \(\Large (a^2)^{-3}(a^3b)^2(b^3)^4\\ \Large =a^{\Box} (a^{\Box}b^{\Box})(b^{\Box})\) What goes in each "box"
anonymous
  • anonymous
im confused
DebbieG
  • DebbieG
Oh, I forgot one other rule you are going to need: \(\Large x^m\cdot x^n=x^{m+n}\)
DebbieG
  • DebbieG
OK, tell me where you're confused. Those rules for exponents that I posted above, you've seen those before, right? Do you understand them?
anonymous
  • anonymous
the box part that confused me
DebbieG
  • DebbieG
\(\Large (x^m)^{n}=x^{mn}\) so, e.g., \(\Large (3^2)^{5}=3^{10}\) \(\Large (z^2)^{-3}=z^{-6}\) etc.
anonymous
  • anonymous
oh ok i understand that
DebbieG
  • DebbieG
ah, ok.. sorry, I was just trying to show you what you needed to do first. One step at a time... you have: \(\Large (a^2)^{-3}(a^3b)^2(b^3)^4\) So tell me what you can do with \(\Large (a^2)^{-3}\) ?? What is \(\Large (a^2)^{-3}=\)?
anonymous
  • anonymous
\[a ^{-6}\]
DebbieG
  • DebbieG
Good! Now how about \(\Large (a^3b)^2=?\)
anonymous
  • anonymous
\[a^{6}b ^{2}\]
DebbieG
  • DebbieG
Perfect. \(\Large (b^3)^4=?\)
anonymous
  • anonymous
\[b ^{12}\]
DebbieG
  • DebbieG
See, you're getting it! Now you have: \(\Large a^{-6}a^6b^2b^{12}\) Which you can mentally think of as grouped according to where you have the base of a, and where you have the base of b, e.g., think of it as: \(\Large (a^{-6}\cdot a^6)(b^2\cdot b^{12})\) And now you'll apply that last rule: \(\Large x^m\cdot x^n=x^{m+n}\) E.g. \(\Large 3^2\cdot 3^5=3^{7}\) \(\Large z^3\cdot z^8=z^{11}\) etc.... Now use that on: \(\Large (a^{-6}\cdot a^6)(b^2\cdot b^{12})\)
DebbieG
  • DebbieG
(you'll use it trice, once on the "a" part and again on the "b" part)
DebbieG
  • DebbieG
*twice lol
anonymous
  • anonymous
\[a ^{0}b ^{14}\]
DebbieG
  • DebbieG
Perfect! Now, do you know the rule for \(x^0\)? What is any base, raised to the power 0?
anonymous
  • anonymous
1
DebbieG
  • DebbieG
Exactly! So the fully simplified form of this expression is...?
anonymous
  • anonymous
1\[1b ^{14}\]
DebbieG
  • DebbieG
That's right! Now, you don't really need the "1"out in front... e.g..... "1x" is really just written as "x" in fully simplified form, follow me?
anonymous
  • anonymous
yes thank you so much
DebbieG
  • DebbieG
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DebbieG
  • DebbieG
no problem, happy to help. :)
anonymous
  • anonymous
yay gold star

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