anonymous
  • anonymous
(x^-3)^-5x^6 a.x^21 b.x^-21 c.x^48 d.x^-48
Algebra
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
@KeithAfasCalcLover can you help me again ?
anonymous
  • anonymous
I might ;)...What is the question?
anonymous
  • anonymous
:) . Thats all they gave me

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anonymous
  • anonymous
@KeithAfasCalcLover
anonymous
  • anonymous
(x^-3)^-5x^6 a.x^21 b.x^-21 c.x^48 d.x^-48
anonymous
  • anonymous
Ehhh...lets see...Se the thing is that I don't know the syntax of the question... I mean it could be: \[(x^{-3})^{{(-5x)}^6}\] or \[(x^{-3})^{-5}*x^6\]
anonymous
  • anonymous
@KeithAfasCalcLover its the second one
anonymous
  • anonymous
Ahh that makes so much sense. Cool. So you know that: \((x^a)^b=x^{ab}\) and \(\frac{x^a}{x^b}=x^{a-b}\) and \((x^a)(x^b)=x^{a+b}\)
anonymous
  • anonymous
So you can use these rules to simplify some terms. For instance: \((x^{-3})^{-5}=x^{-3*-5}=x^{15}\)
anonymous
  • anonymous
And you should in the end get one term with the form: \(x^p\) and Ill let you figure out \(p\) ;-)
anonymous
  • anonymous
@KeithAfasCalcLover Sooooo Maaaaany Leeettttersss :(
anonymous
  • anonymous
I know...isn't it beautiful ;-) The three lines full of letters are called the "laws of exponents"
anonymous
  • anonymous
Not To Me It Isnt :/ i feel like i have dyslexia when i do algebra !
anonymous
  • anonymous
So like if I had something like \(x^2*x\), I could find out that that actually was \(x^2*x^1=x^{2+1}=x^3\).The laws are methods of simplifying exponents. And the higher you go in math, Like calculus and trigonometry and stuff, the more letters are used! Haha. Don't let the numbers trip you up though, they just mean a letter to hold the place of a number Which means that if you put ANY NUMBER in the equation, you can get an answer. For instance, using the example I used above, \(x^2*x=x^3\), Think of ANY NUMBER for x, and it'll be true. Sometimes it helps to put numbers into these rules and see that no matter what number you put, Its always true.
anonymous
  • anonymous
So lets say for: \((x^a)(x^b)=x^{a+b}\) Think of ANY NUMBER for x,a, and b and youll see that it holds true!. So lets say x=5,a=2, and b=8. ______________________________________ \((5^2)(5^8)=5^{10}\) \((25)(390625)=5^{10}\) \(9765625=5^{10}\) ______________________________________ And ill be willing to bet that \(5^{10}=9,765,625\)!
anonymous
  • anonymous
i understand that half but the question they gave me i just dont get @KeithAfasCalcLover
anonymous
  • anonymous
i think the answer is c. \[x^{48}\]
anonymous
  • anonymous
Im not saying its right or wrong but would you be able to show me how you got there? :-)
anonymous
  • anonymous
uhm i uhh lol .added the two exponents that were next to eachother the two negatives and it gave me 8 and i multiplied the 8 by the last exponent which was 6? \[(x ^{-3})^{-5}x ^{6}\] @KeithAfasCalcLover Help Me More ! im a slow learner
anonymous
  • anonymous
Alright well that one wasn't completely bang on but ill help you :-) You have three..."things" being done to x: You have \(x^{-3}\), You have all that \((x^{-3})^{-5}\) and then you have all that multiplied by \(x^6\), right?
anonymous
  • anonymous
yes @KeithAfasCalcLover
anonymous
  • anonymous
Would it be ok with you if I put: \(a=-3\) \(b=-5\) \(c=6\) ??
anonymous
  • anonymous
Ya !
anonymous
  • anonymous
@KeithAfasCalcLover
anonymous
  • anonymous
So then we can rewrite this as: \[(x^a)^b*x^c\] Which is a little bit neater ;) So then we know that \((x^a)^b=x^{ab}\) so we can rewrite the equation: \[(x^a)^b*x^c=x^{ab}*x^c\]. Cool? :-)
anonymous
  • anonymous
Yeah ! @KeithAfasCalcLover |dw:1377712167716:dw|
anonymous
  • anonymous
Great! And we know that \(x^a*x^b=x^{a+b}\) So we can rewrite the whole thing as: \[(x^a)^b*x^c=x^{ab}*x^c=x^{ab+c}\] And since we know what a,b,and c are, we can plug 'em in! Nice so far?
anonymous
  • anonymous
Yes @KeithAfasCalcLover
anonymous
  • anonymous
So we know a and b and c, so lets find out what exactly is : \(ab+c\) \(ab+c=(-3)(-5)+6=15+6=21\) So the answer is \(x^{21}\)! Did all that make sense? :-)
anonymous
  • anonymous
yes ! you shouldve done that earlier lmbo ! @KeithAfasCalcLover
anonymous
  • anonymous
Haha but the thing is using the letters made things so much easier. And those basic laws will take you a long way! Well im glad you got it Deivyneee :-)
anonymous
  • anonymous
Lol :) Thank You @KeithAfasCalcLover
anonymous
  • anonymous
My Pleasure, @Deivyneee

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