Simplify and express with positive indices

  • Thread starter jahaddow
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  • #1
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Ok, so I have this Expression, (6x+2 x 42x-4 x 35-x x 2x-6)/(124x+3 x 92x-3)

But it needs to be simplifyed, and expressed with positive indices. Now as far as my knowledge takes me, this can't be simplifyed, but then I am probably wrong. If anybody could help to solve, that would be great.
 

Answers and Replies

  • #2
lanedance
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maybe you could start by trying breaking things down into prime components eg.
[tex] 6^{x+2} = 3^{x+2}2^{x+2}[/tex]
 
  • #3
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I'm sorry but that example made no sense to me at all? I need more explanation.
 
  • #4
lanedance
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well say you have, separate powers of 2 multiplied toegther, you can simplify by
[tex] 2^{a}2^{b} = 2^{a+b}[/tex]

so break down each number into its prime components, then multiply similar numbers together
6 = 3.2
4 = 2.2 etc.
 
  • #5
HallsofIvy
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The rules of exponents,

[itex]a^xa^y= a^{x+y}[/itex] (which lanedance gave) and [itex](a^x)^y= a^{xy}[/itex], are usually learned long before taking Calculus!

You will also need to know that [itex]4= 2^2[/itex], [itex]6= 3(2)[/itex], [itex]9= 3^2[/itex], and [itex]12= 2^2 (3)[/itex]. Break everything down into powers of 2 and 3 and use the laws of exponents.
 
  • #6
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Could somebody please do this as an example, this is just the first one in the list I have to do.

Thankyou
 
  • #7
lanedance
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you've got all the info, why not try it out?
 
  • #8
All this made no sence to me either. Can someone show an example?
 
  • #9
lanedance
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do for the first terms
[tex] 6^{x+2}.4^{2x-4} = (3.2)^{x+2}.(2.2)^{2x-4}
= (3^{x+2}.2^{x+2})(2^{2x-4}. 2^{2x-4})
= (3^{x+2})(2^{x+2}.2^{2x-4}. 2^{2x-4})
= (3^{x+2})(2^{(x+2) + (2x-4) + (2x-4)})
= (3^{x+2})(2^{5x+6}) [/tex]

then for terms that are divided, the power will be negative
 
  • #10
sorry to be a bother but, that really just confused me even more.
If u can could you please show the whole example, and explain everything?
That would help heeps.
THANKS...:)
 
  • #11
lanedance
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i think you've got way more than enough to attempt something... even though its not your post, have a try & i'll guide you the through

basically all you need to do is the same for the last 4 terms
[itex] 3^{5-x} [/itex] - this is in lowest form
[itex] 2^{x-6} [/itex] - this is in lowest form
[itex] \frac{1}{12^{4x+3}} = 12^{-4x-3} = (2.3.3)^{-4x-3}[/itex]

and then i've leave the last for you... then multiply all the terms together & collect exponents
 
  • #12
i think you've got way more than enough to attempt something...
There may be heeps on info, but it makes no sence to me, thats y i wanted an example so i could attempt questions just like that.
 
  • #13
lanedance
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ok so what doesn't make sense?

post #9 is in effect a simple example
 
  • #14
the whole thing makes sence, can u tell me what answer u got, cus i got 1.5x^3.
Is this correct???
 
  • #15
wait thats not it, never mind.
can u please show me how u did ur working out cus im completly stuck?
 
  • #16
lanedance
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I have in post 9..

part of the confusion may be how its written:
Ok, so I have this Expression, (6x+2 x 42x-4 x 35-x x 2x-6)/(124x+3 x 92x-3)
QUOTE]

I take it to actually mean teh lower crosses are "multlipy", whilst the x's in the exponents are the variable "x", giving below
[tex]
(6^{+2}4^{2x-4} x 3{5-x}x 2{x-6})/(12{4x+3}x 9{2x-3})[/tex]

that's open to interpretation
 
  • #17
sorry but thats even more confusing. can u show me how u worked it out?
PLEASE!!!
 
  • #18
[tex] \frac{6^{(x + 2)} \ \cdot \ 4^{(2x - 4)} \ \cdot \ 3^{(5 - x)} \ \cdot \ 2^{(x - 6)}}{12^{(4x + 3)} \ \cdot \ 9^{(2x - 3)}} [/tex]

Is that the equation??? (the dot means multiplication).

Start off by recognizing that you can do things like;

[tex]6^{(x + 2)} = 6^x \cdot 6^2 [/tex]

Play with this one first, compare the values of the Left Hand Side & The Right Hand side by giving different x values, you'll see something surprising, then generalise this idea.
 
  • #19
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We can continue to try to make sense of what doesn't make sense to you. Which concept is confusing: prime factorization or the laws of exponents alluded to by HallsofIvy and lanedance?
 
  • #20
This will sound dumb but i dont even really know wat that means, thats y i wanted the answer, or the working so i can use it withother similar questions.
 
  • #23
I think you should watch this video

http://khanexercises.appspot.com/video?v=zM_p7tfWvLU [Broken]

It's number 53 in the list underneath it, if you understand it then watch numbers 54 & 55 after it.

I think it will help you out here a bit.

Come back after you watch these and read people's comments again, I bet everything will be clearer.

Then you should watch the videos on logarithms, ( I think I saw you in another thread with logarithm questions)
 
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  • #24
the video helped a little, but not with this question. it just makes me more fustrated. i cant do it no matter how many time u interpret it.
 
  • #25
If you watched 53,54 & 55 and then look at my first post here that should explain everything.

You've also got about 3 other ways to do this problem just given to you from people on this thread.

I can only guess that you haven't studied this stuff before or haven't looked at a good book explaining it.

There are these notes here = http://tutorial.math.lamar.edu/Classes/Alg/Alg.aspx that explicitly taught me how to do these problems ages ago, just read the 'Preliminaries' chapter, but really all you need to do is study this page here http://tutorial.math.lamar.edu/Classes/Alg/IntegerExponents.aspx that is from that link, I've just checked & the method to do this problem is taught explicitly.

If you give studying this page a shot & still can't do it then I'll show you step by step but you have to give this a shot first to learn for yourself.

I thought the video would just remind you of something you'd learned but you must not have ever seen this before, (or just understood it).

The above chapter I gave you should be 100% all you need to master this problem.
 
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