Solving Simple Index Questions with Brackets and Fractions

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Homework Help Overview

The discussion revolves around simplifying expressions involving indices, particularly focusing on the order of operations with brackets and fractions. Participants are attempting to clarify their understanding of index laws and how to apply them correctly in a given problem.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Some participants suggest starting with simplifying the contents of the brackets first. Others express confusion about specific steps in their calculations, particularly regarding the application of index laws and the treatment of negative exponents.

Discussion Status

There is an ongoing exchange of ideas, with some participants providing guidance on systematic approaches to simplifying the expression. Multiple interpretations of the problem are being explored, and while some participants report having solved the problem, there is still discussion about the correctness of their methods and results.

Contextual Notes

Participants mention the challenge of self-teaching and express a desire for clearer understanding, indicating that some foundational concepts may still be under discussion. There are references to specific index laws and the importance of careful application to avoid errors.

lloydowen
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Simple Indicies Question [SOLVED]

Homework Statement


I'm having a little problem with indicies, I know it's simple for someone with a lot of question.

So I'm wondering what to do first in this question, brackets or should I multiply out the fractions inside the brackets?

I have to simplify it that's all :)

Homework Equations



Image%202011-12-03%20at%203.41.54%20PM.png


The Attempt at a Solution



I don't have an attempt yet sorry :(
 
Last edited:
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I'd start by working out the inside of the brackets first. I think that'll be the easiest.
 
Thanks for the reply, so when I worked it out and simplified it first I got (x^2.5 x^2 x^-3)^2 which would equal to something like x^3, but in derive the answer is x^17..

What am I doing wrong :(?
 
lloydowen said:
I got (x^2.5 x^2 x^-3)^2

That X-3 isn't correct, is it??
 
Well I thought it was... What else could it be? I wish I had a good course tutor in College, I literally teach myself almost everything! *try*
 
lloydowen said:
Well I thought it was... What else could it be? I wish I had a good course tutor in College, I literally teach myself almost everything! *try*

You have [tex]F = \left( \frac{X^4 X^5 X}{X^{1.5} X^3 X^{-3}}\right)^2 .[/tex] The first step is to simplify the quantity inside the bracket, to obtain [itex]F = (X^a)^2.[/itex] So, the first order of business is to figure out what is 'a' in the following:
[tex]\frac{X^4 X^5 X}{X^{1.5} X^3 X^{-3}} = X^a.[/tex] After that, the rest is easy: [itex](X^a)^2 = X^{2a} .[/itex]

RGV
 
Sorry, common mistake, so it would be X^3?
 
Here's what I got... That previous post is very complicated :o

Image%202011-12-03%20at%204.43.59%20PM.png
 
You need to slowly apply these rules, you keep making mistakes.

[tex]a^b\cdot a^c=a^{b+c}[/tex]

[tex]\frac{a^b}{a^c}=a^{b-c}[/tex]

[tex]\left(a^b\right)^c=a^{bc}[/tex]
 
  • #10
How is [itex]X^{-3}[/itex] defined?? What is [itex]\frac{X}{X^{-3}}[/itex]??

Are you aware of the identity [itex]\frac{a^n}{a^m}=a^{n-m}[/itex]??
 
  • #11
Thanks guys I have solved this problem now :) I will keep going over and over until I get it perfect.
 
  • #12
Could you show us just to be sure? Because two wrongs can sometimes accidentally make a right :-p

And assuming you used the formulae correctly, just a tip, it'll probably be easier if you simplify the numerator first, then the denominator, then apply the quotient rule.
 
  • #13
What I did first was simplify the insides of the brackets. To do this I applied the 2nd law of indicies and take away the denominator from the numerator for example, first of all I got x^2.5 because 4-1.5 = Positive 2.5... Then the same for the next one in the brackets.

Now the last fraction in the equation at first I forgot the rule of two the same signs make positive and the opposite signs make a negative. So x-(-3) would be equal to x^3.

Then once I got all of them, I added them up to form (x^7.5)^2

(x^7.5)^2
=x^17
 
  • #14
lloydowen said:
What I did first was simplify the insides of the brackets. To do this I applied the 2nd law of indicies and take away the denominator from the numerator for example, first of all I got x^2.5 because 4-1.5 = Positive 2.5... Then the same for the next one in the brackets.

Now the last fraction in the equation at first I forgot the rule of two the same signs make positive and the opposite signs make a negative. So x-(-3) would be equal to x^3.

Then once I got all of them, I added them up to form (x^7.5)^2

(x^7.5)^2
=x^17

OK, that works, but you still have made some errors. However, what people are suggesting is that you do it more systematically, by simplifying the numerator and denominator separately:
[tex]\mbox{numerator} = X^4 X^5 X = X^{4+5+1} = X^{10}[/tex] and
[tex]\mbox{denominator} = X^{1.5} X^3 X^{-3} = X^{1.5 + 3 - 3} = X^{1.5},[/tex] to get [tex]\mbox{ratio} = \frac{\mbox{numerator}}{\mbox{denominator}} = \frac{X^{10}}{X^{1.5}} = X^{10 - 1.5} = X^{8.5}.[/tex] There is less chance of making an error when you do it this way.

RGV
 
  • #15
Oh right I see what you mean! I told you my Tutor was rubish :P I'll get into that routine then, Thank you! :)
 
  • #16
lloydowen said:
(x^7.5)^2
=x^17

How did you get from (x7.5)2=x17?
 
  • #17
Mentallic said:
How did you get from (x7.5)2=x17?

Ah Sorry I must of confused my self somewhere... I meant x^8.5 at least that's what I have on paper..
 
  • #18
Ahh ok just a typo then, because you did it twice :wink:
 
  • #19
Lmao not sure why I did it twice, I was very tired that night :P
 

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