Simple Indicies Question

1. Dec 3, 2011

lloydowen

Simple Indicies Question [SOLVED]

1. The problem statement, all variables and given/known data
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 :)

2. Relevant equations

3. The attempt at a solution

I don't have an attempt yet sorry :(

Last edited: Dec 3, 2011
2. Dec 3, 2011

micromass

Staff Emeritus
I'd start by working out the inside of the brackets first. I think that'll be the easiest.

3. Dec 3, 2011

lloydowen

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 :(?

4. Dec 3, 2011

micromass

Staff Emeritus
That X-3 isn't correct, is it??

5. Dec 3, 2011

lloydowen

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*

6. Dec 3, 2011

Ray Vickson

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

RGV

7. Dec 3, 2011

lloydowen

Sorry, common mistake, so it would be X^3?

8. Dec 3, 2011

lloydowen

Here's what I got... That previous post is very complicated

9. Dec 3, 2011

Mentallic

You need to slowly apply these rules, you keep making mistakes.

$$a^b\cdot a^c=a^{b+c}$$

$$\frac{a^b}{a^c}=a^{b-c}$$

$$\left(a^b\right)^c=a^{bc}$$

10. Dec 3, 2011

micromass

Staff Emeritus
How is $X^{-3}$ defined?? What is $\frac{X}{X^{-3}}$??

Are you aware of the identity $\frac{a^n}{a^m}=a^{n-m}$??

11. Dec 3, 2011

lloydowen

Thanks guys I have solved this problem now :) I will keep going over and over until I get it perfect.

12. Dec 3, 2011

Mentallic

Could you show us just to be sure? Because two wrongs can sometimes accidentally make a right :tongue:

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. Dec 3, 2011

lloydowen

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. Dec 3, 2011

Ray Vickson

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:
$$\mbox{numerator} = X^4 X^5 X = X^{4+5+1} = X^{10}$$ and
$$\mbox{denominator} = X^{1.5} X^3 X^{-3} = X^{1.5 + 3 - 3} = X^{1.5},$$ to get $$\mbox{ratio} = \frac{\mbox{numerator}}{\mbox{denominator}} = \frac{X^{10}}{X^{1.5}} = X^{10 - 1.5} = X^{8.5}.$$ There is less chance of making an error when you do it this way.

RGV

15. Dec 3, 2011

lloydowen

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. Dec 4, 2011

Mentallic

How did you get from (x7.5)2=x17?

17. Dec 4, 2011

lloydowen

Ah Sorry I must of confused my self somewhere... I meant x^8.5 at least thats what I have on paper..

18. Dec 4, 2011

Mentallic

Ahh ok just a typo then, because you did it twice

19. Dec 4, 2011

lloydowen

Lmao not sure why I did it twice, I was very tired that night :P