Solving Simple Index Questions with Brackets and Fractions

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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 :(
 
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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 :(?
 
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]
 
Thanks guys I have solved this problem now :) I will keep going over and over until I get it perfect.
 
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.
 
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
 
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
 
Oh right I see what you mean! I told you my Tutor was rubish :P I'll get into that routine then, Thank you! :)
 
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..
 
Lmao not sure why I did it twice, I was very tired that night :P