Fractional Exponential

  • Thread starter Sirsh
  • Start date
  • #1
267
10
I have been given a Indice. I've been trying to figure it out for awhile and need some assistance, It'd be great if someone could work it out and show me the steps they did and even explain it.

([a-2b-3/ 2a3b-4]2 )/ [ab-15]/a-3b2])


I've attempted the working out in my book, Its abit hard to put the whole working on here as it's time consuming. I'm down to [4a^2b^2]/[a^-12b]

thank you.
 

Answers and Replies

  • #2
35,287
7,140
Your expression is equal to
[tex]\left[{\frac{a^{-2}b^{-3}}{2a^3b^{-4}}\right]^2~\frac{a^{-3}b^2}{ab^{-15}}[/tex]

One approach would be to move all the factors with negative exponents from the top to the bottom or from the bottom to the top in each fraction, which would leave you with positive exponents. The idea is that a-n = 1/an. After that, square each factor in the first fraction. Finally, cancel everything that can be cancelled.
 
  • #3
267
10
Thanks for that :) is this how i do it?

[a-2b-3 / 2a3b-4 ]2 / [a-3b2 / ab-15]

[2a3b4 / a2b3 ]2 / [ab15 / a3b2 ]

[4a6b8 / a4b6 ] = [4a2b2 ]

and [ab15 / a3b2 ] = [a12b13 ]

I am unsure about the second half.
 
  • #4
446
1
[tex](\frac{a^{-2}b^{-3}}{2a^{3}b^{-4}})(\frac{a^{-3}b^{2}}{ab^{-15}})[/tex]
First take a look at the left side;

It will give
[tex]\frac{a^{(-2-3)}b^{[-3-(-4)]}}{2}[/tex]
Which is equal to
[tex]\frac{a^{-5}b}{2}[/tex]

Now look at right side;
[tex]a^{(-3-1)}b^{[2-(-15)]}[/tex]
[tex]a^{-4}b^{13}[/tex]

Combine both LHS and RHS;
[tex]\frac{a^{[-5+(-4)]}b^{(1+1)}}{2}[/tex]

I will leave the rest to you, noting that [tex]a^{-9} = \frac{1}{a^{9}}[/tex]
 
  • #5
267
10
Did you neglect to use the square of the left side? I dont see where you used it.
 
  • #6
446
1
Did you neglect to use the square of the left side? I dont see where you used it.

opps.. missed it.
then i guess you just have to square LHS before combining.
 
  • #7
267
10
That's okay :) and btw, the LFH is divided by the RHS. sorry for confusion I dont know how to work the script stuff on this site.
 
  • #8
267
10
[ [tex]\frac{a^-^2b^-^3}{2a^3b^-^4}[/tex] ]2 / [tex]\frac{ab^-^1^5}{a^-3b^2}[/tex]

[tex]\frac{a^-^4b^-^6}{4a^6b^-8}[/tex] / [tex]\frac{ab^-^1^5}{a^-^3b^2}[/tex]

[tex]\frac{a^(^-^4^-^6^)b^(^-^6^-^(^-^8^)^)}{4}[/tex] / [tex]\frac{a^(^-^1^-^3^)b^(^-^1^5^-^2^)}[/tex]

[tex]\frac{a^-^1^0b^2}{4}[/tex] / [tex]\frac{a^2b^-^1^7}[/tex]

[tex]\frac{a^(^-^1^0^-^2^)b^(^2^-^(^-^1^7^)}{4}[/tex]

[tex]\frac{a^-^1^2b^1^9}{4}[/tex]

I ended up getting [tex]\frac{4}{a^1^2b^1^9}[/tex]

If you ended up trying it, did you also get that?
 
  • #9
446
1
[tex]\frac{a^-^2b^-^3}{2a^3b^-^4}[/tex] ]2 / [tex]\frac{ab^-^1^5}{a^-3b^2}[/tex]



[tex]\frac{a^(^-^4^-^6^)b^(^-^6^-^(^-^8^)^)}{4}[/tex] / [tex]{a^(^-^1^-^3^)b^(^-^1^5^-^2^)}[/tex]

I see mistakes in the second expression :bugeye:
 
  • #10
267
10
Could you point them out please, I'm not very good at indices.
 
  • #11
267
10
[tex]\frac{b^1^9}{4a^1^2}[/tex]

The answer is [tex]\frac{b^1^9}{4a^1^4}[/tex]

So, now the only thing I need to find out is where i lost a 2.
 
  • #12
Hi. I'm just a newbie but would like to give it a try.

Look for any mistakes :approve:!!!

This is what I got after a while working it out:

[tex]
\left[{\frac{a^{-2}b^{-3}}{2a^3b^{-4}}\right]^2*~\frac{a^{-3}b^2}{ab^{-15}}
=
\frac{\left(a^{-2}b^{-3}\right)^2}{\left(2a^3b^{-4}\right)^2}\right]*~\frac{a^{-3}b^2}{ab^{-15}}
=
{\frac{a^{-4}b^{-6}}{4a^6b^{-8}}~*~\frac{a^{-3}b^2}{ab^{-15}}
=
{\frac{a^{-4-3}b^{-6+2}}{4a^{6+1}b^{-8-15}}

=

{\frac{a^{-7}b^{-4}}{4a^{7}b^{-23}}
=
{\frac{b^{-4+23}}{4a^{7+7}}
=
{\frac{b^{19}}{4a^{14}}
[/tex]

By the way, I wonder what class you're taking, 'cause my Pre-Cal class hasn't had this kind of stuff yet.

CN
 
Last edited:
  • #13
Double-posted, sorry! I don't know how to delete doubled posts:confused:.
 
  • #14
48
0
Hi. I'm just a newbie but would like to give it a try.

Look for any mistakes :approve:!!!

This is what I got after a while working it out:

[tex]
\left[{\frac{a^{-2}b^{-3}}{2a^3b^{-4}}\right]^2*~\frac{a^{-3}b^2}{ab^{-15}}
=
\frac{\left(a^{-2}b^{-3}\right)^2}{\left(2a^3b^{-4}\right)^2}\right]*~\frac{a^{-3}b^2}{ab^{-15}}
=
{\frac{a^{-4}b^{-6}}{4a^6b^{-8}}~*~\frac{a^{-3}b^2}{ab^{-15}}
=
{\frac{a^{-4-3}b^{-6+2}}{4a^{6+1}b^{-8-15}}

=

{\frac{a^{-7}b^{-4}}{4a^{7}b^{-23}}
=
{\frac{b^{-4+23}}{4a^{7+7}}
=
{\frac{b^{19}}{4a^{14}}
[/tex]

By the way, I wonder what class you're taking, 'cause my Pre-Cal class hasn't had this kind of stuff yet.

CN

Nope, no mistakes. I'm in Pre-Calc at the moment too, none of this stuff so far. We worked on this kind of stuff in grade 11 if memory serves me right.
 
  • #15
267
10
Thanks guys for the help :) I'm from australia and i'm currently doing the Introcalculus course in year 11. I use to do the Calculus one but my study load was to much.
 

Related Threads on Fractional Exponential

  • Last Post
Replies
11
Views
3K
  • Last Post
Replies
2
Views
755
  • Last Post
Replies
2
Views
328
  • Last Post
Replies
1
Views
725
  • Last Post
Replies
2
Views
525
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
4
Views
934
Top