- #1

- 267

- 10

([a

^{-2}b

^{-3}/ 2a

^{3}b

^{-4}]

^{2})/ [ab

^{-15}]/a

^{-3}b

^{2}])

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.

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- Thread starter Sirsh
- Start date

- #1

- 267

- 10

([a

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.

- #2

Mark44

Mentor

- 35,287

- 7,140

[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

- #3

- 267

- 10

[a

[2a

[4a

and [ab

I am unsure about the second half.

- #4

- 446

- 1

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

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Did you neglect to use the square of the left side? I dont see where you used it.

- #6

- 446

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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

- #8

- 267

- 10

[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

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[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

- #10

- 267

- 10

Could you point them out please, I'm not very good at indices.

- #11

- 267

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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

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- 0

Hi. I'm just a newbie but would like to give it a try.

Look for any mistakes !!!

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

Look for any mistakes !!!

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

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Double-posted, sorry! I don't know how to delete doubled posts.

- #14

- 48

- 0

Hi. I'm just a newbie but would like to give it a try.

Look for any mistakes !!!

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

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