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August 28, 2014

August 28, 2014

Posted by **Alicia** on Monday, June 1, 2009 at 7:48pm.

(-3a^-5 b^6)^4 (a^7b^0)^3

any ideas? thanks in advance!

- Algebra -
**MathMate**, Monday, June 1, 2009 at 8:00pmFirst, I will assume that the -3 at the beginning will not be raised to the -5th power.

A little revision of the law of indices is in order:

(ab)^n = a^n . b^n

this will apply to the ^4 and ^3 outside of the parentheses.

a^-n=1/a^n

This will apply to the term a^-5

b^0 = 1

for all values of b except 0.

(-3a^-5 b^6)^4 (a^7b^0)^3

=(-3b^6/a^5)^4 (a^7 . 1)^3

=(-3)^4 (b^6)^4 /(a^5)^4 (a^7)^3

= (-3)^4 (b^24)/(a^20) (a^21)

= ...

I will let you take it from here.

Post your answer for a check if you wish.

=

- Algebra -
**Alicia**, Monday, June 1, 2009 at 8:18pmOk thanks! I got 81ab^24 Is this correct? Thanks!

- Algebra -
**MathMate**, Monday, June 1, 2009 at 9:56pmExcellent, the answer is correct.

Do start the problem over to understand every step.

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