# Two exponent problems.

• Feb 2nd 2009, 02:45 PM
blastedcody
Two exponent problems.
Ok, here are two more problems involving exponents. I'm not the best mathematician, so my answers may or may not be correct. If they are incorrect, could you please show how one can reach the correct solution?

Thank you.

Here is the first problem:

"simplify:"
$\displaystyle (7x^6y^3)(-8x^2y^6)$

And my solution to same:

$\displaystyle -56x^8y^9$

Note that in this second problem, the "|" represents a bracket that is around the whole problem, NOT the contained numbers' absolute value:

"simplify: "
|-5a^3 b^4 c^0|^-4
|3a^4 b^6 c^3 |

and here are my steps to reaching a solution:

-5a^-12 b^-16 1^-4
3a^-16 b^-24 c^-12

-5 b^-16 1^-4
3a^-4 b^-24 c^-12

-5
3a^-4 b^-40 c^-12
• Feb 2nd 2009, 04:11 PM
Soroban
Hello, blastedcody!

Quote:

Simplify: .$\displaystyle (7x^6y^3)(-8x^2y^6)$

And my solution: .$\displaystyle -56x^8y^9$ . . . . Good!

Quote:

Simplify: .$\displaystyle \left[\frac{\text{-}5a^3 b^4 c^0}{3a^4 b^6 c^3}\right]^{-4}$

and here are my steps to reaching a solution: .$\displaystyle \frac{\text{-}5a^{-12}b^{-16}1^{-4}} {3a^{-16} b^{-24} c^{-12}}$
You forgot about the coefficients . . .

I'd reduce "inside" first: .$\displaystyle \left[\frac{\text{-}5}{3ab^2c^3}\right]^{-4} \;=\;\frac{(\text{-}5)^{-4}}{(3)^{-4}(a)^{-4}(b^2)^{-4}(c^3)^{-4}}$

. . $\displaystyle = \;\frac{3^4}{(\text{-}5)^4a^{-4}b^{-8}c^{-12}} \;=\;\frac{81a^4b^8c^{12}}{625}$

• Feb 2nd 2009, 05:31 PM
blastedcody
Thank you, that was very helpful.
• Feb 15th 2009, 06:33 AM
Aveek Agrawal