# Simplify with negetive exponents

• July 17th 2010, 06:14 PM
kiwisa
Simplify with negetive exponents
How des one simplify this?

(2ab)^-2/2(ab)^3

I am unsure if this correct,

I removed the brackets

-4a^-2b^-2/2a^-3b^-3

to make exponents positive i moved to the top

-4a^-2--3b^-2--3 +2

which is -4ab+2

Is this correct?
• July 17th 2010, 06:30 PM
Gusbob
Quote:

Originally Posted by kiwisa
How des one simplify this?

(2ab)^-2/2(ab)^3

I am unsure if this correct,

I removed the brackets

-4a^-2b^-2/2a^-3b^-3

to make exponents positive i moved to the top

-4a^-2--3b^-2--3 +2

which is -4ab+2

Is this correct?

$\frac{ \left((2ab)^{-2}\right)}{\left(2(ab)^3\right)}$

= $\left(2^{-2} a^{-2} b^{-2}\right) \cdot \left(2^{-1}a^{-3}b^{-3} \right)$

= $2^{-3}a^{-5}b^{-5}$

= $\frac{1}{8a^5b^5}$
• July 17th 2010, 06:43 PM
kiwisa
sorry on i seei have atypo in the original post the exponent should be -3 so it should read as such

(2ab)^-2/2(ab)^-3

But with the equation at the top where does the 2^-1 come from if it was 2^3? The rest i can understand.
Sorry for the typo.
• July 17th 2010, 07:01 PM
Gusbob
Quote:

Originally Posted by kiwisa
sorry on i seei have atypo in the original post the exponent should be -3 so it should read as such

(2ab)^-2/2(ab)^-3

But with the equation at the top where does the 2^-1 come from if it was 2^3? The rest i can understand.
Sorry for the typo.

It's all right. For the equation I posted, you have 2(ab)^3 for the denominator. Using the law of operations the power of 3 only applies to a and b, not on the 2.

As for the fixed equation

$\frac{(2ab)^{-2}}{2(ab)^{-3}}$

= $(2ab)^{-2} \cdot \left(2 (ab)^{-3}\right)^{-1}$

= $2^{-2}a^{-2}b^{-2} \cdot 2^{-1} (ab)^3$

= $2^{-2}a^{-2}b^{-2} \cdot 2^{-1} a^3 b^3$

= $2^{-3} ab$

= $\frac{ab}{8}$
• July 17th 2010, 07:10 PM
kiwisa
Thanks - so to multiply the bottom instead of dividing it I put it all in brackets and multply by -1. The rest i understand.
• July 17th 2010, 08:11 PM
Soroban
Hello, kiwisa!

Quote:

Simplify: . $\dfrac{(2ab)^{-2}}{2(ab)^{-3}}$

We have: . $\dfrac{(2ab)^{-2}}{2(ab)^{-3}} \;=\;\dfrac{2^{-2}\cdot a^{-2}\cdot b^{-2}} {2\cdot a^{-3}\cdot b^{-3}}$

Use the exponent rules: when dividing, subtract exponents.

And we have: . $2^{-2-1}\cdot a^{-2-(-3)}\cdot b^{-1-(-3)}$

. . . . . . . . $=\;2^{-3}\cdot a^1\cdot b^1$

. . . . . . . . $=\; \dfrac{ab}{2^3}$

. . . . . . . . $=\;\dfrac{ab}{8}$
• July 17th 2010, 08:21 PM
kiwisa
Thanks guys got it now (Rock)