Integer Exponents

(-2ab^-2)(4a^-2b)^-2
(-2ab^-2)(-16a^4b^-2)
-2a^5b^-4/-16
a^5b^-4/8
a^5/8b^4

the answer is -a^5/8b^4
also why does -16 drop to the denominator and why does it stay -16 shouldn't it become positive when it is put in the denominator?

Hello, Smith54!

You have created your own rules of exponents . . .

\(\displaystyle \left(-2ab^{-2}\right)\left(4a^{-2}b)^{-2}\)

. . \(\displaystyle =\; \left(-2ab^{-2}\right)\left(-16a^4b^{-2}) \qquad \text{ You claim that: }4^{-2} \,=\,-16\;\;(?)\)

. . \(\displaystyle =\; \frac{-2a^5b^{-4}}{-16}\). . . . . . . . . . . . . . . . . \(\displaystyle \text{ and that: }-16 \,=\,\frac{1}{-16}\;\;(?)\)

Click to expand...

No wonder!

I understand that 4^-2 = -4x-4= 16 but what rule says the cofficient 16 goes into the denominator. Shouldn't you multiply -2x16 and get -32a^5b^-4 If you can explain why the 16 goes in the denominator, I can understand that once it goes into the denominator the sign changes and the answer is a^5/8b^4

(-2ab^-2)(16a^4b^-2)
-2a^5b^-4/-16

Smith54 said:

I understand that 4^(-2) \(\displaystyle \text{not equal to}\) (-4)(-4) = 16 Shouldn't you multiply -2x16 and get -32a^5b^-4 If you can explain why the 16 goes in the denominator, I can understand that once it goes into the denominator the sign changes and the answer is a^5/8b^4

(-2ab^-2)(16a^4b^-2)
-2a^5b^-4/-16

Click to expand...

Smith54,

Look at the use of required, and in certain places, suggested, grouping symbols
with some rules of exponents:

[-2ab^(-2)][4a^(-2)b]^(-2) =

[-2ab^(-2)][4^(-2)]{a^[(-2)(-2)]}b^(-2) =

[-2ab^(-2)]{(1/16)(a^4)[b^(-2)]} =

[-2(1/16)]{(a)(a^4)][(b^(-2)][b^(-2)]} =

(-1/8)(a^5)[b^(-4)] =

-a^5/(8b^4)


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Here are the steps more or less in Latex:


\(\displaystyle (-2ab^{-2})(4a^{-2}b)^{-2} \ = \\)


\(\displaystyle (-2ab^{-2})\{(4^{-2})[a^{(-2)(-2)}](b^{-2}) \} \ = \\)


\(\displaystyle \bigg(-2ab^{-2}\bigg)\bigg[\frac{1}{16}(a^{4})b^{-2}\bigg] \ = \\)


\(\displaystyle \frac{-1}{8}(a)(a^4)(b^{-2})(b^{-2}) \ = \\)


\(\displaystyle \frac{-1}{8}(a^5)(b^{-4}) \ = \\)


\(\displaystyle \frac{-a^5}{8b^4}\)

Smith54 said:

(-2ab^-2)(4a^-2b)^-2
(-2ab^-2)(-16a^4b^-2)
-2a^5b^-4/-16
a^5b^-4/8
a^5/8b^4

the answer is -a^5/8b^4
also why does -16 drop to the denominator and why does it stay -16 shouldn't it become positive when it is put in the denominator?

Click to expand...

Here's a definition that you need to memorize:

a[sup:86kz1fqd]-n[/sup:86kz1fqd] = 1 / a[sup:86kz1fqd]n[/sup:86kz1fqd]

That's why 4[sup:86kz1fqd]-2[/sup:86kz1fqd] becomes 1/4[sup:86kz1fqd]2[/sup:86kz1fqd], or 1/16.