You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an alternative browser.
You should upgrade or use an alternative browser.
Integer Exponents
- Thread starter Smith54
- Start date
Hello, Smith54!
You have created your own rules of exponents . . .
No wonder!
You have created your own rules of exponents . . .
\(\displaystyle \left(-2ab^{-2}\right)\left(4a^{-2}b)^{-2}\)
. . \(\displaystyle =\; \left(-2ab^{-2}\right)\left(-16a^4b^{-2}) \qquad \text{ You claim that: }4^{-2} \,=\,-16\;\;(?)\)
. . \(\displaystyle =\; \frac{-2a^5b^{-4}}{-16}\). . . . . . . . . . . . . . . . . \(\displaystyle \text{ and that: }-16 \,=\,\frac{1}{-16}\;\;(?)\)
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
(-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
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)
-------------------------------------------------------------
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?
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.