[SOLVED] Need help with exponents: simplifying 1

Hello, I was wondering if someone can tell me if I did these exponent problems right? I'm a bit confused.

Convert the following to equivalent forms in which no negative exponents appear:

a. (2/5)^-1 ---------- I said this equals -2/-5, which is 2/5

b. 6/x^-2 (just the x is to the negative 2) ------ I said this was 6^2/x, which is 36/x

c. (-3/2)^-3 --------- I said this was 81/-8, or -81/8

d. 6xy/3x^-1y-2) (that looks weird anyway 6xy over 3 x to the negative one, y to the negative 2) ---- I was confused. I said 2x^2y^3 (2xsquaredycubed)

e. (2x2/3x-1)^-2 -- I said this was -4x-4/-9x2 = 4x/9x^6.

Note that x is the variable, not multiplication.

Thanks!!

Also, is x^3 times x^-3 x^0?

Thank you so much

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Originally Posted by view360

Hello, I was wondering if someone can tell me if I did these exponent problems right? I'm a bit confused.

Convert the following to equivalent forms in which no negative exponents appear:

a. (2/5)^-1 ---------- I said this equals -2/-5, which is 2/5

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that is equals to $\displaystyle \frac{5}{2}$

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b. 6/x^-2 (just the x is to the negative 2) ------ I said this was 6^2/x, which is 36/x

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$\displaystyle =\frac{x^2}{6}$

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c. (-3/2)^-3 --------- I said this was 81/-8, or -81/8

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$\displaystyle =-\frac{8}{27}$

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d. 6xy/3x^-1y-2) (that looks weird anyway 6xy over 3 x to the negative one, y to the negative 2) ---- I was confused. I said 2x^2y^3 (2xsquaredycubed)

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is this the question? $\displaystyle \frac{6xy}{3x^{-1}y^{-2}}$

$\displaystyle =\frac{6x^2y^3}{3}$

$\displaystyle =2x^2y^3$

note: there should be no negative exponents in the denominator

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e. (2x2/3x-1)^-2 -- I said this was -4x-4/-9x2 = 4x/9x^6.

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$\displaystyle (\frac{2x^2}{3x-1})^{-2}$

$\displaystyle \frac{9x^2-6x+1}{4x^4}$

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Also, is x^3 times x^-3 x^0?

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$\displaystyle x^3 * x^-3 * x^0$

$\displaystyle x^3 * \frac{1}{x^3} * 1$

$\displaystyle =1$

Some minor mistakes here

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Originally Posted by princess_21

$\displaystyle =\frac{x^2}{6}$

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$\displaystyle \frac 6{x^{-2}} = 6x^2$

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$\displaystyle =\frac{8}{27}$

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this should be -8/27, you left off the negative sign

also, when superscripts or subscripts have more than one term, you need to put them in { } brackets for the LaTeX to look nice.


[tex]x^-2y^2[/tex] $\displaystyle \longrightarrow x^-2y^2$ :(

[tex]x^{-2}y^2[/tex] $\displaystyle \longrightarrow x^{-2}y^2$ :)


see, we have -2 as the power, that's two things, a minus sign and a 2, were it just a 2, typing x^2 is fine, but for the -2 we need to enclose it x^{-2}

Thanks, Jhevon!!

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Originally Posted by Jhevon

also, this is not right. we can distribute powers across products, not sums

$\displaystyle (a + b)^2 \ne a^2 + b^2$ for instance, but rather $\displaystyle (a + b)^2 = (a + b)(a + b) = a^2 + 2ab + b^2$

similarly, $\displaystyle (x^{-1} + y^{-1})^{-2} = \frac 1{(x^{-1} + y^{-1})^2} = \frac 1{x^{-2} + 2(xy)^{-1} + y^{-2}}$


if you don't like negative powers in the denominator, you could approach it this way

$\displaystyle (x^{-1} + y^{-1})^{-2} = \left( \frac 1x + \frac 1y \right)^{-2} = \left( \frac {x + y}{xy} \right)^{-2} = \frac {x^2y^2}{(x + y)^2} = \frac {x^2y^2}{x^2 + 2xy + y^2}$



also, please post new problems in a new thread. it saves me the work of having to split the thread up :p

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sorry my mistake.

Thanks to both of you, thank you so much for taking the time to help me out. I think I'm finally getting the hang of these now!!