# [SOLVED] Need help with exponents: simplifying 1

• February 3rd 2009, 05:33 PM
view360
[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
• February 3rd 2009, 05:56 PM
princess_21
Quote:

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

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

Quote:

b. 6/x^-2 (just the x is to the negative 2) ------ I said this was 6^2/x, which is 36/x
$=\frac{x^2}{6}$

Quote:

c. (-3/2)^-3 --------- I said this was 81/-8, or -81/8
$=-\frac{8}{27}$

Quote:

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

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

$=2x^2y^3$

note: there should be no negative exponents in the denominator

Quote:

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

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

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

Quote:

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

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

$=1$
• February 3rd 2009, 06:19 PM
Jhevon
Some minor mistakes here
Quote:

Originally Posted by princess_21
$=\frac{x^2}{6}$

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

Quote:

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

$$x^-2y^2$$ $\longrightarrow x^-2y^2$ :(

$$x^{-2}y^2$$ $\longrightarrow x^{-2}y^2$ :)

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}
• February 3rd 2009, 06:21 PM
view360
Thanks, Jhevon!!
• February 3rd 2009, 06:34 PM
princess_21
Quote:

Originally Posted by Jhevon
also, this is not right. we can distribute powers across products, not sums

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

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

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

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

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

sorry my mistake.
• February 3rd 2009, 06:37 PM
view360
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!!