# Math Help - [SOLVED] Need help with exponents: simplifying 1

1. ## [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

2. 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 $\frac{5}{2}$

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

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

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

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

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

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

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

$=1$

3. Some minor mistakes here
Originally Posted by princess_21
$=\frac{x^2}{6}$
$\frac 6{x^{-2}} = 6x^2$

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

4. Thanks, Jhevon!!

5. 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
sorry my mistake.

6. 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!!