# Thread: Negative and zero exponets

1. ## Negative and zero exponets

My teacher assigned me this assignment and it looks totally different that what we did... here are a few of the problems...

X^-2 Y^4

C^0 D^4

R^-3 T^5

There is just a couple of them I just want to get the hang of it so I can do the rest.
P.S I made this thread in the exponents section but i realized that there is this section... I believe it would be better because it is really urgent.

2. Originally Posted by XGBrandon
My teacher assigned me this assignment and it looks totally different that what we did... here are a few of the problems...

X^-2 Y^4

C^0 D^4

R^-3 T^5

There is just a couple of them I just want to get the hang of it so I can do the rest.
P.S I made this thread in the exponents section but i realized that there is this section... I believe it would be better because it is really urgent.
Hi XGBrandon,

What were your instructions? Simplify? Get rid of negative exponents?

$\displaystyle x^{-2}y^4=\frac{y^4}{x^2}$

$\displaystyle c^0d^4=1d^4=d^4$

$\displaystyle r^{-3}t^5=\frac{t^5}{r^3}$

3. Originally Posted by masters
Hi XGBrandon,

What were your instructions? Simplify? Get rid of negative exponents?

$\displaystyle x^{-2}y^4=\frac{y^4}{x^2}$

$\displaystyle c^0d^4=1d^4=d^4$

$\displaystyle r^{-3}t^5=\frac{t^5}{r^3}$
get rid of the negative exponent and it has to be in fraction form.,

4. Also htis one is throwing me off...ab^0 c^-4 would it be c^4 ?

5. Originally Posted by XGBrandon
get rid of the negative exponent and it has to be in fraction form.,
Ok then. That's what I thought. All are in fraction form except $\displaystyle d^4$. If you wanted to, I guess you could write it as $\displaystyle \frac{d^4}{1}$, but I don't know why you would want to.

6. Originally Posted by masters
Ok then. That's what I thought. All are in fraction form except $\displaystyle d^4$. If you wanted to, I guess you could write it as $\displaystyle \frac{d^4}{1}$, but I don't know why you would want to.
Do you got a instant messenger program?

7. Originally Posted by XGBrandon
Also htis one is throwing me off...ab^0 c^-4 would it be c^4 ?
$\displaystyle ab^0c^{-4}=\frac{a}{c^4}$

The variable a remains in the numerator. $\displaystyle c^{-4}$ must go to the denominator to have a positive exponent. $\displaystyle b^0$ is simply 1.

8. another problem I came across is g^7 h^-1 k
would the answer be K/GH ?

9. Originally Posted by XGBrandon
another problem I came across is g^.7 h^-1 k
would the answer be K/GH ?
This one looks a little strange. Is it copied correctly? If so, then

$\displaystyle g^{.7}h^{-1}k=\frac{g^{.7}k}{h} \ \ or \ \ \frac{g^{7/10}k}{h}$

The g and the k term stay in the numerator. The h term moves to the denominator to achieve a positive exponent of 1.

Remember a few rules:

$\displaystyle a^0=1$

$\displaystyle a^{-1}=\frac{1}{a}$

$\displaystyle a^{-n}=\frac{1}{a^n}$

10. Masters, I don't see any decimal 7 in the problem. I see g^7 h^-1 k.

XGBrandon, you have been told, repeatedly, that the negative exponent means it goes into the denominator (with positive exponent). Only the h has a negative exponent. And the "7" exponent on the g doesn't just "disappear".

$\displaystyle g^7h^{-1}k= \frac{gk}{h}$

11. Originally Posted by HallsofIvy
Masters, I don't see any decimal 7 in the problem. I see g^7 h^-1 k.

XGBrandon, you have been told, repeatedly, that the negative exponent means it goes into the denominator (with positive exponent). Only the h has a negative exponent. And the "7" exponent on the g doesn't just "disappear".

$\displaystyle g^7h^{-1}k= \frac{gk}{h}$
Hi HallsofIvy,

I don't know where that decimal point came from. I just quoted XGB's post and there it was. Made it interesting, though. Spooky!