# Negative Exponents question

1. Sep 2, 2008

### LearntIsFun

I'm wondering something when you have a equation:(Of course "^" represents to the power)

4e^4(j^-3)
____________
2d^-3(f^2)
________
2c^2(j^-2)

I know how to get rid of the negative exponents to get:

4e^4j^-3
__________
j^2(f^2)
_________
2d^3(2c^2)

Question comes with bringing the bottom fraction up to multiply and simplify. Do you just have to take the reciprocal of the fraction to multiply or just bring the fraction as is up to multiply because if you take the reciprocal, do the exponents go negative per the rules?

Appreciate the help.

2. Sep 2, 2008

If I understand your original writing, the first fraction is

$$\frac{4e^4 j^{-3}}{\left(\dfrac{2d^{-3}f^2}{2c^2j^{-2}}\right)}$$

If this is correct, your second fraction is only partially correct you wrote

$$\frac{4e^4 j^{-3}}{\left(\dfrac{j^2 f^2}{\left(2d^3\right) \left(2c^2\right)}\right)}$$

Look at your original work and decide whether you really mean to write $$(2c^2)(2d^3)$$ in this expression.

Regardless of that, the answer to the second portion of your question is that you need to perform the classic ''invert and multiply'' to deal with the bottom fraction. As an example,

$$\frac{x^5 y^{-6}}{\left(\dfrac{x^2}{y^2}\right)} = \left(x^5 y^{-6}\right) \cdot \frac{y^2}{x^2} = x^{5-2} y^{-6+2} = x^3 y^{-4} = \frac{x^3}{y^4}$$

Be aware that there are multiple routes that can be taken to simplify fractions like these, and while the workings of two or three different people may seem to be different, if all rules are followed correctly the same answer will result.

3. Sep 2, 2008

### LearntIsFun

So, just to clarify, when taking the reciprocal of the fraction and multiplying it, the exponents still stay positive regardless of the flip, right?

Look at your original work and decide whether you really mean to write in this expression.

Also, not sure about the bottom 2d^3(2c^2) part. Looks correct to me as of right now my knowledge of change negative exponents, but since you're brining it up, it must be incorrect somehow, appreciate your help! :)

4. Sep 2, 2008

has as its answer a resounding YES .

The denominator of the original compound fraction is

$$\frac{2d^{-3} f^2}{2c^2 j^{-2}}$$

If the entire problem were to simplify just this fraction , what would you do with the pair of $$2$$s ?

5. Sep 2, 2008

### LearntIsFun

Ah, I see now said the blind man :D Those two's can be reduced to 1's, totally overlooked it, I just wrote the equation making stuff up to use as an example and didn't take the time to see that. Thanks for the heads up, my mistakes are always overlooking the simpliest things sometimes. I always am striving to get better at that.
-LearntIsFun! :)

Last edited: Sep 2, 2008
6. Sep 2, 2008

$$\frac{2d^{-3} f^2}{2c^2 j^{-2}}$$
If the entire problem were to simplify just this fraction , what would you do with the pair of $$2$$s ?