Question

HELP PLEASE. Evaluate the expression.
(2m^2/n^3)-2 = n^6/4m^4
How did they get this??!?!

Answer 1

The one on the right is the answer, I suppose? :D

Answer 2

\[\Large \left(\frac{2m^2}{n^3}\right)^{-2}=\color{blue}{\frac{n^6}{4m^4}}\]

Answer 3

How did they get the answer though?

Answer 4

Why can't it stay as 4m^4n^6? Why would the n^6 be at the the numerator and the 4m^4 be at the denominator?

Answer 5

Let's do it one step at a time ^_^
First, just like how multiplication distributes over addition and subtraction,
EXPONENTS distribute over multiplication and division.

So, we have to give each multiplied term in the fraction an exponent of -2.
\[\Large \left(\frac{2m^2}{n^3}\right)^{-2}=\color{blue}{\frac{2^{-2}(m^2)^{-2}}{(n^3)^{-2}}}\]

Catch me so far?

Answer 6

Yes

Answer 7

So... your question... you're basically asking why the numerator and denominator switched places, yes? ^_^

Answer 8

Yes D:

Answer 9

Simple... the exponent is negative... :D
\[\Large \left(\frac{2m^2}{n^3}\right)^{\color{red}{-2}}=\color{blue}{\frac{n^6}{4m^4}}\]

Were it POSITIVE 2 instead, we'd have (as you'd expect)
\[\Large \left(\frac{2m^2}{n^3}\right)^{\color{green}{2}}=\color{blue}{\frac{4m^4}{n^6}}\]

Answer 10

Im not sure I get what youre saying... Im sorry :(

Answer 11

Do keep in mind that what THIS
\[\Large \left(\frac{2m^2}{n^3}\right)^{\color{red}{-2}}\]actually means is
\[\Huge \frac1{ \left(\frac{2m^2}{n^3}\right)^{\color{green}{2}}}\]

Answer 12

Right? Laws of exponents for ya ^_^

Answer 13

Yes I know that part

Answer 14

Okay, great :)
We can now distribute the 2-exponent:
\[\Huge \frac1{ \frac{4m^4}{n^6}}\]

Answer 15

And of course, this is just
\[\Large \frac{n^6}{4m^4}\]
Understood? ^_^

Answer 16

Okay, I get it so far

Answer 17

So far? It's already done ^^^

Answer 18

So it would just magically put n^6 to the top and rest at the bottom?

Answer 19

No, ... uhh

Recall that
\[\Huge \frac{\color{white}{ \ \ }1\color{white}{ \ \ }}{\frac{a}b}= \frac{b}a\]

Answer 20

What exponent law is this? Im so sorry! Is this the power of a quotient rule?

Answer 21

This isn't really a law of exponent, it's just evaluating complex fractions.

Answer 22

So once it turns out to 1/a/b the a/b would flip so it would be b/a ?

Answer 23

Put it this way... from elementary school division of fractions:
\[\Huge \frac{ \ 1 \ }{\frac{a}b}= 1 \div \frac{a}b\]

Now, to divide by a fraction, you multiply by its reciprocal:
\[\Large = 1 \times \frac{b}a\]
\[\Huge = \color{blue}{\frac{b}a}\]

Answer 24

Now, the same concept applies readily here:
\[\Huge \frac{ \ \ 1 \ \ }{ \frac{4m^4}{n^6}}\]

Answer 25

Ohhhhh, okay!!! THANK YOU SO MUCH

Answer 26

Another way to look at this and perhaps a more useful way:

Answer 27

\[\Large \left(\frac{2m^2}{n^3}\right)^{-2}=\color{blue}{\frac{2^{-2}(m^2)^{-2}}{(n^3)^{-2}}}\]

Answer 28

So, laws of exponents dictate that this becomes
\[\Large \frac{2^{-2}m^{-4}}{n^{-6}}\]
right?

Answer 29

Well, you can think of negative exponents as a sign that the term 'doesn't belong there'

Meaning, a negative exponent in the numerator means it should be in the denominator (with a positive exponent, of course)
And vice versa.

So, to bring each term 'where they belong', we get
\[\Large = \frac{n^6}{2^2m^4}=\frac{n^6}{4m^4}\]

Answer 30

This one I got for sure, THANK YOU! :) :) :)

Answer 31

Good. Practice ^_^

Answer 32

A fair warning, however.
You probably won't be dealing with this sort until later, but best be prepared...
\[\Large \frac{a^3 + u^{-2}}{x^5}\]
does \(\large \color{red}{not}\) equate to
\[\Large \frac{a^3}{x^5\color{green}{+u^2}}\]

You may ONLY move around the terms if they are multiplied, not if they are added or subtracted.

Just a heads up ^_^

Answer 33

You just saved my life, thank you so much! I appreciate the help, sorry since I'm a slow learner :(

Answer 34

Faster than you think ^_^
Signing off now :D
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Terence out