Question

Did I do this Algebra problem correctly? 7^13/7^3

Answer 1

Simplify. Assume that no denominator is equal to 0.

I got 1^10, correct?

Answer 2

Not exactly

Answer 3

Apply this property of Exponents:

\[\frac{n^a}{n^b} = n^{a - b}\]

Answer 4

Ah, 7^10?

Answer 5

Good job

Answer 6

Oh thank you! Egads, I thought this problem was MUCH harder than it was in reality. Ha ha! Thank you! Thank you so much! :P

Answer 7

I'm a little embarrassed, to be honest. :)

Answer 8

Do you think you could explain this?

(3a/a^2)^-2

Answer 9

To see it a little more clearly:

Answer 10

You cancel factors of 1 leaving just \(7^{10}\)

Answer 11

Alright, thanks, I got it.

Do you think you could explain the next problem I gave you, please? :)

Answer 12

\[\left (\large \frac{3a}{a^2} \right)^{-2}\]

Answer 13

Yes

Answer 14

Basically, the negative exponent you see has nothing to do with negative

Answer 15

That's misleading, but okay.

Answer 16

In reality, it means "inverse"

Answer 17

It means the `INVERSE` of
\[\left (\large \frac{3a}{a^2} \right)^{2}\]

Answer 18

`The inverse of a number is simply one over that number:`

\[a^{-2} = \frac{1}{a^2}\]

Answer 19

The inverse of "a to the negative 2nd power" equals "1 divided by a to the second power".

Answer 20

Now, here's what you do with your problem:

Rewrite it like this:

\[1 \div \left (\large \frac{3a}{a^2} \right)^{2}\]

Answer 21

And then simplify the

\[\left (\large \frac{3a}{a^2} \right)^{2}\] part

Answer 22

3a^2/a^4

Answer 23

You should end up with this:

\[\large 1 \div \large \frac{9a^2}{a^4} \]

Answer 24

Ah

Answer 25

Use the reciprocal rule to rewrite it like this:

\[ \large 1 \times \large \frac{a^4}{9a^2} \]

Answer 26

So.. the same thing. :P

Answer 27

No, NOT the same

Answer 28

1/9a^2!

Answer 29

What did you do with the \(a^4\)?

Answer 30

Let me finish explaining

Answer 31

Ok

Answer 32

Next you use the exponents rule I showed you earlier to re-write it like this:

\[\large \frac{a^{4 -2}}{9}\]

Answer 33

Ah!

a^2/9

Answer 34

I hope I didn't lose you

Answer 35

Not at all! Although, would you mind staying a bit to see if I can understand another problem that's similar? :)

Answer 36

Actually, it's more similar to the first problem I gave you, however, this is different..

12/c^-4d^8

Answer 37

I think it's 12/c^4d^8, correct?

Answer 38

If I were you I would rewrite it like this to avoid confusion:

\[12 \div (c^{-4} \times d^8)\]

Answer 39

This way, you can deal with the \(c^{-4}\) separately

Answer 40

Or you could just leave it as

\[\frac{12}{c^{-4}d^8}\]

and deal with it that way

Answer 41

Now here's the thing. If the negative exponent is in the denominator, move it to the numerator and remove the exponent. That way, it will become:

\[\frac{12c^4}{d^8}\]

Answer 42

Are you lost yet?

Answer 43

A bit. Ha ha! I presume that's the final answer?

Answer 44

Yes. If you follow exactly as I have explained to you, you won't have any problems.

Answer 45

Alright! Now, another similar but different problem :P

Simplify the expression

6k^-6b^6

Answer 46

Okay, you should be able to follow what I've shown you to figure this one out