Question

simplify the expression (a^2)^-3(a^3b)^2(b^3)^4

Answer 1

Solve for b over the real numbers:
b^14 = 0
Eliminate the exponent.
Take the 14^th root of both sides:
Answer: |
| b = 0

Answer 2

how do i do that

Answer 3

Is this the expression?

\(\Large (a^2)^{-3}(a^3b)^2(b^3)^4\)

There is nothing to "solve for b", it isn't an equation... it's just an expression to be simplified.

But is what I have above, correct? It's hard to tell from plain text sometimes. :)

Answer 4

yes

Answer 5

You'll just need a few of the rules for exponents:

\(\Large (x^m)^{n}=x^{mn}\)
\(\Large (xy)^m=x^{m}y^m\)
\(\Large x^{-1}=\dfrac{1}{x} \)
\(\Large x^{-m}=\dfrac{1}{x^m} \)

Answer 6

ok

Answer 7

So start with the exponents outside each set of ( ) and simplify that....

\(\Large (a^2)^{-3}(a^3b)^2(b^3)^4\\
\Large =a^{\Box} (a^{\Box}b^{\Box})(b^{\Box})\)

What goes in each "box"

Answer 8

im confused

Answer 9

Oh, I forgot one other rule you are going to need:

\(\Large x^m\cdot x^n=x^{m+n}\)

Answer 10

OK, tell me where you're confused. Those rules for exponents that I posted above, you've seen those before, right? Do you understand them?

Answer 11

the box part that confused me

Answer 12

\(\Large (x^m)^{n}=x^{mn}\)

so, e.g.,
\(\Large (3^2)^{5}=3^{10}\)
\(\Large (z^2)^{-3}=z^{-6}\)

etc.

Answer 13

oh ok i understand that

Answer 14

ah, ok.. sorry, I was just trying to show you what you needed to do first.

One step at a time... you have:

\(\Large (a^2)^{-3}(a^3b)^2(b^3)^4\)

So tell me what you can do with \(\Large (a^2)^{-3}\) ??

What is \(\Large (a^2)^{-3}=\)?

Answer 15

\[a ^{-6}\]

Answer 16

Good! Now how about

\(\Large (a^3b)^2=?\)

Answer 17

\[a^{6}b ^{2}\]

Answer 18

Perfect.

\(\Large (b^3)^4=?\)

Answer 19

\[b ^{12}\]

Answer 20

See, you're getting it! Now you have:

\(\Large a^{-6}a^6b^2b^{12}\)

Which you can mentally think of as grouped according to where you have the base of a, and where you have the base of b, e.g., think of it as:

\(\Large (a^{-6}\cdot a^6)(b^2\cdot b^{12})\)

And now you'll apply that last rule: \(\Large x^m\cdot x^n=x^{m+n}\)

E.g.
\(\Large 3^2\cdot 3^5=3^{7}\)
\(\Large z^3\cdot z^8=z^{11}\) etc....

Now use that on:
\(\Large (a^{-6}\cdot a^6)(b^2\cdot b^{12})\)

Answer 21

(you'll use it trice, once on the "a" part and again on the "b" part)

Answer 22

*twice lol

Answer 23

\[a ^{0}b ^{14}\]

Answer 24

Perfect! Now, do you know the rule for \(x^0\)? What is any base, raised to the power 0?

Answer 25

1

Answer 26

Exactly! So the fully simplified form of this expression is...?

Answer 27

1\[1b ^{14}\]

Answer 28

That's right! Now, you don't really need the "1"out in front... e.g..... "1x" is really just written as "x" in fully simplified form, follow me?

Answer 29

yes thank you so much

Answer 30

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Answer 31

no problem, happy to help. :)

Answer 32

yay gold star