## anonymous 3 years ago Laws of Exponents..im not understanding this

1. anonymous

Like $a ^{m + n} = a ^{m}a ^{n} ?$

2. anonymous

idk. hang on let me seee lol

3. anonymous

(3a^4)3

4. anonymous

thats one of them problems it gives me.

5. KingGeorge

For that problem, there's two laws that you need to be familiar with. Law 1:$\large (a\cdot b)^n=a^n\cdot b^n$ Law 2: $\large (a^n)^m=a^{n\cdot m}$Using these, can you find the solution to your problem.

6. anonymous

ok that makes a little bit more sense . so could you help me solve that problem that I posted?

7. KingGeorge

Sure. First, the correct formatting is $\large (3a^4)^3$correct?

8. anonymous

yes

9. anonymous

and thankyou!

10. KingGeorge

First, we use the first law I mentioned. So $\large (3a^4)^3=3^3\cdot(a^4)^3$Now, use law 2 to simplify $$(a^4)^3$$. Can you tell me what you get?

11. anonymous

ok give me a sec

12. anonymous

(a)^7 ?? i think thats totaly wrong

13. anonymous

unless is a^12

14. KingGeorge

It's the second one. Good job! That means, we've simplified to $\large 3^3\cdot a^{12} \small .$Now just find $$3^3$$, and you're done.

15. anonymous

with the 3^3 do I multiply that..is it 9?

16. KingGeorge

$$3^3=(3\cdot3)\cdot3=9\cdot3=27$$

17. anonymous

oh i see what you did there, i seee my mistake

18. anonymous

so its (a^12)27?

19. precal

|dw:1356824001142:dw|also, parenthesis between the exponents remind me to multiply the powers, this is one of the ways I remember this law

20. KingGeorge

You could write it as $$27\cdot a^{12}$$ or $$a^{12}\cdot 27$$. And if you have some trouble remembering the law, precal's way to do it an excellent way to remember.

21. precal

|dw:1356824081954:dw|I always associated the laws with things I already know

22. anonymous

i wish I could give u guys both medals :p i appreicate both!!

23. KingGeorge

You're welcome.

24. precal

np yw