## anonymous one year ago Did i do this right? Simplify each expression. Use Positive exponents m3 n^-6 p^0. 1n^-6/1 = 1/n^-6 now we put the positive n^6 back in it's original place. m^3 n^6 p

1. anonymous

uhhhh i think so i belive so

2. SolomonZelman

The rule for negative exponents is: $$\Large\color{black}{ \displaystyle \color{blue}{\rm a}^{-\color{red}{\rm b}}=\frac{1 }{\color{blue}{\rm a}^{\color{red}{\rm b}}} }$$

3. SolomonZelman

so, your expression is: $$m^3n^{-6}p^0$$ is that correct?

4. anonymous

i may be wrong my b

5. anonymous

Okay 1n^-6/1 = 1/n^-6 = 1/n^6 now we put the positive n^6 back in it's original place. m^3 n^6 p How about now

6. SolomonZelman

you are still doing 2 parts incorrectly, not just the n^(-6), but p^0 is incorrect as well-:(

7. SolomonZelman

lets start from $$p^0$$, ok?

8. anonymous

Okay.

9. SolomonZelman

The rule for an exponent of 0 is: $$\Large\color{black}{ \displaystyle \color{blue}{\rm x}^{\color{red}{\rm 0}}=1 }$$ This is true for any value of x, but provided that x is NOT zero. ------------------------------------------------------ ADDITIONALLY: The prove for this rule is: $$\Large\color{black}{ \displaystyle \color{blue}{\rm x}^{\color{red}{\rm 0}}=\color{blue}{\rm x}^{w-w} }$$ (i am using w, but you can replace w with 1, 3, or any other number and the statement will still hold) $$\Large\color{black}{ \displaystyle \color{blue}{\rm x}^{\color{red}{\rm 0}}=\color{blue}{\rm x}^{w-w} =\dfrac{x^w}{x^w}=1}$$ (anything divided by itself gives a result of 1, but if x was 0 then we have 0/0 which is not defined. This is why I said every x but NOT x=0)

10. SolomonZelman

So if any number (except 0) raised to the power of 0, gives a result of 1, then $$p^0={\rm what?}$$

11. SolomonZelman

((Another NOTE: they didn't state that $$p\ne0$$, it is their fault, but you have to make this assumption, to get the answer, or else it would be indeterminate))

12. anonymous

Alright.

13. anonymous

So p^1?

14. SolomonZelman

p^1 ? you don't have that in your expression, do you?

15. anonymous

No, so do i leave the p^0 alone?

16. SolomonZelman

you have p^0, and you know that: (anything)$$^0$$ = ?

17. anonymous

x?

18. SolomonZelman

(anything)$$^0$$=1 (as long as this anything is NOT zero)

19. SolomonZelman

therefore, you can say that p$$^0$$=?

20. anonymous

But p has a zer exponent?

21. SolomonZelman

yes, so p$$^0$$= what?

22. anonymous

Zero* And is it 1

23. SolomonZelman

yes, p$$^0$$=1

24. SolomonZelman

So lets re-write our expression $${\rm m}^3\cdot {\rm n}^{-6} \cdot {\rm p}^0$$ this kis what it is used to be, but we know p$$^0$$=1, so we can say: $${\rm m}^3\cdot {\rm n}^{-6} \cdot 1$$ and multiplying times 1, is not changing the value, so we can leave the "•1" part out. (right?) We have: $${\rm m}^3\cdot {\rm n}^{-6}$$

25. SolomonZelman

is everything making sense till this point?

26. anonymous

Alright. Yeah so far.

27. anonymous

Because the p is = to 1 right?

28. SolomonZelman

yes, because p$$^0$$=1

29. anonymous

So yeah i get it not sure why we need to leave the 1 out after solving the problem though. Is it because it's exponent was 0?

30. SolomonZelman

why we have to leave the "•1" out? Not that we must, but it does matter. For any number, if you multiply it times 1 you still have that same number. Multiplying times 1 is same, and just as good as, not multiplying by anything at all.

31. anonymous

Makes sense. Thanks for your help.

32. SolomonZelman

we aren't done yet.

33. anonymous

Alright

34. SolomonZelman

So our new expression is: $${\rm m}^3\cdot {\rm n}^{-6}$$ And, this can be simplified more. ------------------------------ Now, if we can recall, I mentioned another important property, and it refers to negative exponents. This property is: $$\Large\color{black}{ \displaystyle \color{blue}{\rm a}^{-\color{red}{\rm b}}=\frac{1 }{\color{blue}{\rm a}^{\color{red}{\rm b}}} }$$

35. SolomonZelman

according to this property, $$\Large\color{black}{ \displaystyle \color{blue}{\rm n}^{-\color{red}{\rm 6}}=\frac{1 }{\color{blue}{\rm n}^{\color{red}{\rm 6}}} }$$ right?

36. anonymous

Yeah

37. SolomonZelman

same, but n instead of a, and 6 instead of b. good...

38. anonymous

n^-6/1/n^6

39. SolomonZelman

n$$^{-6}$$ = 1 / n$$^6$$ like this, you mean?

40. anonymous

Yeah forgot to put in my equal.

41. SolomonZelman

lol.

42. anonymous

:P

43. SolomonZelman

so, we had: $${\rm m}^3\cdot {\rm n}^{-6}$$ and now we know that: $${\rm n}^{-6}=\dfrac{1}{{\rm n}^6}$$ So, we can re-write our expression the following way: $${\rm m}^3\cdot\dfrac{1}{{\rm n}^{6}}$$

44. SolomonZelman

then you can make it as 1 fraction, just the following way: $${\rm m}^3\cdot\dfrac{1}{{\rm n}^{6}}$$ $$\dfrac{{\rm m}^3}{1}\cdot\dfrac{1}{{\rm n}^{6}}$$ $$\dfrac{{\rm m}^3\cdot 1}{1\cdot {\rm n}^{6}}$$ $$\dfrac{{\rm m}^3}{{\rm n}^{6}}$$

45. SolomonZelman

if you have any questions regarding any of the rules or steps that were aplied or mentioned, please ask...

46. anonymous

No questions, because you timing it by one which would make it the same number.

47. SolomonZelman

alright. G☼☼d luck

48. anonymous

Thanks :)