anonymous 3 years ago HELP PLEASE!!! Im stuck on how to start(and finish)

1. anonymous

If $\huge(3x ^{n})^{3} \times (3x)^{n-6}= ax ^{2}$ Find the value of n AND a

2. anonymous

$\large (ax^n)^2=ax^{2n}$

3. anonymous

Use that property. Also use this $\large 2x^n \times 2x^{n-1}=2x^{n+(n-1)}$

4. anonymous

how would that help when it is 3x not 2x

5. anonymous

do u play LoL with teemo? hehe

6. anonymous

yea ofc

7. anonymous

$\large ax^n \times ax^{n-1}=ax^{n+(n-1)}$

8. anonymous

i really dont understand hows that gonna help

9. anonymous

Wait I think it's a^2 as the result not just a.

10. campbell_st

look at the 1st part this requires the power of a power law $(ax^n)^m = a^m x^{n \times m}$ the outer power operates on the constant a as well as x^n use this to simplify $(3x^n)^3$

11. anonymous

27x^3n?

12. campbell_st

thats correct... but for this question I'd leave it as 3^3 apply the same rule to the 2nd term... $(3x)^{n -6}$

13. anonymous

3^n-6 x^n-6

14. campbell_st

correct... so do you know about adding powers when mutiplying the same base..? because you have $3^3x^{3n} \times 3^{n - 6} x^{n - 6} = 3^3 \times 3^{n - 6} \times x^{3n} \times x^{n - 6}$

15. anonymous

yea i think i can do that

16. campbell_st

ok... good

17. anonymous

where did the a go?

18. campbell_st

you said you can do it.. use the law $m^a \times m^b = m^{a +b}$

19. campbell_st

but this applies to same base...

20. anonymous

$\huge3^3x^{3n} \times 3^{n - 6} x^{n - 6} = 3^3 \times 3^{n - 6} \times x^{3n} \times x^{n - 6}$ this is it right?

21. anonymous

derp it cut the 6 out :P

22. campbell_st

well you just need to finish it

23. anonymous

im really confused where did the ax^2 went

24. campbell_st

don't worry about ax^2.. all you need to do is simplify the right side of your previous posting $3^{3 + n - 6} \times x^{3n + n - 6} =$

25. anonymous

$\huge3^{n - 3} \times x^{4n - 6}$

26. campbell_st

yep... just write your answer without the multiplication sign

27. anonymous

$\huge3^{n - 3} x^{4n - 6}$

28. campbell_st

thats a correct solution

29. anonymous

mhm not sure if that helps lol

30. campbell_st

ok... so lets look at the power of x 4n - 6 = 2 solve for n....

31. anonymous

x=2?

32. anonymous

then we plug x into the equation?

33. campbell_st

not x... n = 2 so you have all you need to do is evaluate $3^{n - 3}$ when n = 2 to find a

34. campbell_st

does that make sense... when you equate the powers of x... you have $x^{4n - 6} = x^2$ so 4n - 6 = 2 so n = 2 use this for the powers of 3 3$3^{2 - 3} = 3^{-1}.... or..... \frac{1}{3}$ so a = 1/3

35. anonymous

ahh ok :) thanks for putting up with me :)

36. campbell_st

glad to help... lots of index laws in simplifying before you can solve for n and a

37. anonymous

your my fave maths teach lol well so far...