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TeemoTheTerific

HELP PLEASE!!! Im stuck on how to start(and finish)

  • one year ago
  • one year ago

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  1. TeemoTheTerific
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    If \[\huge(3x ^{n})^{3} \times (3x)^{n-6}= ax ^{2}\] Find the value of n AND a

    • one year ago
  2. Azteck
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    \[\large (ax^n)^2=ax^{2n}\]

    • one year ago
  3. Azteck
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    Use that property. Also use this \[\large 2x^n \times 2x^{n-1}=2x^{n+(n-1)}\]

    • one year ago
  4. TeemoTheTerific
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    how would that help when it is 3x not 2x

    • one year ago
  5. theanonymous27
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    do u play LoL with teemo? hehe

    • one year ago
  6. TeemoTheTerific
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    yea ofc

    • one year ago
  7. Azteck
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    \[\large ax^n \times ax^{n-1}=ax^{n+(n-1)}\]

    • one year ago
  8. TeemoTheTerific
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    i really dont understand hows that gonna help

    • one year ago
  9. Azteck
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    Wait I think it's a^2 as the result not just a.

    • one year ago
  10. campbell_st
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    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 \[(3x^n)^3\]

    • one year ago
  11. TeemoTheTerific
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    27x^3n?

    • one year ago
  12. campbell_st
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    thats correct... but for this question I'd leave it as 3^3 apply the same rule to the 2nd term... \[(3x)^{n -6}\]

    • one year ago
  13. TeemoTheTerific
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    3^n-6 x^n-6

    • one year ago
  14. campbell_st
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    correct... so do you know about adding powers when mutiplying the same base..? because you have \[3^3x^{3n} \times 3^{n - 6} x^{n - 6} = 3^3 \times 3^{n - 6} \times x^{3n} \times x^{n - 6}\]

    • one year ago
  15. TeemoTheTerific
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    yea i think i can do that

    • one year ago
  16. campbell_st
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    ok... good

    • one year ago
  17. TeemoTheTerific
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    where did the a go?

    • one year ago
  18. campbell_st
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    you said you can do it.. use the law \[m^a \times m^b = m^{a +b}\]

    • one year ago
  19. campbell_st
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    but this applies to same base...

    • one year ago
  20. TeemoTheTerific
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    \[\huge3^3x^{3n} \times 3^{n - 6} x^{n - 6} = 3^3 \times 3^{n - 6} \times x^{3n} \times x^{n - 6}\] this is it right?

    • one year ago
  21. TeemoTheTerific
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    derp it cut the 6 out :P

    • one year ago
  22. campbell_st
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    well you just need to finish it

    • one year ago
  23. TeemoTheTerific
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    im really confused where did the ax^2 went

    • one year ago
  24. campbell_st
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    don't worry about ax^2.. all you need to do is simplify the right side of your previous posting \[3^{3 + n - 6} \times x^{3n + n - 6} = \]

    • one year ago
  25. TeemoTheTerific
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    \[\huge3^{n - 3} \times x^{4n - 6} \]

    • one year ago
  26. campbell_st
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    yep... just write your answer without the multiplication sign

    • one year ago
  27. TeemoTheTerific
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    \[\huge3^{n - 3} x^{4n - 6}\]

    • one year ago
  28. campbell_st
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    thats a correct solution

    • one year ago
  29. TeemoTheTerific
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    mhm not sure if that helps lol

    • one year ago
  30. campbell_st
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    ok... so lets look at the power of x 4n - 6 = 2 solve for n....

    • one year ago
  31. TeemoTheTerific
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    x=2?

    • one year ago
  32. TeemoTheTerific
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    then we plug x into the equation?

    • one year ago
  33. campbell_st
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    not x... n = 2 so you have all you need to do is evaluate \[3^{n - 3}\] when n = 2 to find a

    • one year ago
  34. campbell_st
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    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

    • one year ago
  35. TeemoTheTerific
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    ahh ok :) thanks for putting up with me :)

    • one year ago
  36. campbell_st
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    glad to help... lots of index laws in simplifying before you can solve for n and a

    • one year ago
  37. TeemoTheTerific
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    your my fave maths teach lol well so far...

    • one year ago
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