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TeemoTheTerific

  • 3 years ago

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

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

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

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

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

  5. theanonymous27
    • 3 years ago
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    do u play LoL with teemo? hehe

  6. TeemoTheTerific
    • 3 years ago
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    yea ofc

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

  8. TeemoTheTerific
    • 3 years ago
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    i really dont understand hows that gonna help

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

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

  11. TeemoTheTerific
    • 3 years ago
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    27x^3n?

  12. campbell_st
    • 3 years ago
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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... \[(3x)^{n -6}\]

  13. TeemoTheTerific
    • 3 years ago
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    3^n-6 x^n-6

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

  15. TeemoTheTerific
    • 3 years ago
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    yea i think i can do that

  16. campbell_st
    • 3 years ago
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    ok... good

  17. TeemoTheTerific
    • 3 years ago
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    where did the a go?

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

  19. campbell_st
    • 3 years ago
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    but this applies to same base...

  20. TeemoTheTerific
    • 3 years ago
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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?

  21. TeemoTheTerific
    • 3 years ago
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    derp it cut the 6 out :P

  22. campbell_st
    • 3 years ago
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    well you just need to finish it

  23. TeemoTheTerific
    • 3 years ago
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    im really confused where did the ax^2 went

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

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

  26. campbell_st
    • 3 years ago
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    yep... just write your answer without the multiplication sign

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

  28. campbell_st
    • 3 years ago
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    thats a correct solution

  29. TeemoTheTerific
    • 3 years ago
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    mhm not sure if that helps lol

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

  31. TeemoTheTerific
    • 3 years ago
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    x=2?

  32. TeemoTheTerific
    • 3 years ago
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    then we plug x into the equation?

  33. campbell_st
    • 3 years ago
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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

  34. campbell_st
    • 3 years ago
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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

  35. TeemoTheTerific
    • 3 years ago
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    ahh ok :) thanks for putting up with me :)

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

  37. TeemoTheTerific
    • 3 years ago
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    your my fave maths teach lol well so far...

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