anonymous
  • anonymous
HELP PLEASE!!! Im stuck on how to start(and finish)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
If \[\huge(3x ^{n})^{3} \times (3x)^{n-6}= ax ^{2}\] Find the value of n AND a
anonymous
  • anonymous
\[\large (ax^n)^2=ax^{2n}\]
anonymous
  • anonymous
Use that property. Also use this \[\large 2x^n \times 2x^{n-1}=2x^{n+(n-1)}\]

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anonymous
  • anonymous
how would that help when it is 3x not 2x
anonymous
  • anonymous
do u play LoL with teemo? hehe
anonymous
  • anonymous
yea ofc
anonymous
  • anonymous
\[\large ax^n \times ax^{n-1}=ax^{n+(n-1)}\]
anonymous
  • anonymous
i really dont understand hows that gonna help
anonymous
  • anonymous
Wait I think it's a^2 as the result not just a.
campbell_st
  • 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 \[(3x^n)^3\]
anonymous
  • anonymous
27x^3n?
campbell_st
  • campbell_st
thats correct... but for this question I'd leave it as 3^3 apply the same rule to the 2nd term... \[(3x)^{n -6}\]
anonymous
  • anonymous
3^n-6 x^n-6
campbell_st
  • campbell_st
correct... so do you know about adding powers when mutiplying the same base..? because you have \[3^3x^{3n} \times 3^{n - 6} x^{n - 6} = 3^3 \times 3^{n - 6} \times x^{3n} \times x^{n - 6}\]
anonymous
  • anonymous
yea i think i can do that
campbell_st
  • campbell_st
ok... good
anonymous
  • anonymous
where did the a go?
campbell_st
  • campbell_st
you said you can do it.. use the law \[m^a \times m^b = m^{a +b}\]
campbell_st
  • campbell_st
but this applies to same base...
anonymous
  • 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?
anonymous
  • anonymous
derp it cut the 6 out :P
campbell_st
  • campbell_st
well you just need to finish it
anonymous
  • anonymous
im really confused where did the ax^2 went
campbell_st
  • campbell_st
don't worry about ax^2.. all you need to do is simplify the right side of your previous posting \[3^{3 + n - 6} \times x^{3n + n - 6} = \]
anonymous
  • anonymous
\[\huge3^{n - 3} \times x^{4n - 6} \]
campbell_st
  • campbell_st
yep... just write your answer without the multiplication sign
anonymous
  • anonymous
\[\huge3^{n - 3} x^{4n - 6}\]
campbell_st
  • campbell_st
thats a correct solution
anonymous
  • anonymous
mhm not sure if that helps lol
campbell_st
  • campbell_st
ok... so lets look at the power of x 4n - 6 = 2 solve for n....
anonymous
  • anonymous
x=2?
anonymous
  • anonymous
then we plug x into the equation?
campbell_st
  • 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
campbell_st
  • 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
anonymous
  • anonymous
ahh ok :) thanks for putting up with me :)
campbell_st
  • campbell_st
glad to help... lots of index laws in simplifying before you can solve for n and a
anonymous
  • anonymous
your my fave maths teach lol well so far...

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