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

\[\large (ax^n)^2=ax^{2n}\]

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

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

do u play LoL with teemo? hehe

yea ofc

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

i really dont understand hows that gonna help

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

27x^3n?

3^n-6 x^n-6

yea i think i can do that

ok... good

where did the a go?

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

but this applies to same base...

derp it cut the 6 out :P

well you just need to finish it

im really confused where did the ax^2 went

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

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

\[\huge3^{n - 3} x^{4n - 6}\]

thats a correct solution

mhm not sure if that helps lol

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

x=2?

then we plug x into the equation?

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

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

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

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