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

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

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

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

6 years agohow would that help when it is 3x not 2x

6 years agodo u play LoL with teemo? hehe

6 years agoyea ofc

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

6 years agoi really dont understand hows that gonna help

6 years agoWait I think it's a^2 as the result not just a.

6 years agolook 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 \[(3x^n)^3\]

6 years ago27x^3n?

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

6 years ago3^n-6 x^n-6

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

6 years agoyea i think i can do that

6 years agook... good

6 years agowhere did the a go?

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

6 years agobut this applies to same base...

6 years ago\[\huge3^3x^{3n} \times 3^{n - 6} x^{n - 6} = 3^3 \times 3^{n - 6} \times x^{3n} \times x^{n - 6}\] this is it right?

6 years agoderp it cut the 6 out :P

6 years agowell you just need to finish it

6 years agoim really confused where did the ax^2 went

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

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

6 years agoyep... just write your answer without the multiplication sign

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

6 years agothats a correct solution

6 years agomhm not sure if that helps lol

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

6 years agox=2?

6 years agothen we plug x into the equation?

6 years agonot x... n = 2 so you have all you need to do is evaluate \[3^{n - 3}\] when n = 2 to find a

6 years agodoes 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

6 years agoahh ok :) thanks for putting up with me :)

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

6 years agoyour my fave maths teach lol well so far...

6 years ago