Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

do u play LoL with teemo? hehe

Answer 6

yea ofc

Answer 7

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

Answer 8

i really dont understand hows that gonna help

Answer 9

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

Answer 10

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

\[(3x^n)^3\]

Answer 11

27x^3n?

Answer 12

thats correct...

but for this question I'd leave it as 3^3

apply the same rule to the 2nd term...

\[(3x)^{n -6}\]

Answer 13

3^n-6 x^n-6

Answer 14

correct... so do you know about adding powers when mutiplying the same base..?

because you have

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

Answer 15

yea i think i can do that

Answer 16

ok... good

Answer 17

where did the a go?

Answer 18

you said you can do it.. use the law

\[m^a \times m^b = m^{a +b}\]

Answer 19

but this applies to same base...

Answer 20

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

Answer 21

derp it cut the 6 out :P

Answer 22

well you just need to finish it

Answer 23

im really confused where did the ax^2 went

Answer 24

don't worry about ax^2..

all you need to do is simplify the right side of your previous posting

\[3^{3 + n - 6} \times x^{3n + n - 6} = \]

Answer 25

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

Answer 26

yep...

just write your answer without the multiplication sign

Answer 27

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

Answer 28

thats a correct solution

Answer 29

mhm not sure if that helps lol

Answer 30

ok... so lets look at the power of x

4n - 6 = 2

solve for n....

Answer 31

x=2?

Answer 32

then we plug x into the equation?

Answer 33

not x... n = 2

so you have

all you need to do is evaluate

\[3^{n - 3}\]

when n = 2 to find a

Answer 34

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

Answer 35

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

Answer 36

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

Answer 37

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