OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

$\large (ax^n)^2=ax^{2n}$

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

do u play LoL with teemo? hehe

5 years ago
OpenStudy (anonymous):

yea ofc

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

i really dont understand hows that gonna help

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (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 $(3x^n)^3$

5 years ago
OpenStudy (anonymous):

27x^3n?

5 years ago
OpenStudy (campbell_st):

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

5 years ago
OpenStudy (anonymous):

3^n-6 x^n-6

5 years ago
OpenStudy (campbell_st):

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

5 years ago
OpenStudy (anonymous):

yea i think i can do that

5 years ago
OpenStudy (campbell_st):

ok... good

5 years ago
OpenStudy (anonymous):

where did the a go?

5 years ago
OpenStudy (campbell_st):

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

5 years ago
OpenStudy (campbell_st):

but this applies to same base...

5 years ago
OpenStudy (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?

5 years ago
OpenStudy (anonymous):

derp it cut the 6 out :P

5 years ago
OpenStudy (campbell_st):

well you just need to finish it

5 years ago
OpenStudy (anonymous):

im really confused where did the ax^2 went

5 years ago
OpenStudy (campbell_st):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (campbell_st):

5 years ago
OpenStudy (anonymous):

$\huge3^{n - 3} x^{4n - 6}$

5 years ago
OpenStudy (campbell_st):

thats a correct solution

5 years ago
OpenStudy (anonymous):

mhm not sure if that helps lol

5 years ago
OpenStudy (campbell_st):

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

5 years ago
OpenStudy (anonymous):

x=2?

5 years ago
OpenStudy (anonymous):

then we plug x into the equation?

5 years ago
OpenStudy (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

5 years ago
OpenStudy (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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (campbell_st):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago