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):

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

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
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