TeemoTheTerific
HELP PLEASE!!!
Im stuck on how to start(and finish)
Delete
Share
This Question is Closed
TeemoTheTerific
Best Response
You've already chosen the best response.
0
If \[\huge(3x ^{n})^{3} \times (3x)^{n-6}= ax ^{2}\]
Find the value of n AND a
Azteck
Best Response
You've already chosen the best response.
0
\[\large (ax^n)^2=ax^{2n}\]
Azteck
Best Response
You've already chosen the best response.
0
Use that property. Also use this
\[\large 2x^n \times 2x^{n-1}=2x^{n+(n-1)}\]
TeemoTheTerific
Best Response
You've already chosen the best response.
0
how would that help when it is 3x not 2x
theanonymous27
Best Response
You've already chosen the best response.
0
do u play LoL with teemo? hehe
TeemoTheTerific
Best Response
You've already chosen the best response.
0
yea ofc
Azteck
Best Response
You've already chosen the best response.
0
\[\large ax^n \times ax^{n-1}=ax^{n+(n-1)}\]
TeemoTheTerific
Best Response
You've already chosen the best response.
0
i really dont understand hows that gonna help
Azteck
Best Response
You've already chosen the best response.
0
Wait I think it's a^2 as the result not just a.
campbell_st
Best Response
You've already chosen the best response.
2
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
\[(3x^n)^3\]
TeemoTheTerific
Best Response
You've already chosen the best response.
0
27x^3n?
campbell_st
Best Response
You've already chosen the best response.
2
thats correct...
but for this question I'd leave it as 3^3
apply the same rule to the 2nd term...
\[(3x)^{n -6}\]
TeemoTheTerific
Best Response
You've already chosen the best response.
0
3^n-6 x^n-6
campbell_st
Best Response
You've already chosen the best response.
2
correct... so do you know about adding powers when mutiplying the same base..?
because you have
\[3^3x^{3n} \times 3^{n - 6} x^{n - 6} = 3^3 \times 3^{n - 6} \times x^{3n} \times x^{n - 6}\]
TeemoTheTerific
Best Response
You've already chosen the best response.
0
yea i think i can do that
campbell_st
Best Response
You've already chosen the best response.
2
ok... good
TeemoTheTerific
Best Response
You've already chosen the best response.
0
where did the a go?
campbell_st
Best Response
You've already chosen the best response.
2
you said you can do it.. use the law
\[m^a \times m^b = m^{a +b}\]
campbell_st
Best Response
You've already chosen the best response.
2
but this applies to same base...
TeemoTheTerific
Best Response
You've already chosen the best response.
0
\[\huge3^3x^{3n} \times 3^{n - 6} x^{n - 6} = 3^3 \times 3^{n - 6} \times x^{3n} \times x^{n - 6}\]
this is it right?
TeemoTheTerific
Best Response
You've already chosen the best response.
0
derp it cut the 6 out :P
campbell_st
Best Response
You've already chosen the best response.
2
well you just need to finish it
TeemoTheTerific
Best Response
You've already chosen the best response.
0
im really confused where did the ax^2 went
campbell_st
Best Response
You've already chosen the best response.
2
don't worry about ax^2..
all you need to do is simplify the right side of your previous posting
\[3^{3 + n - 6} \times x^{3n + n - 6} = \]
TeemoTheTerific
Best Response
You've already chosen the best response.
0
\[\huge3^{n - 3} \times x^{4n - 6} \]
campbell_st
Best Response
You've already chosen the best response.
2
yep...
just write your answer without the multiplication sign
TeemoTheTerific
Best Response
You've already chosen the best response.
0
\[\huge3^{n - 3} x^{4n - 6}\]
campbell_st
Best Response
You've already chosen the best response.
2
thats a correct solution
TeemoTheTerific
Best Response
You've already chosen the best response.
0
mhm not sure if that helps lol
campbell_st
Best Response
You've already chosen the best response.
2
ok... so lets look at the power of x
4n - 6 = 2
solve for n....
TeemoTheTerific
Best Response
You've already chosen the best response.
0
x=2?
TeemoTheTerific
Best Response
You've already chosen the best response.
0
then we plug x into the equation?
campbell_st
Best Response
You've already chosen the best response.
2
not x... n = 2
so you have
all you need to do is evaluate
\[3^{n - 3}\]
when n = 2 to find a
campbell_st
Best Response
You've already chosen the best response.
2
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
TeemoTheTerific
Best Response
You've already chosen the best response.
0
ahh ok :) thanks for putting up with me :)
campbell_st
Best Response
You've already chosen the best response.
2
glad to help... lots of index laws in simplifying before you can solve for n and a
TeemoTheTerific
Best Response
You've already chosen the best response.
0
your my fave maths teach lol well so far...