How were these numbers transformed? (-3^(3n+6) * 3^(3n+3) + 6) / (3^(3n+4))

[Zulgok]

Hello, I can't understand how this was transformed from :

(-3^(3n+6) * 3^(3n+3) + 6) / (3^(3n+4))

to

(3^(3n+3)*(-3^(3) + 6)) / (3^(3n+4))

I have been struggling on this particular case for quite a while now, and I think I need your help. Sorry for no formatting.

 

[ksdhart2]

Okay, so it looks like the starting point was as follows:

\(\displaystyle \dfrac{-3^{3n+6}\cdot 3^{3n+3}+6}{3^{3n+4}}\)

And it looks like the ending point was as follows:

\(\displaystyle \dfrac{3^{3n+3}\cdot \left(-3^3+6\right)}{3^{3n+4}}\)

However, I think at least one of the above has an error, as if these two expressions are meant to be equal, it's not true for all values of n. Please reply with any necessary corrections, and/or consult with your instructor. Thank you.

 

[Ishuda]

Hello, I can't understand how this was transformed from :

(-3^(3n+6) * 3^(3n+3) + 6) / (3^(3n+4))

to

(3^(3n+3)*(-3^(3) + 6)) / (3^(3n+4))

I have been struggling on this particular case for quite a while now, and I think I need your help. Sorry for no formatting.

Are you familiar with exponents? For example
\(\displaystyle \dfrac{3^a}{3^b}\, =\, 3^{a-b}\)

If a were 17n+8 and b were 15n+7 then we have
\(\displaystyle \dfrac{3^{17n+8}}{3^{15n+7}}\,\,\, =\, 3^{17n+8-(15n+7)}\)
= 3²ⁿ⁺¹

 

[Zulgok]

The original form is

(-27^(n+2) + 6 * 3^(3n+3)) / (3^(n) * 9^(n+2))

Is this form equal to the the first form I gave you ?

 

[Zulgok]

Are you familiar with exponents? For example
\(\displaystyle \dfrac{3^a}{3^b}\, =\, 3^{a-b}\)

If a were 17n+8 and b were 15n+7 then we have
\(\displaystyle \dfrac{3^{17n+8}}{3^{15n+7}}\,\,\, =\, 3^{17n+8-(15n+7)}\)
= 3²ⁿ⁺¹

Yes, but I just dont understand how teacher from this form

\(\displaystyle \dfrac{-3^{3n+6}\cdot 3^{3n+3}+6}{3^{3n+4}}\)

Made this

\(\displaystyle \dfrac{3^{3n+3}\cdot \left(-3^3+6\right)}{3^{3n+4}}\)[FONT=MathJax_Main][/FONT]

 

[ksdhart2]

The original form is

(-27^(n+2) + 6 * 3^(3n+3)) / (3^(n) * 9^(n+2))

Is this form equal to the the first form I gave you ?

No, not for all values of n, only for one specific value. In general:

\(\displaystyle \dfrac{-27^{n+2}+6\cdot 3^{3n+3}}{3^n\cdot 9^{n+2}} \ne \dfrac{-3^{3n+6}\cdot 3^{3n+3}+6}{3^{3n+4}}\)

As I suggested earlier, I'd speak with your instructor for clarification regarding this problem, as I'm also at a loss as to explain this "transformation."

 

[Ishuda]

Yes, but I just dont understand how teacher from this form

\(\displaystyle \dfrac{-3^{3n+6}\cdot 3^{3n+3}+6}{3^{3n+4}}\)

Made this

\(\displaystyle \dfrac{3^{3n+3}\cdot \left(-3^3+6\right)}{3^{3n+4}}\)

Well, according to your post above,the
\(\displaystyle \dfrac{-3^{3n+6}\cdot 3^{3n+3}+6}{3^{3n+4}}\)
isn't correct as indicated by ksdhart2 so that is not what the instructor did.

You first need to get the transformation of the original formula correct. You will need
(a^b)^c = a^b*c
For example,
4²ⁿ = (2²)²ⁿ = 2⁴ⁿ