Question

Simplify using exponent laws
3^(n+2) x 27^(n+1)/9^(n+1) x 3^(2n+1)

Answer 1

\(\large\frac{3^{n+1}27^{n+1}}{9^{n+1}3^{2n+1}}=\frac{3^{n+1}*(3^3)^{n+1}}{(3^2)^{n+1}3^{2n+1}}=\frac{3^{n+1}3^{3n+3}}{3^{2n+2}3^{2n+1}}=\frac{3^{n+1+3n+3}}{3^{2n+2+2n+1}}\\=\frac{3^{4n+4}}{3^{4n+3}}=\frac{3^{4n}*3^4}{3^{4n}*3^3}=\frac{3^4}{3^3}=3\)

Answer 2

awesome! thanks so much, it looks right

Answer 3

wait

Answer 4

\(\large\frac{3^{n+2}27^{n+1}}{9^{n+1}3^{2n+1}}=\frac{3^{n+2}*(3^3)^{n+1}}{(3^2)^{n+1}3^{2n+1}}=\frac{3^{n+2}3^{3n+3}}{3^{2n+2}3^{2n+1}}=\frac{3^{n+2+3n+3}}{3^{2n+2+2n+1}}\\=\frac{3^{4n+5}}{3^{4n+3}}=\frac{3^{4n}*3^5}{3^{4n}*3^3}=\frac{3^5}{3^3}=3^2=9\)

Answer 5

I had 3^(n+1) we wanted 3^(n+2)

Answer 6

https://www.wolframalpha.com/input/?i=3^%28n%2B2%29+x+27^%28n%2B1%29%2F%289^%28n%2B1%29+x+3^%282n%2B1%29%29

Answer 7

I am a little confused on the part where you get 3^4n+5/3^4n+5.

Answer 8

How did you get 5? or rather 2+3 before that

Answer 9

Do you multiply (3^3) to both n and +1?