#### Question

Simplify the following:

`(3^nxx9^(n+1))/(3^(n-1)xx9^(n-1))`

#### Solution

`(3^nxx9^(n+1))/(3^(n-1)xx9^(n-1))`

`=(3^nxx(3^2)^(n+1))/(3^(n-1)xx(3^2)^(n-1))`

`=(3^nxx3^(2n+2))/(3^(n-1)xx3^(2n-2))`

`=3^(n+2n+2)/3^(n-1+2n-2)`

`=3^(3n + 2)/3^(3n-3)`

`=3^(3n+2-3n+3)`

= 3^{5}

= 243

Is there an error in this question or solution?

Solution Simplify the Following: `(3^Nxx9^(N+1))/(3^(N-1)Xx9^(N-1))` Concept: Laws of Exponents for Real Numbers.