Certainly! Let's simplify the given expression step by step:
Given:
[tex]\[
\left(\frac{3^n \times 9^{n+1}}{3^{n-1} \times 9^{n-1}}\right)
\][/tex]
First, we recall that [tex]\(9\)[/tex] can be written as [tex]\(3^2\)[/tex]. Therefore, we can rewrite the expression with base [tex]\(3\)[/tex]:
[tex]\[
9 = 3^2
\][/tex]
Substituting [tex]\(9\)[/tex] in the expression:
[tex]\[
\left(\frac{3^n \times (3^2)^{n+1}}{3^{n-1} \times (3^2)^{n-1}}\right)
\][/tex]
Next, we simplify the exponents. Using the property of exponents [tex]\((a^m)^n = a^{mn}\)[/tex]:
[tex]\[
(3^2)^{n+1} = 3^{2(n+1)} = 3^{2n + 2}
\][/tex]
[tex]\[
(3^2)^{n-1} = 3^{2(n-1)} = 3^{2n - 2}
\][/tex]
Substituting these back into the expression, we get:
[tex]\[
\left(\frac{3^n \times 3^{2n + 2}}{3^{n-1} \times 3^{2n - 2}}\right)
\][/tex]
Using the property of exponents [tex]\(a^m \times a^n = a^{m+n}\)[/tex], we can combine the exponents in the numerator and the denominator:
In the numerator:
[tex]\[
3^n \times 3^{2n + 2} = 3^{n + 2n + 2} = 3^{3n + 2}
\][/tex]
In the denominator:
[tex]\[
3^{n-1} \times 3^{2n-2} = 3^{(n-1) + (2n-2)} = 3^{n - 1 + 2n - 2} = 3^{3n - 3}
\][/tex]
So, the expression simplifies to:
[tex]\[
\left(\frac{3^{3n + 2}}{3^{3n - 3}}\right)
\][/tex]
Using the property [tex]\(\frac{a^m}{a^n} = a^{m-n}\)[/tex]:
[tex]\[
3^{(3n + 2) - (3n - 3)} = 3^{3n + 2 - 3n + 3} = 3^{5}
\][/tex]
Finally, calculating [tex]\(3^5\)[/tex]:
[tex]\[
3^5 = 3 \times 3 \times 3 \times 3 \times 3 = 243
\][/tex]
Thus, the simplified expression is:
[tex]\[
\boxed{243}
\][/tex]
