Sure! Let's solve the given expression step-by-step:
Given expression:
[tex]\[ \frac{3^x \times 9^{x+1}}{3^{x-1} \times 9^{x-1}} \][/tex]
First, let's rewrite the terms involving 9 in terms of powers of 3. Recall that \( 9 = 3^2 \), so:
[tex]\[ 9^{x+1} = (3^2)^{x+1} = 3^{2(x+1)} = 3^{2x+2} \][/tex]
[tex]\[ 9^{x-1} = (3^2)^{x-1} = 3^{2(x-1)} = 3^{2x-2} \][/tex]
Now, let's substitute these back into the original expression:
[tex]\[ \frac{3^x \times 3^{2x+2}}{3^{x-1} \times 3^{2x-2}} \][/tex]
Next, combine the exponents in the numerator and the denominator. Using the property of exponents \(a^m \times a^n = a^{m+n}\):
[tex]\[ \text{Numerator: } 3^x \times 3^{2x+2} = 3^{x + 2x + 2} = 3^{3x + 2} \][/tex]
[tex]\[ \text{Denominator: } 3^{x-1} \times 3^{2x-2} = 3^{x - 1 + 2x - 2} = 3^{3x - 3} \][/tex]
Now, we have the expression:
[tex]\[ \frac{3^{3x + 2}}{3^{3x - 3}} \][/tex]
To divide powers with the same base, subtract the exponents:
[tex]\[ 3^{(3x + 2) - (3x - 3)} = 3^{3x + 2 - 3x + 3} = 3^{5} \][/tex]
Therefore, the expression simplifies to:
[tex]\[ 3^5 \][/tex]
And we know that:
[tex]\[ 3^5 = 243 \][/tex]
So, the final answer is [tex]\( 243 \)[/tex].
