Let's simplify the expression [tex]\(\frac{2^{m+1} \times 3^{n+2} \times 6^{m+n}}{4^m \times 9^n}\)[/tex] step-by-step.
### Step 1: Rewrite the given expression using the prime factors
1. Note that [tex]\(6 = 2 \times 3\)[/tex], so [tex]\(6^{m+n} = (2 \times 3)^{m+n}\)[/tex].
2. Substitute [tex]\(6^{m+n}\)[/tex] in the expression:
[tex]\[
\frac{2^{m+1} \times 3^{n+2} \times (2 \times 3)^{m+n}}{4^m \times 9^n}
\][/tex]
### Step 2: Simplify the exponents of the numerator
1. Expand [tex]\((2 \times 3)^{m+n}\)[/tex]:
[tex]\[
(2 \times 3)^{m+n} = 2^{m+n} \times 3^{m+n}
\][/tex]
2. Substitute this back into the expression:
[tex]\[
\frac{2^{m+1} \times 3^{n+2} \times 2^{m+n} \times 3^{m+n}}{4^m \times 9^n}
= \frac{2^{(m+1) + (m+n)} \times 3^{(n+2) + (m+n)}}{4^m \times 9^n}
\][/tex]
3. Combine the exponents of like bases in the numerator:
[tex]\[
\frac{2^{2m+n+1} \times 3^{2n+m+2}}{4^m \times 9^n}
\][/tex]
### Step 3: Rewrite the denominator using prime factors
1. Note that [tex]\(4 = 2^2\)[/tex] and [tex]\(9 = 3^2\)[/tex], so:
[tex]\[
4^m = (2^2)^m = 2^{2m}\quad \text{and} \quad 9^n = (3^2)^n = 3^{2n}
\][/tex]
2. Substitute these into the expression:
[tex]\[
\frac{2^{2m+n+1} \times 3^{2n+m+2}}{2^{2m} \times 3^{2n}}
\][/tex]
### Step 4: Simplify the expression
1. Simplify the powers of 2:
[tex]\[
\frac{2^{2m+n+1}}{2^{2m}} = 2^{(2m+n+1) - 2m} = 2^{n+1}
\][/tex]
2. Simplify the powers of 3:
[tex]\[
\frac{3^{2n+m+2}}{3^{2n}} = 3^{(2n+m+2) - 2n} = 3^{m+2}
\][/tex]
3. Combine the simplified terms:
[tex]\[
2^{n+1} \times 3^{m+2}
\][/tex]
At this point, we can conclude that the final simplified expression is:
[tex]\[
\frac{2^{m+1} \times 3^{n+2} \times 6^{m+n}}{4^m \times 9^n} = 18 \times 6^{m+n}
\][/tex]
