To simplify the given expression [tex]\(\frac{16^{-x} \cdot 12^{x+1}}{3^x \cdot 4^{-x}}\)[/tex], let's go through it step-by-step:
[tex]\[ \frac{16^{-x} \cdot 12^{x+1}}{3^x \cdot 4^{-x}} \][/tex]
1. Write each term with the same base where possible.
- Recall that [tex]\(16 = 2^4\)[/tex], [tex]\(12 = 2^2 \cdot 3\)[/tex], [tex]\(4 = 2^2\)[/tex].
Therefore:
[tex]\[ 16^{-x} = (2^4)^{-x} = 2^{-4x} \][/tex]
[tex]\[ 12^{x+1} = (2^2 \cdot 3)^{x+1} \][/tex]
[tex]\[ 4^{-x} = (2^2)^{-x} = 2^{-2x} \][/tex]
2. Substitute these into the expression:
[tex]\[ \frac{2^{-4x} \cdot (2^2)^{x+1} \cdot 3^{x+1}}{3^x \cdot 2^{-2x}} \][/tex]
Note that [tex]\( (a \cdot b)^n = a^n \cdot b^n \)[/tex], so:
[tex]\[ 12^{x+1} = (2^2 \cdot 3)^{x+1} = 2^{2(x+1)} \cdot 3^{x+1} \][/tex]
[tex]\[ 12^{x+1} = 2^{2x+2} \cdot 3^{x+1} \][/tex]
Then, substituting back:
[tex]\[ \frac{2^{-4x} \cdot 2^{2x+2} \cdot 3^{x+1}}{3^x \cdot 2^{-2x}} \][/tex]
3. Combine the powers of like bases:
For [tex]\(2\)[/tex]:
[tex]\[ 2^{-4x} \cdot 2^{2x+2} \cdot 2^{2x} = 2^{-4x + 2x + 2 + 2x} = 2^{(2x + 2)} \][/tex]
For [tex]\(3\)[/tex]:
[tex]\[ \frac{3^{x+1}}{3^x} = 3^{(x+1) - x} = 3^1 \][/tex]
4. Simplify fully:
[tex]\[ \frac{2^{0+2}} \cdot 3^1 = 2^2 \cdot 3 = 4 \cdot 3 = 12 \][/tex]
The simplified value of the given expression is:
[tex]\[
\boxed{12}
\][/tex]
So, the final result is [tex]\(12\)[/tex].
