Answer:
The final answer is [tex]\frac{8}{27}[/tex] .
Step-by-step explanation:
Solve the following equation:
[tex](2^{2} \cdot 3^{-1} \cdot 4^{-1})^{-1} \over 2^{-3} \cdot 3^{4}[/tex]
-Simplify [tex]2^{2}[/tex] :
[tex](2^{2} \cdot 3^{-1} \cdot 4^{-1})^{-1} \over 2^{-3} \cdot 3^{4}[/tex]
[tex](4 \times 3^{-1} \times 4^{-1})^{-1} \over 2^{-3} \times 3^{4}[/tex]
-Simplify [tex]3^{-1}[/tex] :
[tex](4 \times 3^{-1} \times 4^{-1})^{-1} \over 2^{-3} \times 3^{4}[/tex]
[tex](4 \times( \frac{1}{3}) \times 4^{-1})^{-1} \over 2^{-3} \times 3^{4}[/tex]
-Multiply both [tex]4[/tex] and [tex]\frac{1}{3}[/tex] :
[tex](4 \times( \frac{1}{3}) \times 4^{-1})^{-1} \over 2^{-3} \times 3^{4}[/tex]
[tex](\frac{4}{3} \times 4^{-1})^{-1} \over 2^{-3} \times 3^{4}[/tex]
-Simplify [tex]4^{-1}[/tex] :
[tex](\frac{4}{3} \times 4^{-1})^{-1} \over 2^{-3} \times 3^{4}[/tex]
[tex](\frac{4}{3} \times (\frac{1}{4}) )^{-1} \over 2^{-3} \times 3^{4}[/tex]
-Multiply both [tex]\frac{4}{3}[/tex] and [tex]\frac{1}{4}[/tex] together:
[tex](\frac{4}{3} \times (\frac{1}{4}) )^{-1} \over 2^{-3} \times 3^{4}[/tex]
[tex](\frac{1}{3})^{-1} \over 2^{-3} \times 3^{4}[/tex]
-Simplify [tex]\frac{1}{3}^{-1}[/tex] :
[tex](\frac{1}{3})^{-1} \over 2^{-3} \times 3^{4}[/tex]
[tex]3 \over 2^{-3} \times 3^{4}[/tex]
-Then, on the denominator, simplify [tex]2^{-3}[/tex] :
[tex]3 \over 2^{-3} \times 3^{4}[/tex]
[tex]3 \over \frac{1}{8} \times 3^{4}[/tex]
Simplify [tex]3^{4}[/tex] :
[tex]3 \over \frac{1}{8} \times 3^{4}[/tex]
[tex]3 \over \frac{1}{8} \times 81[/tex]
Multiply both [tex]\frac{1}{8}[/tex] and [tex]81[/tex] together:
[tex]3 \over \frac{1}{8} \times 81[/tex]
[tex]3 \over \frac{81}{8}[/tex]
Divide [tex]3[/tex] and [tex]\frac{81}{8}[/tex] by multiplying [tex]3[/tex] by the reciprocal of [tex]\frac{81}{8}[/tex] :
[tex]3 \over \frac{81}{8}[/tex]
[tex]3 \times (\frac{81}{8})[/tex]
-And you multiply [tex]3[/tex] and [tex]\frac{81}{8}[/tex] together:
[tex]3 \times (\frac{81}{8})[/tex]
[tex]\frac{8}{27}[/tex]
So, the final answer would be [tex]\frac{8}{27}[/tex] .
