Answer:
12
Step-by-step explanation:
Given the expression:
[tex](\dfrac{(2^{-1}\times 3^2)}{(2^{2}\times 3^{-4})})^\frac{7}{2} \times (\dfrac{(2^{-2}\times 3^3)}{(2^{3}\times 3^{-5})})^\frac{-5}{2}[/tex]
We have to simplify the above expression.
Let us have a look at a few formulas:
[tex]1.\ a^b = \dfrac{1}{a^{-b}}[/tex]
[tex]2.\ a^b\times a^c = a^{b+c}[/tex]
[tex]3.\ (a^b ) ^c = a^{bc}[/tex]
Now, let us try to solve the given above expression using the formulas:
[tex]\Rightarrow (\dfrac{(3^{4}\times 3^2)}{(2^{2}\times 2^{1})})^\frac{7}{2} \times (\dfrac{(3^{5}\times 3^3)}{(2^{3}\times 2^{2})})^\frac{-5}{2}[/tex]
[tex]\Rightarrow (\dfrac{(3^{4+2})}{(2^{2+1})})^\frac{7}{2} \times (\dfrac{(3^{5+3})}{(2^{3+2})})^\frac{-5}{2}\\\Rightarrow (\dfrac{3^{6}}{2^{3}})^\frac{7}{2} \times (\dfrac{3^{8}}{2^{5}})^\frac{-5}{2}\\\Rightarrow (\dfrac{3^{\frac{42}{2}}}{2^{\frac{3\times 7}{2}}})\times (\dfrac{2^{5}}{3^{8}})^\frac{5}{2}\\\Rightarrow (\dfrac{3^{21}}{2^{\frac{21}{2}}})\times (\dfrac{2^{\frac{5\times 5}{2}}}{3^{\frac{8\times 5}{2}}})[/tex]
[tex]\Rightarrow (\dfrac{3^{21}}{2^{\frac{21}{2}}})\times (\dfrac{2^{\frac{25}{2}}}{3^{20}})\\\Rightarrow {3^{1}}\times {2^{\frac{25-21}{2}}}\\\Rightarrow 3\times 4 \\\Rightarrow \bold{12}[/tex]