Answer:
[tex]2^2=4[/tex]
Step-by-step explanation:
[tex](3^8 \cdot 2^{-5} \cdot 9^0)^{-2} \cdot \left(\dfrac{2^{-2}}{3^3}\right)^4 \cdot3^{28}[/tex]
Using exponent rule [tex]a^0=1[/tex]
[tex]\implies (3^8 \cdot 2^{-5} )^{-2} \cdot \left(\dfrac{2^{-2}}{3^3}\right)^4 \cdot3^{28}[/tex]
Using exponent rule [tex](a^b \cdot a^c)^d=(a^{bd} \cdot a^{cd})[/tex]
[tex]\implies 3^{(8\times-2)} \cdot 2^{(-5\times-2)} \cdot \left(\dfrac{2^{-2}}{3^3}\right)^4 \cdot3^{28}[/tex]
[tex]\implies 3^{-16} \cdot 2^{10} \cdot \left(\dfrac{2^{-2}}{3^3}\right)^4 \cdot3^{28}[/tex]
Using exponent rule [tex]\left(\dfrac{a^b}{a^c}\right)^d=\left(\dfrac{a^{bd}}{a^{cd}}\right)[/tex]
[tex]\implies 3^{-16} \cdot 2^{10} \cdot \left(\dfrac{2^{(-2\times4)}}{3^{(3\times4)}}\right) \cdot3^{28}[/tex]
[tex]\implies 3^{-16} \cdot 2^{10} \cdot \left(\dfrac{2^{-8}}{3^{12}}\right) \cdot3^{28}[/tex]
Rewrite as one fraction:
[tex]\implies \dfrac{3^{-16} \cdot 2^{10} \cdot2^{-8}\cdot3^{28}}{3^{12}}[/tex]
Using exponent rule [tex]a^b \cdot a^c=a^{b+c}[/tex]
[tex]\implies \dfrac{3^{(-16+28)} \cdot 2^{(10-8)}}{3^{12}}[/tex]
[tex]\implies \dfrac{3^{12} \cdot 2^{2}}{3^{12}}[/tex]
Cancel the common factor [tex]3^{12}[/tex]
[tex]\implies 2^2[/tex]
[tex]\implies 4[/tex]
