Answer: 4
=========================================================
Work Shown:
Part 1
[tex]\left(3^8*2^{-5}*9^0\right)^{-2}*\left(\frac{2^{-2}}{3^3}\right)^4*3^{28}\\\\\\\left(3^8*\frac{1}{2^5}*1\right)^{-2}*\left(\frac{1}{3^3}\frac{1}{2^2}\right)^4*3^{28}\\\\\\\left(\frac{3^8}{2^5}\right)^{-2}*\left(\frac{1}{3^3*2^2}\right)^4*3^{28}\\\\\\\left(\frac{2^5}{3^8}\right)^{2}*\left(\frac{1}{3^3*2^2}\right)^4*3^{28}\\\\\\\frac{\left(2^5\right)^2}{\left(3^8\right)^{2}}*\frac{1}{\left(3^3\right)^4*\left(2^2\right)^4}*3^{28}\\\\\\\frac{2^{5*2}}{3^{8*2}}*\frac{1}{3^{3*4}*2^{2*4}}*3^{28}\\\\\\[/tex]
Part 2
[tex]\frac{2^{10}}{3^{16}}*\frac{1}{3^{12}*2^{8}}*3^{28}\\\\\\\frac{2^{10}*3^{28}}{3^{16}*3^{12}*2^{8}}\\\\\\\frac{2^{10}*3^{28}}{3^{16+12}*2^{8}}\\\\\\\frac{2^{10}*3^{28}}{3^{28}*2^{8}}\\\\\\\frac{2^{10}}{2^{8}}\\\\\\2^{10-8}\\\\\\2^2\\\\\\4\\\\\\[/tex]
------------
The exponent rules I used were
- a^(-b) = 1/(a^b)
- (a/b)^(-c) = (b/a)^c
- a^b*a^c = a^(b+c)
- (a^b)/(a^c) = a^(b-c)
- (a^b)^c = a^(b*c)
