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gmany

[tex](2^8\cdot3^{-5}\cdot6^0)^{-2}\cdot\left(\dfrac{3^{-2}}{2^3}\right)^4\cdot2^{28}\\\\=(2^8)^{-2}\cdot(3^{-5})^{-2}\cdot1^{-2}\cdot\dfrac{(3^{-2})^4}{(2^3)^4}\cdot2^{28}\\\\=2^{-16}\cdot3^{10}\cdot\dfrac{3^{-8}}{2^{12}}\cdot2^{28}\\\\=2^{-16}\cdot2^{28}\cdot2^{-12}\cdot3^{10}\cdot3^{-8}\\\\=2^{-16+28+(-12)}\cdot3^{10+(-8)}\\\\=2^0\cdot3^2=1\cdot9=\boxed{9}\\\\Used:\\\\(a\cdot b)^n=a^n\cdot b^n\\\\(a^n)^m=a^{n\cdot m}\\\\a^n\cdot a^m=a^{n+m}\\\\a^{-n}=\dfrac{1}{a^n}[/tex]

GoddessofDork

Here are a few rules about exponents:

  1. Powering a power: [tex](x^m)^n=x^{m*n}[/tex]
  2. Zeroth power: [tex]x^0=1[/tex]
  3. Multiplying exponents of the same base: [tex]x^m*x^n=x^{m+n}[/tex]
  4. Dividing exponents of the same base: [tex]\frac{x^m}{x^n}=x^{m-n}[/tex]

Firstly, cancel out the 6^0 and solve the power of powers:

[tex](2^8*3^{-5})^{-2}=2^{8*-2}3^{-5*-2}=2^{-16}3^{10}\\\\(\frac{3^{-2}}{2^3})^4=\frac{3^{-2*4}}{2^{3*4}}=\frac{3^{-8}}{2^{12}}\\\\2^{-16}3^{10}*\frac{3^{-8}}{2^{12}}*2^{28}[/tex]

Next, multiply:

[tex]\frac{2^{-16}3^{10}}{1}*\frac{3^{-8}}{2^{12}}*\frac{2^{28}}{1}=\frac{2^{-16+28}3^{10+(-8)}}{2^{12}}=\frac{2^{12}3^{2}}{2^{12}}[/tex]

Next, divide:

[tex]\frac{2^{12}3^{2}}{2^{12}}=2^{12-12}3^{2-0}=2^03^2=3^2[/tex]

Your final answer is 3^2, or 9.

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