Question

(2^8 x 3^−5 x 6^0)−2 x 3 to the power of negative 2 over 2 to the power of 3, whole to the power of 4 x 2^28

Answer 1

Here are a few rules about exponents:

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

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

$(2^8*3^{-5})^{-2}=2^{8*-2}3^{-5*-2}=2^{-16}3^{10}\\\\(\frac{3^{-2}}{2^3})^4=\frac{3^{-2*4}}{2^{3*4}}=\frac{3^{-8}}{2^{12}}\\\\2^{-16}3^{10}*\frac{3^{-8}}{2^{12}}*2^{28}$

Next, multiply:

$\frac{2^{-16}3^{10}}{1}*\frac{3^{-8}}{2^{12}}*\frac{2^{28}}{1}=\frac{2^{-16+28}3^{10+(-8)}}{2^{12}}=\frac{2^{12}3^{2}}{2^{12}}$

Next, divide:

$\frac{2^{12}3^{2}}{2^{12}}=2^{12-12}3^{2-0}=2^03^2=3^2$

Your final answer is 3^2, or 9.

Answer 2

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