Question

Help this is really hard

Answer 1

(2^8 *3^-5* 6^0)^-2 * ((3^-2)/(2^3))^4 * 2^28

anything to the 0 power is 1

(2^8 *3^-5* 1)^-2 * ((3^-2)/(2^3))^4 * 2^28

using the power of power property to take the power inside

(2^(8*-2) *3^(-5* -2) * (3^-2*4)/(2^3*4) * 2^28

simplify

2^ -16 * 3^10 * 3^-8 /2*12   * 2^28

get rid of the division by making the exponent negative

2^-16 * 3^10 * 3^-8 *2*-12   * 2^28

combine exponents with like bases

2^(-16-12+28) * 3^(10-8)

2^(0) *3^2

anything to the 0 power is 1

1*9

9

Answer 2

$(2^8\cdot3^{-5}\cdot6^0)^{-2}\cdot\left(\dfrac{3^{-2}}{2^3}\right)^4\cdot2^{28}\\\\\text{use}\\(a\cdot b)^n=a^n\cdot b^n\\(a^n)^m=a^{nm}\\\left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}\\ a^0=1\ for\ a\neq0\\\\=(2^8)^{-2}\cdot(3^{-5})^{-2}\cdot1^{-2}\cdot\dfrac{(3^{-2})^4}{(2^3)^4}\cdot2^{28}=2^{(8)(-2)}\cdot3^{(-5)(-2)}\cdot1\cdot\dfrac{3^{(-2)(4)}}{2^{(3)(4)}}\cdot2^{28}$

$=2^{-16}\cdot3^{10}\cdot\dfrac{3^{-8}}{2^{12}}\cdot2^{28}=\dfrac{2^{-16}\cdot2^{28}}{2^{12}}\cdot3^{10}\cdot3^{-8}\\\\\text{use}\ a^n\cdot a^m=a^{nm}\\\\=\dfrac{2^{-16+28}}{2^{12}}\cdot3^{10+(-8)}=\dfrac{2^{12}}{2^{12}}\cdot3^2=1\cdot3^2=3^2=9\\\\Answer:\ \boxed{9}$