Question

Help this is really hard

Answer 1

`(2^8\cdot3^(-5)\cdot6^0)^(-2)\cdot\left((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((a)/(b)\right)^n=(a^n)/(b^n)\\ a^0=1\ for\ a\\eq0\\\\=(2^8)^(-2)\cdot(3^(-5))^(-2)\cdot1^(-2)\cdot((3^(-2))^4)/((2^3)^4)\cdot2^(28)=2^((8)(-2))\cdot3^((-5)(-2))\cdot1\cdot(3^((-2)(4)))/(2^((3)(4)))\cdot2^(28)`

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

Answer 2

(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