Solve (2^frac{1{2}})^-4+(-4)^3*16^-2

$$( 2 ^ { \frac { 1 } { 2 } } ) ^ { - 4 } + ( - 4 ) ^ { 3 } \times 16 ^ { - 2 }$$

Evaluate

$$0$$

Solution Steps

To raise a power to another power, multiply the exponents. Multiply \frac{1}{2} and -4 to get -2.

$$2^{-2}+\left(-4\right)^{3}\times 16^{-2}$$

Calculate 2 to the power of -2 and get \frac{1}{4}.

$$\frac{1}{4}+\left(-4\right)^{3}\times 16^{-2}$$

Calculate -4 to the power of 3 and get -64.

$$\frac{1}{4}-64\times 16^{-2}$$

Calculate 16 to the power of -2 and get \frac{1}{256}.

$$\frac{1}{4}-64\times \frac{1}{256}$$

Multiply -64 and \frac{1}{256} to get -\frac{1}{4}.

$$\frac{1}{4}-\frac{1}{4}$$

Subtract \frac{1}{4} from \frac{1}{4} to get 0.

$$0$$

Factor

$$0$$