Solve (1/2)^-3*(-1/2)^4-(3^2)^0=

$$( \frac { 1 } { 2 } ) ^ { - 3 } \times ( - \frac { 1 } { 2 } ) ^ { 4 } - ( 3 ^ { 2 } ) ^ { 0 } =$$

Evaluate

$$-\frac{1}{2}=-0.5$$

Solution Steps

To raise a power to another power, multiply the exponents. Multiply 2 and 0 to get 0.

$$\left(\frac{1}{2}\right)^{-3}\left(-\frac{1}{2}\right)^{4}-3^{0}$$

Calculate \frac{1}{2} to the power of -3 and get 8.

$$8\left(-\frac{1}{2}\right)^{4}-3^{0}$$

Calculate -\frac{1}{2} to the power of 4 and get \frac{1}{16}.

$$8\times \frac{1}{16}-3^{0}$$

Multiply 8 and \frac{1}{16} to get \frac{1}{2}.

$$\frac{1}{2}-3^{0}$$

Calculate 3 to the power of 0 and get 1.

$$\frac{1}{2}-1$$

Subtract 1 from \frac{1}{2} to get -\frac{1}{2}.

$$-\frac{1}{2}$$

Factor

$$-\frac{1}{2} = -0.5$$