Solve 16^{-1}*(4^-2)^-1div(2^-5)^0

$$16 ^ { - 1 } \times ( 4 ^ { - 2 } ) ^ { - 1 } \div ( 2 ^ { - 5 } ) ^ { 0 }$$

Evaluate

$$1$$

Solution Steps

To raise a power to another power, multiply the exponents. Multiply -2 and -1 to get 2.

$$\frac{16^{-1}\times 4^{2}}{\left(2^{-5}\right)^{0}}$$

To raise a power to another power, multiply the exponents. Multiply -5 and 0 to get 0.

$$\frac{16^{-1}\times 4^{2}}{2^{0}}$$

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

$$\frac{\frac{1}{16}\times 4^{2}}{2^{0}}$$

Calculate 4 to the power of 2 and get 16.

$$\frac{\frac{1}{16}\times 16}{2^{0}}$$

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

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

Calculate 2 to the power of 0 and get 1.

$$\frac{1}{1}$$

Anything divided by one gives itself.

$$1$$

Factor

$$1$$