Solve [32^0,5*(sqrt{8})^-1]^-2=

$$[ 32 ^ { 0,5 } \cdot ( \sqrt { 8 } ) ^ { - 1 } ] ^ { - 2 } =$$

Evaluate

$$0,25$$

Solution Steps

Expand \left(32^{0,5}\left(\sqrt{8}\right)^{-1}\right)^{-2}.

$$\left(32^{0,5}\right)^{-2}\left(\left(\sqrt{8}\right)^{-1}\right)^{-2}$$

To raise a power to another power, multiply the exponents. Multiply 0,5 and -2 to get -1.

$$32^{-1}\left(\left(\sqrt{8}\right)^{-1}\right)^{-2}$$

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

$$32^{-1}\left(\sqrt{8}\right)^{2}$$

Calculate 32 to the power of -1 and get \frac{1}{32}.

$$\frac{1}{32}\left(\sqrt{8}\right)^{2}$$

The square of \sqrt{8} is 8.

$$\frac{1}{32}\times 8$$

Multiply \frac{1}{32} and 8 to get \frac{1}{4}.

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

Factor

$$\frac{1}{2 ^ {2}} = 0.25$$