Solve (2^-2-4^-2)*(2^-1+4^-1)

$$( 2 ^ { - 2 } - 4 ^ { - 2 } ) \cdot ( 2 ^ { - 1 } + 4 ^ { - 1 } )$$

Evaluate

$$\frac{9}{64}=0.140625$$

Solution Steps

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

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

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

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

Subtract \frac{1}{16} from \frac{1}{4} to get \frac{3}{16}.

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

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

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

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

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

Add \frac{1}{2} and \frac{1}{4} to get \frac{3}{4}.

$$\frac{3}{16}\times \frac{3}{4}$$

Multiply \frac{3}{16} and \frac{3}{4} to get \frac{9}{64}.

$$\frac{9}{64}$$

Factor

$$\frac{3 ^ {2}}{2 ^ {6}} = 0.140625$$