Solve (2^{-5}div2^-3)^2*2^-7*2^-1

$$( 2 ^ { - 5 } \div 2 ^ { - 3 } ) ^ { 2 } \cdot 2 ^ { - 7 } \cdot 2 ^ { - 1 }$$

Evaluate

$$\frac{1}{4096}=0.000244141$$

Solution Steps

Rewrite 2^{-3} as 2^{-5}\times 2^{2}. Cancel out 2^{-5} in both numerator and denominator.

$$\left(\frac{1}{2^{2}}\right)^{2}\times 2^{-7}\times 2^{-1}$$

To multiply powers of the same base, add their exponents. Add -7 and -1 to get -8.

$$\left(\frac{1}{2^{2}}\right)^{2}\times 2^{-8}$$

Calculate 2 to the power of 2 and get 4.

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

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

$$\frac{1}{16}\times 2^{-8}$$

Calculate 2 to the power of -8 and get \frac{1}{256}.

$$\frac{1}{16}\times \frac{1}{256}$$

Multiply \frac{1}{16} and \frac{1}{256} to get \frac{1}{4096}.

$$\frac{1}{4096}$$

Factor

$$\frac{1}{2 ^ {12}} = 0.000244140625$$