Solve frac{1/2^{-3}-1/2^-1}{1^-3/2^9-1-3/2^3}

$$\frac { \frac { 1 } { 2 ^ { - 3 } } - \frac { 1 } { 2 ^ { - 1 } } } { \frac { 1 ^ { - 3 } } { 2 ^ { 9 } } - \frac { 1 - 3 } { 2 ^ { 3 } } }$$

Evaluate

$$\frac{1024}{43}\approx 23.813953488$$

Solution Steps

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

$$\frac{\frac{1}{\frac{1}{8}}-\frac{1}{2^{-1}}}{\frac{1^{-3}}{2^{9}}-\frac{1-3}{2^{3}}}$$

Divide 1 by \frac{1}{8} by multiplying 1 by the reciprocal of \frac{1}{8}.

$$\frac{1\times 8-\frac{1}{2^{-1}}}{\frac{1^{-3}}{2^{9}}-\frac{1-3}{2^{3}}}$$

Multiply 1 and 8 to get 8.

$$\frac{8-\frac{1}{2^{-1}}}{\frac{1^{-3}}{2^{9}}-\frac{1-3}{2^{3}}}$$

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

$$\frac{8-\frac{1}{\frac{1}{2}}}{\frac{1^{-3}}{2^{9}}-\frac{1-3}{2^{3}}}$$

Divide 1 by \frac{1}{2} by multiplying 1 by the reciprocal of \frac{1}{2}.

$$\frac{8-1\times 2}{\frac{1^{-3}}{2^{9}}-\frac{1-3}{2^{3}}}$$

Multiply 1 and 2 to get 2.

$$\frac{8-2}{\frac{1^{-3}}{2^{9}}-\frac{1-3}{2^{3}}}$$

Subtract 2 from 8 to get 6.

$$\frac{6}{\frac{1^{-3}}{2^{9}}-\frac{1-3}{2^{3}}}$$

Calculate 1 to the power of -3 and get 1.

$$\frac{6}{\frac{1}{2^{9}}-\frac{1-3}{2^{3}}}$$

Calculate 2 to the power of 9 and get 512.

$$\frac{6}{\frac{1}{512}-\frac{1-3}{2^{3}}}$$

Subtract 3 from 1 to get -2.

$$\frac{6}{\frac{1}{512}-\frac{-2}{2^{3}}}$$

Calculate 2 to the power of 3 and get 8.

$$\frac{6}{\frac{1}{512}-\frac{-2}{8}}$$

Reduce the fraction \frac{-2}{8} to lowest terms by extracting and canceling out 2.

$$\frac{6}{\frac{1}{512}-\left(-\frac{1}{4}\right)}$$

The opposite of -\frac{1}{4} is \frac{1}{4}.

$$\frac{6}{\frac{1}{512}+\frac{1}{4}}$$

Add \frac{1}{512} and \frac{1}{4} to get \frac{129}{512}.

$$\frac{6}{\frac{129}{512}}$$

Divide 6 by \frac{129}{512} by multiplying 6 by the reciprocal of \frac{129}{512}.

$$6\times \frac{512}{129}$$

Multiply 6 and \frac{512}{129} to get \frac{1024}{43}.

$$\frac{1024}{43}$$

Factor

$$\frac{2 ^ {10}}{43} = 23\frac{35}{43} = 23.813953488372093$$