Solve (frac{2^{-3}}{2^{-7}}*(4^2)^-3/2^-9)^2

$$( \frac { 2 ^ { - 3 } } { 2 ^ { - 7 } } \cdot \frac { ( 4 ^ { 2 } ) ^ { - 3 } } { 2 ^ { - 9 } } ) ^ { 2 }$$

Evaluate

$$4$$

Solution Steps

To divide powers of the same base, subtract the denominator's exponent from the numerator's exponent. Subtract -7 from -3 to get 4.

$$\left(2^{4}\times \frac{\left(4^{2}\right)^{-3}}{2^{-9}}\right)^{2}$$

To raise a power to another power, multiply the exponents. Multiply 2 and -3 to get -6.

$$\left(2^{4}\times \frac{4^{-6}}{2^{-9}}\right)^{2}$$

Calculate 2 to the power of 4 and get 16.

$$\left(16\times \frac{4^{-6}}{2^{-9}}\right)^{2}$$

Calculate 4 to the power of -6 and get \frac{1}{4096}.

$$\left(16\times \frac{\frac{1}{4096}}{2^{-9}}\right)^{2}$$

Calculate 2 to the power of -9 and get \frac{1}{512}.

$$\left(16\times \frac{\frac{1}{4096}}{\frac{1}{512}}\right)^{2}$$

Divide \frac{1}{4096} by \frac{1}{512} by multiplying \frac{1}{4096} by the reciprocal of \frac{1}{512}.

$$\left(16\times \frac{1}{4096}\times 512\right)^{2}$$

Multiply \frac{1}{4096} and 512 to get \frac{1}{8}.

$$\left(16\times \frac{1}{8}\right)^{2}$$

Multiply 16 and \frac{1}{8} to get 2.

$$2^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4$$

Factor

$$2^{2}$$