Solve {left(1/4right)}^-2+sqrt[8]{left(2^4right)^2}+256^0*25

$${ \left( \frac{ 1 }{ 4 } \right) }^{ -2 } + \sqrt[ 8 ]{ { \left( { 2 }^{ 4 } \right) }^{ 2 } } + { 256 }^{ 0 \cdot 25 }$$

Evaluate

$$19$$

Solution Steps

To raise a power to another power, multiply the exponents. Multiply 4 and 2 to get 8.

$$\left(\frac{1}{4}\right)^{-2}+\sqrt[8]{2^{8}}+256^{0\times 25}$$

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

$$16+\sqrt[8]{2^{8}}+256^{0\times 25}$$

Calculate 2 to the power of 8 and get 256.

$$16+\sqrt[8]{256}+256^{0\times 25}$$

Calculate \sqrt[8]{256} and get 2.

$$16+2+256^{0\times 25}$$

Add 16 and 2 to get 18.

$$18+256^{0\times 25}$$

Multiply 0 and 25 to get 0.

$$18+256^{0}$$

Calculate 256 to the power of 0 and get 1.

$$18+1$$

Add 18 and 1 to get 19.

$$19$$

Factor

$$19$$