Solve 4^5:4^4*4^2-sqrt{81}-(7-3)^0-24*8=

$$4 ^ { 5 } : 4 ^ { 4 } \cdot 4 ^ { 2 } - \sqrt { 81 } - ( 7 - 3 ) ^ { 0 } - 24 \cdot 8 =$$

Evaluate

$$-138$$

Solution Steps

To divide powers of the same base, subtract the denominator's exponent from the numerator's exponent. Subtract 4 from 5 to get 1.

$$4^{1}\times 4^{2}-\sqrt{81}-\left(7-3\right)^{0}-24\times 8$$

To multiply powers of the same base, add their exponents. Add 1 and 2 to get 3.

$$4^{3}-\sqrt{81}-\left(7-3\right)^{0}-24\times 8$$

Calculate 4 to the power of 3 and get 64.

$$64-\sqrt{81}-\left(7-3\right)^{0}-24\times 8$$

Calculate the square root of 81 and get 9.

$$64-9-\left(7-3\right)^{0}-24\times 8$$

Subtract 9 from 64 to get 55.

$$55-\left(7-3\right)^{0}-24\times 8$$

Subtract 3 from 7 to get 4.

$$55-4^{0}-24\times 8$$

Calculate 4 to the power of 0 and get 1.

$$55-1-24\times 8$$

Subtract 1 from 55 to get 54.

$$54-24\times 8$$

Multiply 24 and 8 to get 192.

$$54-192$$

Subtract 192 from 54 to get -138.

$$-138$$

Factor

$$-138$$