Solve (-7)^5*49^2/7^0*7*(-49)^3

$$\frac { ( - 7 ) ^ { 5 } \cdot 49 ^ { 2 } } { 7 ^ { 0 } \cdot 7 \cdot ( - 49 ) ^ { 3 } }$$

Evaluate

$$49$$

Solution Steps

To multiply powers of the same base, add their exponents. Add 0 and 1 to get 1.

$$\frac{\left(-7\right)^{5}\times 49^{2}}{7^{1}\left(-49\right)^{3}}$$

Calculate -7 to the power of 5 and get -16807.

$$\frac{-16807\times 49^{2}}{7^{1}\left(-49\right)^{3}}$$

Calculate 49 to the power of 2 and get 2401.

$$\frac{-16807\times 2401}{7^{1}\left(-49\right)^{3}}$$

Multiply -16807 and 2401 to get -40353607.

$$\frac{-40353607}{7^{1}\left(-49\right)^{3}}$$

Calculate 7 to the power of 1 and get 7.

$$\frac{-40353607}{7\left(-49\right)^{3}}$$

Calculate -49 to the power of 3 and get -117649.

$$\frac{-40353607}{7\left(-117649\right)}$$

Multiply 7 and -117649 to get -823543.

$$\frac{-40353607}{-823543}$$

Divide -40353607 by -823543 to get 49.

$$49$$

Factor

$$7^{2}$$