Solve (13^2/7^5div26/7^3)^-1

$$( \frac { 13 ^ { 2 } } { 7 ^ { 5 } } \div \frac { 26 } { 7 ^ { 3 } } ) ^ { - 1 }$$

Evaluate

$$\frac{98}{13}\approx 7.538461538$$

Solution Steps

Divide \frac{13^{2}}{7^{5}} by \frac{26}{7^{3}} by multiplying \frac{13^{2}}{7^{5}} by the reciprocal of \frac{26}{7^{3}}.

$$\left(\frac{13^{2}\times 7^{3}}{7^{5}\times 26}\right)^{-1}$$

Cancel out 7^{3} in both numerator and denominator.

$$\left(\frac{13^{2}}{26\times 7^{2}}\right)^{-1}$$

Calculate 13 to the power of 2 and get 169.

$$\left(\frac{169}{26\times 7^{2}}\right)^{-1}$$

Calculate 7 to the power of 2 and get 49.

$$\left(\frac{169}{26\times 49}\right)^{-1}$$

Multiply 26 and 49 to get 1274.

$$\left(\frac{169}{1274}\right)^{-1}$$

Reduce the fraction \frac{169}{1274} to lowest terms by extracting and canceling out 13.

$$\left(\frac{13}{98}\right)^{-1}$$

Calculate \frac{13}{98} to the power of -1 and get \frac{98}{13}.

$$\frac{98}{13}$$

Factor

$$\frac{2 \cdot 7 ^ {2}}{13} = 7\frac{7}{13} = 7.538461538461538$$