Solve (14^3)^4*(14^-7)^2

$$( 14 ^ { 3 } ) ^ { 4 } \cdot ( 14 ^ { - 7 } ) ^ { 2 }$$

Evaluate

$$\frac{1}{196}\approx 0.005102041$$

Solution Steps

To raise a power to another power, multiply the exponents. Multiply 3 and 4 to get 12.

$$14^{12}\times \left(14^{-7}\right)^{2}$$

To raise a power to another power, multiply the exponents. Multiply -7 and 2 to get -14.

$$14^{12}\times 14^{-14}$$

To multiply powers of the same base, add their exponents. Add 12 and -14 to get -2.

$$14^{-2}$$

Calculate 14 to the power of -2 and get \frac{1}{196}.

$$\frac{1}{196}$$

Factor

$$\frac{1}{2 ^ {2} \cdot 7 ^ {2}} = 0.00510204081632653$$