Solve (2^{2}*2^-3*3^-1*3^2)^2

$$( 2 ^ { 2 } \cdot 2 ^ { - 3 } \cdot 3 ^ { - 1 } \cdot 3 ^ { 2 } ) ^ { 2 }$$

Evaluate

$$\frac{9}{4}=2.25$$

Solution Steps

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

$$\left(2^{-1}\times 3^{-1}\times 3^{2}\right)^{2}$$

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

$$\left(2^{-1}\times 3^{1}\right)^{2}$$

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

$$\left(\frac{1}{2}\times 3^{1}\right)^{2}$$

Calculate 3 to the power of 1 and get 3.

$$\left(\frac{1}{2}\times 3\right)^{2}$$

Multiply \frac{1}{2} and 3 to get \frac{3}{2}.

$$\left(\frac{3}{2}\right)^{2}$$

Calculate \frac{3}{2} to the power of 2 and get \frac{9}{4}.

$$\frac{9}{4}$$

Factor

$$\frac{3 ^ {2}}{2 ^ {2}} = 2\frac{1}{4} = 2.25$$