Solve (198*4^-2*3^4*6^-3)^2

$$( 198 \times 4 ^ { - 2 } \times 3 ^ { 4 } \times 6 ^ { - 3 } ) ^ { 2 }$$

Evaluate

$$\frac{88209}{4096}\approx 21.535400391$$

Solution Steps

Calculate 4 to the power of -2 and get \frac{1}{16}.

$$\left(198\times \frac{1}{16}\times 3^{4}\times 6^{-3}\right)^{2}$$

Multiply 198 and \frac{1}{16} to get \frac{99}{8}.

$$\left(\frac{99}{8}\times 3^{4}\times 6^{-3}\right)^{2}$$

Calculate 3 to the power of 4 and get 81.

$$\left(\frac{99}{8}\times 81\times 6^{-3}\right)^{2}$$

Multiply \frac{99}{8} and 81 to get \frac{8019}{8}.

$$\left(\frac{8019}{8}\times 6^{-3}\right)^{2}$$

Calculate 6 to the power of -3 and get \frac{1}{216}.

$$\left(\frac{8019}{8}\times \frac{1}{216}\right)^{2}$$

Multiply \frac{8019}{8} and \frac{1}{216} to get \frac{297}{64}.

$$\left(\frac{297}{64}\right)^{2}$$

Calculate \frac{297}{64} to the power of 2 and get \frac{88209}{4096}.

$$\frac{88209}{4096}$$

Factor

$$\frac{3 ^ {6} \cdot 11 ^ {2}}{2 ^ {12}} = 21\frac{2193}{4096} = 21.535400390625$$