What is (2^5*(2^2)^3*9^2)/(4^3*18*3^2) ?

The solution to (2^5*(2^2)^3*9^2)/(4^3*18*3^2) is 16

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simplify (2^5(2^2)^39^2)/(4^3183^2)

simplify

\frac{2 ^{5} \cdot ( 2 ^{2} ) ^{3} \cdot 9 ^{2}}{4 ^{3} \cdot 18 \cdot 3 ^{2}}

16

Solution steps

\frac{2 ^{5} ( 2 ^{2} ) ^{3} \cdot 9 ^{2}}{4 ^{3} \cdot 18 \cdot 3 ^{2}}

(2^{2})^{3} = 64

= \frac{2 ^{5} \cdot 64 \cdot 9 ^{2}}{4 ^{3} \cdot 18 \cdot 3 ^{2}}

Factor the number:

64 = 2 \cdot 32

= \frac{2 ^{5} \cdot 2 \cdot 32 \cdot 9 ^{2}}{4 ^{3} \cdot 18 \cdot 3 ^{2}}

Factor the number:

18 = 2 \cdot 9

= \frac{2 ^{5} \cdot 2 \cdot 32 \cdot 9 ^{2}}{4 ^{3} \cdot 2 \cdot 9 \cdot 3 ^{2}}

Cancel the common factor:

2

= \frac{2 ^{5} \cdot 32 \cdot 9 ^{2}}{4 ^{3} \cdot 9 \cdot 3 ^{2}}

\frac{9 ^{2}}{9} = 9

= \frac{2 ^{5} \cdot 32 \cdot 9}{4 ^{3} \cdot 3 ^{2}}

Factor

4^{3} : 2^{6}

= \frac{2 ^{5} \cdot 32 \cdot 9}{2 ^{6} \cdot 3 ^{2}}

\frac{2 ^{5}}{2 ^{6}} = \frac{1}{2}

= \frac{32 \cdot 9}{2 \cdot 3 ^{2}}

Factor the number:

32 = 2 \cdot 16

= \frac{2 \cdot 16 \cdot 9}{2 \cdot 3 ^{2}}

Cancel the common factor:

2

= \frac{16 \cdot 9}{3 ^{2}}

Factor the number:

9 = 3^{2}

= \frac{16 \cdot 3 ^{2}}{3 ^{2}}

Cancel the common factor:

3^{2}

= 16

(\frac{5}{9} \cdot x^{4 n + \frac{3}{4}})^{4} \cdot (1 \frac{4}{5} \cdot x^{\frac{3}{2}})^{4}

x - x^{- 1}

3 x^{n} - 15 x^{n - 2} - 12 x^{n - 1}

\frac{40 e ^{9}}{5 e ^{8}}

i^{117}

What is (2^5*(2^2)^3*9^2)/(4^3*18*3^2) ?
The solution to (2^5*(2^2)^3*9^2)/(4^3*18*3^2) is 16