What is (2^3\times 3^2\times 5^{-2})^2 ?

The solution to (2^3\times 3^2\times 5^{-2})^2 is 5184/625

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simplify (2^3\times 3^2\times 5^{-2})^2

simplify

(2^{3} \cdot 3^{2} \cdot 5^{- 2})^{2}

\frac{5184}{625}

Decimal Notation

8.2944

Solution steps

(2^{3} \cdot 3^{2} \cdot 5^{- 2})^{2}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

(2^{3} \cdot 3^{2} \cdot 5^{- 2})^{2} = (2^{3})^{2} (3^{2})^{2} (5^{- 2})^{2}

= (2^{3})^{2} (3^{2})^{2} (5^{- 2})^{2}

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

= 64 (3^{2})^{2} (5^{- 2})^{2}

(3^{2})^{2} = 81

= 64 \cdot 81 (5^{- 2})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}, a \geq 0

(5^{- 2})^{2} = 5^{- 2 \cdot 2}

= 64 \cdot 81 \cdot 5^{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= 64 \cdot 81 \cdot 5^{- 4}

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

5^{- 4} = \frac{1}{5 ^{4}}

= 64 \cdot 81 \cdot \frac{1}{5 ^{4}}

5^{4} = 625

= 64 \cdot 81 \cdot \frac{1}{625}

64 \cdot 81 \cdot \frac{1}{625} = \frac{5184}{625}

= \frac{5184}{625}

\frac{π}{4}

(- 3 e^{- 3 x})^{2}

9 b^{2} - 100

6 (2 x - 3) - 2 (2 x + 1)

\frac{4}{5} + 6

What is (2^3\times 3^2\times 5^{-2})^2 ?
The solution to (2^3\times 3^2\times 5^{-2})^2 is 5184/625