What is (2^22^{-3}3^{-2}3^34^{-2}4^3)^2 ?

The solution to (2^22^{-3}3^{-2}3^34^{-2}4^3)^2 is 36

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simplify (2^22^{-3}3^{-2}3^34^{-2}4^3)^2

simplify

(2^{2} 2^{- 3} 3^{- 2} 3^{3} 4^{- 2} 4^{3})^{2}

36

Solution steps

(2^{2} \cdot 2^{- 3} \cdot 3^{- 2} \cdot 3^{3} \cdot 4^{- 2} \cdot 4^{3})^{2}

Apply exponent rule:

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

(2^{2} \cdot 2^{- 3} \cdot 3^{- 2} \cdot 3^{3} \cdot 4^{- 2} \cdot 4^{3})^{2} = (2^{2})^{2} (2^{- 3})^{2} (3^{- 2})^{2} (3^{3})^{2} (4^{- 2})^{2} (4^{3})^{2}

= (2^{2})^{2} (2^{- 3})^{2} (3^{- 2})^{2} (3^{3})^{2} (4^{- 2})^{2} (4^{3})^{2}

(2^{2})^{2} = 16

= 16 (2^{- 3})^{2} (3^{- 2})^{2} (3^{3})^{2} (4^{- 2})^{2} (4^{3})^{2}

(2^{- 3})^{2} = \frac{1}{64}

= 16 \cdot \frac{1}{64} (3^{- 2})^{2} (3^{3})^{2} (4^{- 2})^{2} (4^{3})^{2}

(3^{- 2})^{2} = \frac{1}{81}

= 16 \cdot \frac{1}{64} \cdot \frac{1}{81} (3^{3})^{2} (4^{- 2})^{2} (4^{3})^{2}

(3^{3})^{2} = 729

= 16 \cdot \frac{1}{64} \cdot \frac{1}{81} \cdot 729 (4^{- 2})^{2} (4^{3})^{2}

(4^{- 2})^{2} = \frac{1}{256}

= 16 \cdot \frac{1}{64} \cdot \frac{1}{81} \cdot 729 \cdot \frac{1}{256} (4^{3})^{2}

Apply exponent rule:

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

(4^{3})^{2} = 4^{3 \cdot 2}

= 16 \cdot \frac{1}{64} \cdot \frac{1}{81} \cdot 729 \cdot \frac{1}{256} \cdot 4^{3 \cdot 2}

Multiply the numbers:

3 \cdot 2 = 6

= 16 \cdot \frac{1}{64} \cdot \frac{1}{81} \cdot 729 \cdot \frac{1}{256} \cdot 4^{6}

4^{6} = 4096

= 16 \cdot \frac{1}{64} \cdot \frac{1}{81} \cdot 729 \cdot \frac{1}{256} \cdot 4096

16 \cdot \frac{1}{64} \cdot \frac{1}{81} \cdot 729 \cdot \frac{1}{256} \cdot 4096 = 36

= 36

15 y^{2} - 30 = 22 y + 7

3 x + 23 = 2 x + 59

x = x + \frac{x}{3} + 1

3 (x - 4) = 8 - 5 (3 x + 1)

6 - 3 ∣ 2 x + 6 ∣ = 0

What is (2^22^{-3}3^{-2}3^34^{-2}4^3)^2 ?
The solution to (2^22^{-3}3^{-2}3^34^{-2}4^3)^2 is 36