What is [(2^3)^{-2}]^{0.5}*[(0.5)^{0.75}]^{-4} ?

The solution to [(2^3)^{-2}]^{0.5}*[(0.5)^{0.75}]^{-4} is 1

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simplify [(2^3)^{-2}]^{0.5}*[(0.5)^{0.75}]^{-4}

simplify

[(2^{3})^{- 2}]^{0.5} \cdot [(0.5)^{0.75}]^{- 4}

1

Solution steps

[(2^{3})^{- 2}]^{0.5} [(0.5)^{0.75}]^{- 4}

(2^{3})^{- 2} : 2^{- 6}

= (2^{- 6})^{0.5} [0. 5^{0.75}]^{- 4}

[0. 5^{0.75}]^{- 4} : 0. 5^{- 3}

= (2^{- 6})^{0.5} \cdot 0. 5^{- 3}

(2^{- 6})^{\frac{1}{2}} = \frac{1}{2 ^{3}}

= 0. 5^{- 3} \cdot \frac{1}{2 ^{3}}

Apply exponent rule:

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

0. 5^{- 3} = \frac{1}{0. 5 ^{3}}

= \frac{1}{2 ^{3}} \cdot \frac{1}{0. 5 ^{3}}

Multiply fractions:

\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}

= \frac{1 \cdot 1}{2 ^{3} \cdot 0. 5 ^{3}}

Multiply the numbers:

1 \cdot 1 = 1

= \frac{1}{2 ^{3} \cdot 0. 5 ^{3}}

2^{3} \cdot 0. 5^{3} = 1

= \frac{1}{1}

Apply rule

\frac{a}{1} = a

= 1

e^{- 3 (t - 2)}

- \frac{1}{2} e^{- 3}

(e^{t}) (e^{- t^{2}}) (e^{t})

\frac{39}{61}

What is [(2^3)^{-2}]^{0.5}*[(0.5)^{0.75}]^{-4} ?
The solution to [(2^3)^{-2}]^{0.5}*[(0.5)^{0.75}]^{-4} is 1