What is (3e^{-2j})^3(4e^{-3j})^{-2} ?

The solution to (3e^{-2j})^3(4e^{-3j})^{-2} is 27/16

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simplify (3e^{-2j})^3(4e^{-3j})^{-2}

simplify

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

\frac{27}{16}

Decimal Notation

1.6875

Solution steps

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

Apply exponent rule:

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

= \frac{1}{( 4 e ^{- 3 j} ) ^{2}} (3 e^{- 2 j})^{3}

(3 e^{- 2 j})^{3} = 3^{3} e^{- 6 j}

= 3^{3} e^{- 6 j} \frac{1}{( 4 e ^{- 3 j} ) ^{2}}

\frac{1}{( 4 e ^{- 3 j} ) ^{2}} = \frac{1}{4 ^{2} e ^{- 6 j}}

= 3^{3} e^{- 6 j} \frac{1}{4 ^{2} e ^{- 6 j}}

Multiply fractions:

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

= \frac{1 \cdot 3 ^{3} e ^{- 6 j}}{4 ^{2} e ^{- 6 j}}

Cancel the common factor:

e^{- 6 j}

= \frac{1 \cdot 3 ^{3}}{4 ^{2}}

Refine

= \frac{27}{16}

6 (3.43) (1 0^{- 7})

(12 x^{2}) \cdot (3 x)

e^{- 2 x} (\frac{e ^{x} - cos ( x )}{x})

((- \frac{1}{7})^{2})^{- 1}^{- 1}

What is (3e^{-2j})^3(4e^{-3j})^{-2} ?
The solution to (3e^{-2j})^3(4e^{-3j})^{-2} is 27/16