What is (3j^{i0.3})(2j^{-i0.2}) ?

The solution to (3j^{i0.3})(2j^{-i0.2}) is 6j^{0.1i}

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simplify (3j^{i0.3})(2j^{-i0.2})

simplify

(3 j^{i 0.3}) (2 j^{- i 0.2})

6 j^{0.1 i}

Solution steps

(3 j^{i \cdot 0.3}) (2 j^{- i \cdot 0.2})

Simplify

2 j^{- i \cdot 0.2} : \frac{2}{j ^{0.2 i}}

= 3 j^{i \cdot 0.3} \frac{2}{j ^{0.2 i}}

Apply the fraction rule:

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

= \frac{3 j ^{i \cdot 0.3} \cdot 2}{j ^{0.2 i}}

Simplify

\frac{j ^{i \cdot 0.3}}{j ^{0.2 i}} : j^{0.1 i}

= 3 j^{0.1 i} \cdot 2

Multiply the numbers:

3 \cdot 2 = 6

= 6 j^{0.1 i}

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What is (3j^{i0.3})(2j^{-i0.2}) ?
The solution to (3j^{i0.3})(2j^{-i0.2}) is 6j^{0.1i}