What is (-1/2 e^{2x}+1/6 e^{3x})(1/3 e^{-4x}) ?

The solution to (-1/2 e^{2x}+1/6 e^{3x})(1/3 e^{-4x}) is (1/6 e^{3x}-1/2 e^{2x})/(3e^{4x)}

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simplify (-1/2 e^{2x}+1/6 e^{3x})(1/3 e^{-4x})

simplify

(- \frac{1}{2} e^{2 x} + \frac{1}{6} e^{3 x}) (\frac{1}{3} e^{- 4 x})

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

Solution steps

(- \frac{1}{2} e^{2 x} + \frac{1}{6} e^{3 x}) (\frac{1}{3} e^{- 4 x})

Apply exponent rule:

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

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

Multiply fractions:

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

= \frac{1 \cdot 1 \cdot ( - \frac{1}{2} e ^{2 x} + \frac{1}{6} e ^{3 x} )}{3 e ^{4 x}}

1 \cdot 1 \cdot (- \frac{1}{2} e^{2 x} + \frac{1}{6} e^{3 x}) = - \frac{1}{2} e^{2 x} + \frac{1}{6} e^{3 x}

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

e^{- 0.0001216}

50 e^{- 0.223 \cdot 3}

e^{- a t} sin (ω t u)

b^{2} \cdot 4 a^{2}

e^{- \frac{120}{144}}

What is (-1/2 e^{2x}+1/6 e^{3x})(1/3 e^{-4x}) ?
The solution to (-1/2 e^{2x}+1/6 e^{3x})(1/3 e^{-4x}) is (1/6 e^{3x}-1/2 e^{2x})/(3e^{4x)}