What is e^{-3s}L{(e^{(t+3)-2})/5 } ?

The solution to e^{-3s}L{(e^{(t+3)-2})/5 } is (Le^{t+1-3s})/5

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simplify e^{-3s}L{(e^{(t+3)-2})/5 }

simplify

e^{- 3 s} L {\frac{e ^{(t + 3) - 2}}{5}}

\frac{L e ^{t + 1 - 3 s}}{5}

Solution steps

e^{- 3 s} L {\frac{e ^{(t + 3) - 2}}{5}}

Apply exponent rule:

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

= L \frac{1}{e ^{3 s}} {\frac{e ^{t + 3 - 2}}{5}}

Remove parentheses:

(a) = a

= \frac{1}{e ^{3 s}} L \frac{e ^{t + 3 - 2}}{5}

Add/Subtract the numbers:

3 - 2 = 1

= L \frac{1}{e ^{3 s}} \cdot \frac{e ^{t + 1}}{5}

Multiply fractions:

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

= \frac{1 \cdot e ^{t + 1} L}{e ^{3 s} \cdot 5}

Multiply:

1 \cdot e^{t + 1} = e^{t + 1}

= \frac{L e ^{t + 1}}{5 e ^{3 s}}

Apply exponent rule:

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

\frac{e ^{t + 1}}{e ^{3 s}} = e^{(t + 1) - 3 s}

= \frac{L e ^{t + 1 - 3 s}}{5}

e^{- λ}

(2 + 2 x + e^{2 x})^{- 1}

e^{2 t} \cdot 3 e^{3 t}

2^{12}

What is e^{-3s}L{(e^{(t+3)-2})/5 } ?
The solution to e^{-3s}L{(e^{(t+3)-2})/5 } is (Le^{t+1-3s})/5