What is e^{-ax}*cos(bt+c) ?

The solution to e^{-ax}*cos(bt+c) is (cos(c+bt))/(e^{ax)}

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simplify e^{-ax}*cos(bt+c)

simplify

e^{- a x} \cdot cos (b t + c)

\frac{cos ( c + b t )}{e ^{a x}}

Solution steps

e^{- a x} cos (b t + c)

Apply exponent rule:

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

= \frac{1}{e ^{a x}} cos (c + b t)

Multiply fractions:

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

= \frac{1 \cdot cos ( b t + c )}{e ^{a x}}

Multiply:

1 \cdot cos (b t + c) = cos (b t + c)

= \frac{cos ( b t + c )}{e ^{a x}}

e^{- 0.198}

- \frac{2}{2} \cdot \frac{2}{2}

4 e^{- 0.5 x}

100 e^{- 2}

100 e^{- 3}

What is e^{-ax}*cos(bt+c) ?
The solution to e^{-ax}*cos(bt+c) is (cos(c+bt))/(e^{ax)}