What is (b/a+ce^{-at})^{-1} ?

The solution to (b/a+ce^{-at})^{-1} is a/(b+ace^{-at)}

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simplify (b/a+ce^{-at})^{-1}

simplify

(\frac{b}{a} + c e^{- a t})^{- 1}

\frac{a}{b + a c e ^{- a t}}

Solution steps

(\frac{b}{a} + c e^{- a t})^{- 1}

Apply exponent rule:

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

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

Join

\frac{b}{a} + c e^{- a t} : \frac{b + a c e ^{- a t}}{a}

= \frac{1}{\frac{b + a c e ^{- a t}}{a}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{a}{b + c e ^{- a t} a}

- 1.6 x 1 0^{- 19}

t \cdot e^{- 3 t}

(5 x^{6} y^{7} z^{3}) (4 x yz) (2 x^{4} y^{3} z^{5})

2 e^{- 3 \cdot 2}

π^{2}

What is (b/a+ce^{-at})^{-1} ?
The solution to (b/a+ce^{-at})^{-1} is a/(b+ace^{-at)}