 # Laplace transform of a exponential. we know that Laplace Erika Bernard 2022-04-15 Answered
Laplace transform of a exponential.
we know that Laplace transform of ${x}^{n}$ is
$\mathcal{L}\left[{x}^{n}\right]$
provided n is a positive integer
but what is Laplace transform of
${a}^{x}$ where a is some constant number
$\mathcal{L}\left[{x}^{n}\right]={\int }_{0}^{\mathrm{\infty }}{a}^{x}{e}^{-sx}dx$
but how to solve further ( i am using x insted of t here)
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${\int }_{0}^{\mathrm{\infty }}{a}^{x}{e}^{-sx}dx={\int }_{0}^{\mathrm{\infty }}{e}^{x\mathrm{ln}a}{e}^{-sx}dx={\int }_{0}^{\mathrm{\infty }}{e}^{-\left(s-\mathrm{ln}a\right)x}dx=\frac{1}{s-\mathrm{ln}a}$
###### Not exactly what you’re looking for? Drantumcem0
${a}^{x}{e}^{-sx}={\left(a{e}^{-s}\right)}^{x}$
$\mathcal{L}{\left(a{e}^{-s}\right)}^{x}={\int }_{0}^{\mathrm{\infty }}{\left(a{e}^{-s}\right)}^{x}=\frac{1}{\mathrm{ln}\left(a{e}^{-s}\right)}{\left(a{e}^{-s}\right)}^{x}{\mid }_{0}^{\mathrm{\infty }}=\frac{1}{\mathrm{ln}a-s}\left(0-1\right)=\frac{1}{s-\mathrm{ln}a}$