Please check my work. Did I calculate the following Laplace Transform correctly?
$$f(t)=e^{kt}u(t-a)$$
My solution:
Use the following corollary from the second shifting theorem (t-shift)...
$$\mathcal{L}\{g(t)u(t-a)\}=e^{-as}\mathcal{L}\{g(t+a)\}$$
where...
$$f(t)=g(t+a)$$
with $g(t)=e^{kt}$ and $a=a$
$$g(t+a)=e^{k(t+a)}=e^{kt}e^{ka}$$
$$\mathcal{L}\{e^{kt}u(t-a)\}=e^{-as}\cdot \mathcal{L}\{e^{ka}e^{kt}\}$$
$$e^{-as}\cdot e^{ka}\mathcal{L}\{e^{kt}=e^{-a(s-k)}\cdot \mathcal{L}\{e^{kt}\}$$
$$F(s)=\frac{e^{-a(s-k)}}{s-k}$$