the transform of e^(at)
The exponential function e^(at) is the natural language of growth and decay — bank interest, radioactive material, a cooling cup of coffee, the natural modes of a circuit. So its Laplace transform is one of the most-used entries in the whole table, and pleasingly it is just a single shift away from the transform of the constant.
The rule is L{e^(at)} = 1 / (s - a). To see where it comes from, plug into the definition: integral from 0 to infinity of e^(-st) e^(at) dt = integral from 0 to infinity of e^(-(s - a) t) dt = 1/(s - a), valid for s > a. The two exponentials combine into one, e^(-(s - a) t), which is just like the constant case but with s replaced by s - a — and that is exactly why this transform is the seed of the whole first shifting theorem. Notice the answer has a 'pole' at s = a: the denominator vanishes there, marking the growth rate of the original exponential right in the s-domain.
This single fact quietly powers the whole constant-coefficient story. The characteristic-equation method tells you that solutions of linear equations are built from exponentials e^(at), and 1/(s - a) is the s-domain fingerprint of each one. When you invert and see a factor 1/(s - 3), you read it instantly as e^(3t). The location a of the pole tells you the growth (a > 0) or decay (a < 0) of that mode.
L{e^(-2t)} = 1/(s - (-2)) = 1/(s + 2). Conversely, inverting 5/(s - 4) gives 5 e^(4t). The number you subtract inside the denominator IS the rate in the exponent.
e^(at) becomes 1/(s - a): the pole sits exactly at the growth rate a.
Mind the sign: e^(-2t) gives 1/(s + 2), not 1/(s - 2). The denominator is s minus the exponent's coefficient, so a negative rate flips to a plus inside the parentheses.