Question

\[\int_0^{\infty } 2^t e^{-s t} \, dt\]

Answer 1

2^t= e^Log[2^t]= e^2Log[2]
\[\int_0^{\infty } e^{-s t} e^{t \log (2)} \, dt\]

\[\int _0^{\infty } e^{t \log (2)-s t}dt\]

Answer 2

is this = 1 / (log2 - s) * e ^ (tlog2 - st)

im not very confident in integration

Answer 3

just
\[\frac{1}{s-\text{Log}[2]}\]

Answer 4

- this is the result after applying the limits?

Answer 5

your first line should be
= e^tLog[2] right?

Answer 6

turning 2^t to an exponential function is pretty clever!

Answer 7

yes, after limit

Answer 8

gret

Answer 9

cwrw238 are you jimmyrep? :-)