is it?
i just want to make sure.
Might it help to move the exponential to get a positive power...?
. . . . .$\displaystyle P\, +\, 56.5Pe^{-0.73t}\, =\, 230$
. . . . .$\displaystyle P\, +\, \frac{56.5P}{e^{0.73t}}\, =\, 230$
. . . . .$\displaystyle e^{0.73t}P\, +\, 56.5P\, =\, 230 e^{0.73t}$
. . . . .$\displaystyle 56.5P\, =\, 230 e^{0.73t}\, -\, e^{0.73t}P$
. . . . .$\displaystyle 56.5P\, =\, \left(230\, -\, P\right) e^{0.73t}$
. . . . .$\displaystyle \frac{56.5P}{230\, -\, P}\, =\, e^{0.73t}$
. . . . .$\displaystyle \ln{\left(\frac{56.5P}{230\, -\, P}\right)}\, =\, 0.73t$
...and so forth.
P.S. To review why "ln(e^(some power))" simplifies to just the (some power), try here.