Question

Isolate P in:
ln|P| - ln|L-P| = kt + C

I get to the point where |P/(L-P)| = e^(kt+C) and I drop the absolute value since e^(anything) should be > 0.

In the end I get:
ALe^(kt)
------
1+Ae^(kt) where A = e^C

but my prof says A = +/- e^C.. why do we keep the negative? Isolate P in:
ln|P| - ln|L-P| = kt + C

I get to the point where |P/(L-P)| = e^(kt+C) and I drop the absolute value since e^(anything) should be > 0.

In the end I get:
ALe^(kt)
------
1+Ae^(kt) where A = e^C

but my prof says A = +/- e^C.. why do we keep the negative? @Mathematics

Answer 1

by converting ln |P| - ln|L-P| to ln (|P|/|L-P|) it is on the condition that L-P != 0

Answer 2

Does != mean not equal 0?

Answer 3

yes

Answer 4

since you can't divide by 0

Answer 5

Right. But, if my prof says we can get A = +/- e^C, doesn't that mean that when he is solving for P, when removing the absolute value, he says P/(L-P) = +/- e^C*e^kt?
Is it not just equal to + e^c*e^kt since e^(C+kt) must always be positive?

Answer 6

no since you're putting (+/-1) in front of e^C*e^kt. e^C*e^kt will be > 0, but then you can multiply by -1 after

Answer 7

btw, my first answer didn't really answer your question, but you can add it as a note

Answer 8

Or put another way. Only e^C*e^kt has to be positive, not everything that's on the right side of the equal sign.

Answer 9

Ohhh I see, right it's true since we don't know if P/(L-P) will be negative or not.. ahh ok thanks!! :)

Answer 10

You're welcome :)