# fun with natural logs....

• Apr 4th 2011, 10:31 AM
slapmaxwell1
fun with natural logs....
ok so we have the problem...

ln N - ln(N-2000)= t + C

so i tried to say that ln a = ln B = ln (a/B)

so i got ln(N/N-2000)= t + c

so then i raised both sides to e and got

N/N-2000 = e^(t+c)

my book got the same thing except their answer was positive, i dont see how this could be possible? can someone tell me where i went wrong
• Apr 4th 2011, 10:41 AM
Quacky
Quote:

Originally Posted by slapmaxwell1
ok so we have the problem...

ln N - ln(N-2000)= t + C

so i tried to say that ln a = ln B = ln (a/B)

so i got ln(N/N-2000)= t + C

so then i raised both sides to e and got

N/N-2000 = e^(t+C)

my book got the same thing except their answer was positive, i dont see how this could be possible? can someone tell me where i went wrong

You haven't done anything wrong at this stage; the error lies elsewhere with your working or with the textbook. What is the original question?

$N=e^{t+C}(N-2000)$

$N=Ne^{t+C}-2000e^{t+C}$

$N-Ne^{t+C}=-2000e^{t+C}$

$N(1-e^{t+C})=-2000e^{t+C}$

$N=\dfrac{-2000e^{t+C}}{1-e^{t+C}}$
• Apr 5th 2011, 06:39 AM
HallsofIvy
Notice that $e^{t+ C}= ce^y$ where $c= e^C$.

Also, you can multiply both numerator and denominator by -1 to get
$\frac{2000ce^t}{ce^t- 1}$

Perhaps that is the "positive" answer you refer to.