# Exponential with E^-kt, solving for t

• Aug 9th 2011, 08:02 AM
Aluminium
Exponential with E^-kt, solving for t
I hope this image will show through nicely:

Attachment 22010

I've done so far:

1/100 = e^-kt
ln (1/100) = ln e^-kt
-kt = ln(1/100)/ln e
t = ln(1/100)/ln e * -k

Am I right? Maybe ln e needs to be cancelled to something? Please help me out here.
• Aug 9th 2011, 08:10 AM
skeeter
Re: Exponential with E^-kt, solving for t
Quote:

Originally Posted by Aluminium
I hope this image will show through nicely:

Attachment 22010

I've done so far:

1/100 = e^-kt
ln (1/100) = ln e^-kt
-kt = ln(1/100)/ln e
t = ln(1/100)/ln e * -k

Am I right? Maybe ln e needs to be cancelled to something? Please help me out here.

$\ln(e) = 1$
• Aug 9th 2011, 08:15 AM
Plato
Re: Exponential with E^-kt, solving for t
Quote:

Originally Posted by Aluminium
I hope this image will show through nicely:

Attachment 22010

Do you see that $1=100e^{-kt}$ is the same as $e^{kt}=100~?$
• Aug 9th 2011, 08:41 AM
Aluminium
Re: Exponential with E^-kt, solving for t
I don't see that yet :(
• Aug 9th 2011, 08:48 AM
Siron
Re: Exponential with E^-kt, solving for t
Well:
$1=100\cdot e^{-kt} \Leftrightarrow \frac{1}{100}=e^{-kt} \Leftrightarrow 100=\frac{1}{e^{-kt}} \Leftrightarrow 100=e^{kt}$

Can you solve it now for $t$?