Exponential with E^-kt, solving for t

**Aluminium** · August 9th 2011, 07:02 AM

I hope this image will show through nicely:

Exponential with E^-kt, solving for t-solve-t.gif

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.

**skeeter** · August 9th 2011, 07:10 AM

Originally Posted by Aluminium

I hope this image will show through nicely:

Click image for larger version.

Name: solve for t.gif
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ID: 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.

$\ln(e) = 1$

**Plato** · August 9th 2011, 07:15 AM

Originally Posted by Aluminium

I hope this image will show through nicely:

Do you see that $1=100e^{-kt}$ is the same as $e^{kt}=100~?$

**Aluminium** · August 9th 2011, 07:41 AM

I don't see that yet

**Siron** · August 9th 2011, 07:48 AM

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 $t$ ?

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