More log questions

**Paymemoney** · March 31st 2010, 04:46 PM

Hi
I am having trouble with the following question:
1)Make the independent varaible the subject of the equation:
$x(t) = \frac{e^t - 1}{e^t + 1}$

P.S

**Sudharaka** · March 31st 2010, 04:58 PM

Originally Posted by Paymemoney

Hi
I am having trouble with the following question:
1)Make the independent varaible the subject of the equation:
$x(t) = \frac{e^t - 1}{e^t + 1}$

P.S

Dear Paymemoney,

$x(t) = \frac{e^t - 1}{e^t + 1}$

$x(t)(e^t+1)=e^t-1$

$x(t)e^t-e^t+x(t)+1=0$

$e^t[x(t)-1]=-[x(t)+1]$

$e^t=\frac{-[x(t)+1]}{x(t)-1}$

$t=ln\left(\frac{-[x(t)+1]}{x(t)-1}\right)$

Hope this helps you.

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