Hi
I am having trouble with the following question:
1)Make the independent varaible the subject of the equation:
$\displaystyle x(t) = \frac{e^t - 1}{e^t + 1}$
P.S
Dear Paymemoney,
$\displaystyle x(t) = \frac{e^t - 1}{e^t + 1}$
$\displaystyle x(t)(e^t+1)=e^t-1$
$\displaystyle x(t)e^t-e^t+x(t)+1=0$
$\displaystyle e^t[x(t)-1]=-[x(t)+1]$
$\displaystyle e^t=\frac{-[x(t)+1]}{x(t)-1}$
$\displaystyle t=ln\left(\frac{-[x(t)+1]}{x(t)-1}\right)$
Hope this helps you.