Inverse function of $f(t)=5 +\frac{75}{1 + e^{-((t-50)/10)}}$

Question

i need to find the inverse function of $$ v= f(t)=5 + \frac{75}{1 + e^{-\frac{t-50}{10}}} $$ so far i have $$ v - 5 = \frac{75}{1 + e^{-\frac{t-50}{10}}} $$ $$ (v-5) \left(1 + e^{-\frac{t-50}{10}}\right) = 75 $$ $$ 1 + e^{-\frac{t-50}{10}} =\frac{75}{v-5} $$ $$ e^{-\frac{t-50}{10}} =\frac{75}{v-5}-1 $$ and I'm stuck i need to find t in terms of v. Is this right or have i made a mistake.

Answer 1

$e^{-(t-50)/10}=\frac{75}{v-5}-1=\frac{80-v}{v-5}$ then taking logs we get \begin{align} -\frac{t-5}{10}&=\ln\left(\frac{80-v}{v-5}\right)\\ t-5&=-10\ln\left(\frac{80-v}{v-5}\right)\\ t&=5-10\ln\left(\frac{80-v}{v-5}\right).\\ \end{align}