The wording "the voltage is expected to be set" is a little peculiar but I think it is reasonably clear. You are given the formula \(\displaystyle V= 210(1- e^{-\frac{R}{LT}})\). You are told that L= 3 ohms and R= 200 micro-Hertz so the formula becomes \(\displaystyle V= 210(1- e^{-\frac{200}{3T}})\).
You are then given a list of 9 voltages and asked to find the time, T, when V is each of those. So you are asked to solve 9 equations for T:
\(\displaystyle 210(1- e^{-\frac{200}{3T}})= 110\)
\(\displaystyle 210(1- e^{-\frac{200}{3T}})= 120\)
etc.
For the first one, \(\displaystyle 210(1- e^{-\frac{200}{3T}})= 110\), an obvious first step is to divide both sides by 210:
\(\displaystyle 1- e^{-\frac{200}{3T}}= \frac{110}{210}= 0.5238\) (rounded to four decimal places). Then subtract 1 from both sides:
\(\displaystyle -e^{-\frac{200}{3T}}=0.5238- 1= -0.4762\).
Multiply both sides by -1:
\(\displaystyle e^{-\frac{200}{3T}}= 0.4762\).
Take the natural logarithm of both sides:
\(\displaystyle - \frac{200}{3T}= ln(0.4762)= -0.7419\)
and continue to solve for T.
