While there is solution being added to the tank at a rate of $4$ liters per minute, the tank is also being drained at a rate of $4$ liters per minute, so the volume stays constant at $100$ liters, throughout the process.
Say volume of water at $t = 0$ is $V_0$.
Volume of water at time $t = 4n$ where $n \in 0, \mathbb{Z+}$ is then given by
$\displaystyle V_{4n} = \bigg(\frac{3}{5}\bigg)^n V_0 + 10I \bigg(1 - \bigg(\frac{3}{5}\bigg)^n \bigg)$
And volume of water at time $4n+t$ is given by,
$\displaystyle V_{4n+t} = V_{4n} \bigg(1 - \frac{t}{10}\bigg) + I \, t$ where $0 \leq t \leq 4$
Here is how the expression comes (wherever I write $t$, it is $0 \leq t \leq 4$) -
$V_t = V_0 - \frac{V_0}{10}t + I \, t$
$V_4 = V_0 - \frac{4V_0}{10} + 4I = \frac{3}{5}V_0 + 4I$
$\displaystyle V_{4+t} = V_4 - \frac{V_4}{10}t + I \, t$
i.e. $\displaystyle V_{4+t} = \frac{3}{5}V_0 + 4I - \frac{t}{10}(\frac{3}{5}V_0 + 4I) + I \, t$
i.e. $\displaystyle V_{4+t} = \frac{3}{5}V_0 (1 - \frac{t}{10})+ (4 + \frac{3t}{5})I$
$\therefore \displaystyle V_{8} = (\frac{3}{5})^2 V_0+ 4(1 + \frac{3}{5})I$
Similarly, $\displaystyle V_{12} = (\frac{3}{5})^3 V_0 + 4(1 + \frac{3}{5} + (\frac{3}{5})^2)I$
At $t = 4n$, the geometric series $\, 1 + \frac{3}{5} + ... + (\frac{3}{5})^{n-1} = \displaystyle \frac{1 - (\frac{3}{5})^{n}}{1 - \frac{3}{5}} = \frac{5}{2} (1 - (\frac{3}{5})^{n})$
And this leads to expression at time $4n$ and $4n + t$.
Best Answer
You forgot the integration constant when solving the DE. The general solution to $\frac{dV}{dt}=-kV$ is $$V(t)=e^{-kt+c}=e^ce^{-kt}$$ where the constant $e^c=V_0$. You then have the equation $$\frac{1}{2}V_0=V_0 e^{-7/V_0}$$
Let me know if the path from here isn't clear, and I'll elaborate.