Answer
$\approx 147.41 \ hours$
Work Step by Step
Our aim is to compute the net change in the quantity of water that has flown out of the tank with respect to the time.
$\bf{Calculations:}$
$\text{Net change}=\int_0^t V'(x) \ dx $
or, $=\int_0^t \dfrac{15}{t+1} \ dx$
or, $=15[\ln(t+1)]_0^t$
or, $=15[\ln(t+1) -\ln 1]$
or, $=15 \ln(t+1)$
Next, we will find the time at which the net change is equal to $75$ at that time when the tank gets empty.Thus, we have
$15 \ln(t+1)=75 \implies t=e^5-1 \approx 147.41 \ hours$
