A country has a food deficit of 10%. Its population grows continuously at a rate of 3% per year. Its annual food production every year is 4% more than that of the year. Assuming that the average food requirement per person remains constant, then the country will become self-sufficient in food after n years, where n is the smallest integer bigger than or equal to  $\frac{{\log }_{e}a-{\log }_{e}b}{\left({\log }_e\frac{c}{d}\right)-0.03}$ where the  $\text{a},\text{b},\text{c},\text{d}\in N$ and  $(a,b)$ are coprime,  $(c,d)$ are coprime then:

Let  $P_0$ be the initial population  $Q_0$ be its initial food production.
Let P be the population of the country in year t and Q be its food production in year t.
 $\Rightarrow \frac{dP}{dt}=\frac{3P}{100}or\frac{dP}{P}=\frac{3}{100}dt$
Integrating, we get  $\log P=\frac{3}{100}t+C$
At  $t=0,$ we have  $P=P_0$
$\Rightarrow C=\log P_0$ 
$\Rightarrow P=P_0{e}^{0.03t}$ ......(i)
It is given that the annual food production every year is 4% more than that of last year.
 $\Rightarrow Q=Q_0{(1+\frac{4}{100})}^t$
Let the average consumption per person be k units.
 $\Rightarrow Q_0=k{P}_0\left(\frac{99}{100}\right)=0.9k{P}_0$
$\therefore Q=0.9k{P}_0{\left(1.04\right)}^t$ .....(ii)
This gives quantity of food available in year t. The population in year t is,
$P=P_0{e}^{0.03t}$  [from Eq. (i)]
 Consumption in year,  $t=k{P}_0{e}^{0.03t}$ .........(iii)
The country will be self sufficient, if  $Q\underset{\_}{>}P$
$$\begin{gathered} \Rightarrow 0.9k{P}_0{\left(1.04\right)}^t\underset{\_}{>}k{P}_0{e}^{0.03t} \Rightarrow \frac{9}{10}{\left(1.04\right)}^t\underset{\_}{>}{e}^{0.03t} \\ \Rightarrow {\left(1.04\right)}^t{e}^{-0.03t}\underset{\_}{>}\frac{10}{9} \Rightarrow t\log \left(1.04\right)-0.03t\underset{\_}{>}\log \left(\frac{10}{9}\right) \\ \Rightarrow t\{10g\left(1.04\right)-0.03\}\underset{\_}{>}\log \left(\frac{10}{9}\right)\Rightarrow t\underset{\_}{>}\frac{\log 10-\log 9}{\log \left(1.04\right)-0.03} \end{gathered}$$

Thus, the least number of year in which country becomes self sufficient.

$\Rightarrow t=\frac{\log 10-\log 9}{\log \left(1.04\right)-0.03}$