In a bank, principal increases continuously at the rate of $r\%$ per year. Find the value of $r$ if Rs. 100 double itself in 10 years$({\log _e}2 = 0.6931)$.

: Let $P$ be the principal at any time $t$.

Now, $\cfrac{{dP}}{{dt}} = \cfrac{r}{{100}}.P \Rightarrow \cfrac{{dP}}{P} = \cfrac{r}{{100}}dt$

…(1)
Integrating (1) both sides,

we get
$\int {\cfrac{{dP}}{P}} = \int {\cfrac{r}{{100}}} dt \Rightarrow \log P = \cfrac{r}{{100}}t + {C_1} \Rightarrow P = {e^{\cfrac{{\prime \prime }}{{100}}}}.{e^{{C_1}}}$

$\Rightarrow P = C{e^{\cfrac{{}}{{100}}}}$

(where ${e^{{C_1}}} = C$)
…(2)

Now, $P = 100$, when $t = 0$

Substituting the values of $P$ and $t$ in (2),

we get
$100 = C{e^0} \Rightarrow C = 100$
$\therefore$ Equation (2) becomes, $P = 100{e^{\cfrac{{}}{{100}}}}$
…(3)

When $P = 200,t = 10$

Substituting the values of $P$ and $t$ in (3),

we get
$200 = {e^{\cfrac{{10r}}{{100}}}} \times 100 \Rightarrow 2 = {e^{\cfrac{r}{{10}}}} \Rightarrow \log 2 = \cfrac{r}{{10}}$

$\Rightarrow r = 10\log 2 \Rightarrow r = 10 \times 0.6931 \Rightarrow r = 6.931$

Hence, $r = 6.93\%$ per annum