Question

In a bank, principal increases continuously at the rate of $5\%$ per year. In how many years $\text{Rs}1000$ double itself?

Answer 1

It is given that
Rate of change of principal = $5\%$ of Principal

$\frac{dP}{dt}=5\%\times P$

(Where $P$ is Principal at any time $t$)

$\frac{dP}{dt}=\frac{5}{100}\times P$

$\frac{dP}{dt}=\frac{P}{20}$

$\frac{dp}{p}=\frac{dt}{20}$

Integrating both sides
We get,

$\int \frac{dp}{p}=\int \frac{dt}{20}$

$\log |P|=\frac{t}{2}+c$

$P=e^{\frac{t}{20}+c}$

$P=e^{\frac{t}{20}}\times e^c$

Putting $e^c=r$
$P=r.e^{\frac{t}{20}}$ (i)

Now, we need to find the time $\left(t\right)$ when principal is double $\text{Rs}2000$

$$\begin{array}{ccc}\text{Time}& t=0& t=?\\ \text{Principal}& 1000& 2\times 1000=2000\end{array}$$

At $t=0,P=1000$

Putting in $\left(i\right)$ then we get,

$1000=r.e^{\frac{0}{20}}$

$1000=r.1$

putting $r=1000$ in (i) then we get,

$P=1000.e^{\frac{t}{20}}$

Now,
we need to find time $t$ when Principal is
$\text{Rs}2000i.e.,$
$P=2000,t=t$

$\frac{2000}{1000}=e^{\frac{t}{20}}$

$e^{\frac{t}{20}}=2$

Integrating both sides
We get,

${\log }_e{e}^{\frac{t}{20}}={\log }_{e}2$

$\frac{t}{2}{\log }_{e}e={\log }_{e}2$

$(\log x^n=n\log x)$

$\frac{t}{20}\times 1={\log }_{e}2(\log e=1)$

$t=20{\log }_{e}2$

Final Answer:
Hence, the required time is

$t=20{\log }_{e}2$