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# In a bank,principal increases continuously at the rate of r% per year.Find the value of r if Rs100 doubles itself in 10years. $\quad (log_e2=0.6931)$

Toolbox:
• If the principle increases continuously at the rate of r% per year, then $\large\frac{dp}{dt}=(\large\frac{r}{100})$$p Step 1: From the information given,let p be the principle, t be the time and r be the rate of interest respectively Using the information in the tool box \large\frac{dp}{dt}= (\large\frac{r}{100})$$p$
Now seperating the variables
we get $\large\frac{dp}{p}=(\large\frac{r}{100})$$dt Step 2: Integrating on both sides we get \int\large\frac{dp}{p}= (\large\frac{r}{100})\int dt \log p=\large\frac{rt}{100}$$+k$
Or $p=e^{(\Large\frac{rt}{100})+k}$
Step 3:
But it is given when $t=0,p=100$.Hence $100= e^k$
If $t=10$, then $p= 2(100) = 200$
Therefore $200= e^{[(\large\frac{r}{10})+k]}$
Or $e^{(\large\frac{r}{10}) }. e^k$
but $e^k = 100$.
Hence $200 = e^{(\Large\frac{r}{10})}$$. 100 e^{(\large\frac{r}{10})} = 2 \large\frac{r}{10}$$ =\log_e 2$