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The sum of Rs $1000$ is compounded continuously, the nominal rate of interest being four percent per annum. In how many years will the amount be twice the original principal ? $(\log_{e}$$2=0.6931)$

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Toolbox:
  • First order,first degree DE variable separatable : Variable of a DE are rearranged to separate there (i.e.) $f_1(x)g_2(y)dx+f_2(x)g_1(y)dy=0$
  • It can be written as $\large\frac{g_1(y)}{g_2(y)}$$dy=\large\frac{f_1(x)}{f_2(x)}$$dx$
  • The solution is therefore $\int \large\frac{g_1(y)}{g_2(y)}$$dy=-\int \large\frac{f_1(x)}{f_2(x)}$$dx+c$
Step 1:
The sum of Rs1000 is the principal or initial amount.($x_0)$
Step 2:
Let $x$ be the amount at time $t$ years,the amount being compounded continuously,at an annual rate of 4% per annum.
Step 3:
$\therefore$ rate of change of $x$ is 0.04
$\large\frac{dx}{dt}$$=0.04x$
$\large\frac{dx}{x}$$=0.04dt$
$\int \large\frac{dx}{x}$$=\int 0.04 dt+\log c$
$\log x=0.04t+\log c$
$\log\large\frac{x}{c}$$=0.04 t$
$\therefore x=ce^{0.04t}$
Step 4:
When $t=0,x_0=1000$
$x=1000e^{0.04t}$
Step 5:
When the amount doubles ,$x=2000$
$\therefore 2000=1000e^{0.04t}$
$\Rightarrow e^{0.04t}=2$
$0.04t\log_e e=\log _e 2$
$0.04t=0.6931$
$t=\large\frac{0.6931}{0.04}$
$t=17.3275$years
The time taken for the amount to double =17 years approximately.
answered Sep 6, 2013 by sreemathi.v
 

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