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Suppose that I have 100 dollars and this amount has increased up to 150 within 5 years, in order to get the growth rate we solve the following equation $\ln(150/100)×100/5= 0.081=8.1$%

The question is what if this amount has increased within seconds or minutes or days how do I calculate it then?!

I guess we must divide $5$ to convert it to minutes or seconds or days, is that right?!

K. M.
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  • This is compound continuous interest? – fleablood Jul 03 '18 at 19:41
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    @Fleablood, Yes it is. – K. M. Jul 03 '18 at 19:43
  • The number of years need not be an integer greater than $1$. It could be a real number less than $1$. But you should express it as years. So if for $5$ minutes. There are $6024365$ minutes in a year so $5 minute = \frac 5{6024365}$. So instead of dividing by $5 years$ divide by $\frac {5}{6024365} years$. (which is the same thing as multiplying by $\frac {6024365}5$. – fleablood Jul 03 '18 at 19:52
  • @fleablood, please can you re-write this in an answer – K. M. Jul 03 '18 at 19:54
  • I already did write it as an answer. – fleablood Jul 03 '18 at 19:56
  • Sorry, I mean the comment you said an example with the minutes please – K. M. Jul 03 '18 at 19:59

2 Answers2

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Do it the same way. Let $a$ be the number of years. $a$ can be any positive real number and it can be less than $1$. So do $\ln \frac {\text{end ammount}}{\text{initial amount}}*\frac 1a \times 100\%$

So if $a = 57$ days $=\frac {57}{365}years$ then and the end amount is $\$102$ then the rate is $\frac{\ln \frac {102}{100}\times 100\%}{\frac {57}{365}} = 12.7\%$

fleablood
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I'd like to write Fleablood's comment here+ thanks for him

The number of years need not be an integer greater than $1$. It could be a real number less than $1$. But you should express it as years. So if for $5$ minutes. There are $60*24*365$ minutes in a year so $5 minute = \frac 5{60*24*365}$. So instead of dividing by $5 years$ divide by $\frac {5}{60*24*365} years$. (which is the same thing as multiplying by $\frac {60*24*365}5$.

K. M.
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