(a) Suppose you deposit $P$ dollars into a bank that pays an interest rate $r$ compounded continuously. How long does it take to double your original deposit $P$.

(b) Suppose you deposit $P$ dollars into a bank that compounds interest continuously. What is the interest rate $r$ that doubles your original investment $P$ after the first year.

The equation for continuous compound interest is $A = Pe^{rt}$ where $P$ = principal value, $r$ = interest rate per year, $t$ = time in years, $A$ = amount, and $e$ = the mathematical constant $e$

I've been working on this question for hours now, but I don't know if the answers I got are correct so I would appreciate some confirmation

The answer I got for a. is $t = (\ln2A/P)/r$

The answer I got for b. is $r = (\ln2A/P)/t$

  • $\begingroup$ The amount A is just 2P for case (a). You're solving Pe^(rt) = 2P so t = (ln 2)/r. $\endgroup$ – TheMathemagician Apr 21 '17 at 16:10
  • 1
    $\begingroup$ Indeed, you should start by setting $A = 2P$ to represent doubling your investment. Then $A/P$ simplifies to just $2$. You should also learn how to format your questions so that they are more readable. $\endgroup$ – badjohn Apr 21 '17 at 16:19
  • $\begingroup$ Presumably in (b) you have $t=1$ $\endgroup$ – Henry Apr 21 '17 at 16:38


a. Time to double the principal P:

The results you obtained after infinitesimal time compounding in exponential form is correct. Now after cancelling $(A,P)$ magnitudes its quotient is 2,

$$e^{rt}=\dfrac{A}{P} =2$$

or simply (you need to plug in A/P value!)

$$ rt =\ln2,\, t=\dfrac {\ln 2}{r} $$

b. Humongous interest rate to double the principal in one year itself

Now doubling the principal in one year needs an exorbitant charging (so least it deserves is representation with a capital $R$!)

$$ e^{\,1\cdot R}= \dfrac{A}{P}= 2 \rightarrow R = ln\,2 \approx 0.69315 $$ so the rate of interest would be a whopping

$$ R \approx 69.315\text{ % }.$$

  • $\begingroup$ In question a) the question is time, not interest rate, so in the last step you need t = ..., not r = ... . $\endgroup$ – oszkar Apr 21 '20 at 10:30
  • $\begingroup$ Thanks, fixed it, $\endgroup$ – Narasimham Apr 21 '20 at 10:32

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