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This question already has an answer here:

How long does it take an investment to double if continuously compounded at a 6% rate? Is this right?

\begin{align*} e^{0.06 \cdot t} & = 2\\ 0.06t & = \ln(2)\\ t & = \frac{\ln 2}{0.06}\\ & \approx 11.55 \end{align*}

Why does $e$ equal the $\lim_{n \to \infty} (1+\frac{1}{m})^m$?

I know that if $r$ is the rate and $n$ is the number of compounding periods in $t$ years, then the value of an investment is:

$$A_0\left(1+\frac{r}{n}\right)^{nt}$$

But how do we get from here to the definition with $e$ in continuous compounding?

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marked as duplicate by user296602, N. F. Taussig, Misha Lavrov, Saad, Shailesh Feb 19 '18 at 0:33

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ The question of why $e = \lim_{n \to \infty} (1 + 1/n)^n$ has been asked here a very large number of times. $\endgroup$ – user296602 Feb 12 '18 at 1:31
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$lim_{n\rightarrow \infty}(1+\frac{r}{n})^n = e^r$ is proved in calculus. If you don't remember this, you can prove it easily by proving that the logarithm goes to $r$. As to your second question, $(1+\frac{r}{n})^{nt}=((1+\frac{r}{n})^n)^t,$ by the laws of exponents.

As to you calculations, they're correct.

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