Why the rate is part of the exponent on continuous compounding formula?

Question

Could someone explain how the rate is part of the exponent on the continuous compounding formula

$A= p e^{rt}$

while that is not the case in the simple compound interest formula?

$A= p (1+\frac{r}{n})^{ nt}$

Thank you.

Answer 1

Upon taking the logarithm in your 2nd equation yields $\ln(A) = \ln(p) + n t \ln(1+r/n)$. In the limit of $n \rightarrow \infty$ then $\ln(1+r/n)$ tends to $r/n$. Hence $\ln(A) \rightarrow \ln(p) + r t$ which is your 1st equation.