Answer
$ 6.9\%$
Work Step by Step
After $t$ years, the balance, $A$, in an account with principal $P$ and annual
interest rate $r$ (in decimal form) is given by one of the following formulas:
1. For $n$ compoundings per year: $A=P(1+\displaystyle \frac{r}{n})^{nt}$
2. For continuous compounding: $A=Pe^{rt}$.
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We insert A=2P (the principal doubles), t=10
into formula (1), and solve for r:
$2P=2e^{r\cdot 10}$
$ 2=e^{10r}\qquad$... apply ln( ) to both sides...
$\ln 2=10r\qquad/\div 10$
$ r=\displaystyle \frac{\ln 2}{10}\approx$0.069314718056$\approx 6.9\%$
