Answer :
An account with a starting balance of P accruing interest with rate r for time t, componunded continuously, ends up with a balance A according to
[tex]A=Pe^{rt}[/tex]
Plug in everything you know and solve for r :
[tex]8624.97=6139e^{17r}[/tex]
Divide both sides by 6139:
[tex]\dfrac{8624.97}{6139}=e^{17r}[/tex]
Take the natural logarithm of both sides:
[tex]\ln\left(\dfrac{8624.97}{6139}\right)=\ln e^{17r}[/tex]
Recall that [tex]\ln x^y=y=\ln x[/tex], so that
[tex]\ln\left(\dfrac{8624.97}{6139}\right)=17r\ln e[/tex]
and ln(e) = 1, so
[tex]\ln\left(\dfrac{8624.97}{6139}\right)=17r[/tex]
Finally, divide both sides by 17:
[tex]r=\dfrac1{17}\ln\left(\dfrac{8624.97}{6139}\right)=\boxed{0.02}[/tex]
So the account has an interest rate of 2%.