For questions 5 and 6, recall that, when interest is compounded continuously, the balance in an account after t years is given by
A = Pe^rt,
where P is the initial investment and r is the interest rate.

5. How long will it take for $2000 to double if it is invested at 6.25% interest compounded continuously?

6. What rate of interest compounded continuously is needed for an investment of $500 to grow to $900 in 10 years?

Question

For questions 5 and 6, recall that, when interest is compounded continuously, the balance in an account after t years is given by 
A = Pe^rt,
where P is the initial investment and r is the interest rate. 

5. How long will it take for $2000 to double if it is invested at 6.25% interest compounded continuously? 

6. What rate of interest compounded continuously is needed for an investment of $500 to grow to $900 in 10 years?

anonymous · Answer

problem 5, solve $e^{.065t}=2$ for t in two steps 
a) $.0625t=\ln(2)$
b) $ t=\frac{\ln(2)}{.0625}$

anonymous · Answer

second one solve $900=500e^{10r}$ for r , steps are very similar

anonymous · Answer

divide by 500 get 
$$1.8=e^{10r}$$
$$\ln(1.8)=10r$$
$$r=\frac{\ln(1.8)}{10}$$

anonymous · Answer

That all looks great but I'm so confused.. I need to find a resource to help explain it to me..Online classes are so tough!

anonymous · Answer

So would the answer be 0.05877?

anonymous · Answer

I got the first one figured out. :) Just double checking the second one