SOLUTION: suppose that $15000 is invested in 2010 in an educational savings account. The investment is at an interest rate k compounded continuously. After 18 years the investment has grown

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Question 429064: suppose that $15000 is invested in 2010 in an educational savings account. The investment is at
an interest rate k compounded continuously. After 18 years the investment has grown to $55000.
What is the interest rate k ?
*Could you please help me solve this, I would also like to know you got to the solution for future math problems like this.
Found 2 solutions by Gogonati, ewatrrr:
Answer by Gogonati(855) About Me  (Show Source):
You can put this solution on YOUR website!
Solution:Since the investment was compounding continuously we can write:
, Divide both sides by 15000
11%2F3=e%5E%2818%2Ak%29Taking the ln of both sides we have:
ln%2811%2F3%29=18k, since lne=1
k=ln%2811%2F3%29%2F%2818%29
k=0.072
Answer: The interest rate is 7.2%






Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!


Hi

Note:  A = Pe^(r*t) Compounding continously

Question states***

 55,000 = 15,000*e^18r

 55,000/15000 = e^18r

 ln(55,000/15000) = 18r

   1.299 = 18r

   1.299/18 = r

     .07218 = r  OR 7.218 %