Compound Interest Problem

Kathy deposits $10,000 in her retirement account, which compounds interest continually. If she has $19,600 in the account after four years, how much money will she have in the account after another two years?

I'm not familiar with the compound interest formula.

Re: Compound Interest Problem

The compound interest formula is $\displaystyle \displaystyle \begin{align*} A = P \left( 1 + R \right) ^n \end{align*}$, where P is your principal, R is your interest rate for each time period of compounding, and n is the number of time periods that the interest compounds over.

Now attempt your problem.

Re: Compound Interest Problem

Alright, so I solved for R, which got me (sqrt(35)/5)-1 and plugging it back into the equation and setting n to 6, I got $27,440. That's right, right?

Re: Compound Interest Problem

Actually, I apologise, I have made an error. What I have given you is the compound interest formula over a discrete number of time intervals, not over a continuous period. If the interest compounds continually, then the formula is $\displaystyle \displaystyle \begin{align*} A = P \, e^{r\,t} \end{align*}$, where P is the principle, e is Euler's number, r is the annual interest rate, and t is the number of years.

Re: Compound Interest Problem

I don't see how I would solve that second equation (and get a natural number, approximately) without the use of some sort of calculator. The studying material I've been using has all natural numbers for answers and by using the discrete compound interest formula, the answer $27,440 is an option.

Re: Compound Interest Problem

You are given the amount after four years and the principle, so substitute into the equation to get the yearly interest rate. Then you can substitute the value of 6. You should find that if you keep everything in terms of exponentials and logarithms, that things will simplify, and so you won't need a calculator.

Re: Compound Interest Problem

A=P*e^rt

19600=10000e^4r

ln 1.96=ln e^4r

0.673=4r

r=0.168

change P to 19600 and calculate the new amount after 2 years