Question

Alex invests $1000 at an interest rate of 4.5% compounded monthly. When will the value of Alex's investment reach $3000?

Answer 1

This is a compound interest question. \[A = P \left ( 1 + {r \over n} \right)^{nt}\]where r is the decimal form of the interest rate, n is the number of times it is compounding in a year, t is the number of years, P is the principle amount, and A is the balance after t years.

Here A = 3000, P = 1000, r = 0.045, n = 12, t = ?

We can solve for t.

Answer 2

Thank you, that's what I was looking for. I have all of the work it's just plugging it into the calculator where I mess up. I don't know if I'm not adding enough parenthesis or where to put them exactly.

Answer 3

Yes, t is supposed to evaluated by ln(3) / (12ln)(1 +.45/12). I'm just having trouble putting it into my graphing calculator to get the number of years.

Answer 4

ok... then I'll erase my post then.. thanks for correcting me :)

Answer 5

@Kreshnik Yes. That is for simple interest, where interest is calculated every period based on the initial investment value. In a compounded interest case, the interest earned from the previous compounding period is added to the principle for the next compound period's interest amount. Simple interest is linear, while compounded interest is exponential, as we can observe from our equations.

@JTLA223 You should have obtained the following solution when solving for t\[t = {\log({A \over P}) \over \log \left ( 1 + {r \over n} \right)}\]This can be typed into a calculator as log(A/P)/log(1 +(r/n))