1. ## Solving Exponential Problem

Avery invests $2000 in saving certificates at an interest rate of 12% per year compounded quarterly. How long will it take for this sum to double? A = P(1 + i)^n 4000 = 2000((1 + 0.12) / 4)^4n 2 = 0.28^4n log2 = 4nlog0.28 log2 / log0.28 = 4n -0.5445 = 4n -0.13612 = n Now a year can't be negative so something fishy is going on here. 2. Originally Posted by Devi09 Avery invests$2000 in saving certificates at an interest rate of 12% per year compounded quarterly. How long will it take for this sum to double?
A = P(1 + i)^n
4000 = 2000((1 + 0.12) / 4)^4n
It been years & years since I taught this.
But you have some elements of the formula wrong.
It should be $P\left( {1 + \frac{r}{n}} \right)^{nt}$ where r is the annual rate; n is the number of pay periods per year; and t is the number of years.

So solve this $2 = \left( {1.03} \right)^{4t}$.

3. Yeh I see now. I divided 1 + r by n instead of just the r. Now I get the number I'm suppose to be getting

4. Just remember that if a^p = x, then p = log(x) / log(a)