7000 dollars is invested in a bank account at an interest rate of 9 per cent per year, compounded continuously. Meanwhile, 20000 dollars is invested in a bank account at an interest rate of 3 percent compounded annually.

To the nearest year, when will the two accounts have the same balance?

Question

7000 dollars is invested in a bank account at an interest rate of 9 per cent per year, compounded continuously. Meanwhile, 20000 dollars is invested in a bank account at an interest rate of 3 percent compounded annually.

To the nearest year, when will the two accounts have the same balance?

anonymous · Answer

x(1+i)^n is the compund interest formula so 
7000(1+0.09)^n=20000(1+0.03)^n 
7(1.09)^n=20(1.03)^n 
1.09^n-(20(1.03)^n)/7
 I don't know where to go from here.

turingtest · Answer

continuously compounded the formula is $$Pe^{tr}=7000e^{0.9t}$$we want to know when that equals our other formula:$$P(1+{r \over n})^{nt}=2000(1+0.03)^{t}$$because n=1 once per year
Set equal and solve for t.