3000 dollars is invested in a bank account at an interest rate of 10 percent per year,

Question

3000 dollars is invested in a bank account at an interest rate of 10 percent per year, compounded continuously. Meanwhile, 34000 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? 

The two accounts will have the same balance after    ???????  years.

Answer 1

3000(1 +0.1)^t = 34000(1 + 0.03)^t; 3· 1.1^t = 34·1.03^t; ln3 + tln1.1 = ln34 + tln1.03;

t(ln1.1 - ln1.03) = ln34 - ln3; t = (ln34 - ln3)/(ln1.1 - ln1.03).

t = 36.92 years

Answer 2

3000e^0.10t = 34000(1.03)^t

0.088235294 ≈ 1.03^t / e^0.10t

1.03 ≈ e^0.029558802

0.088235294 ≈ e^0.029558802t / e^0.10t

0.088235294 ≈ e^{-0.070441199t}

-2.427748237 ≈ -0.070441199t

34.46 ≈ t