Interest Rates

Question

8000 dollars is invested in a bank account at an interest rate of 8 per cent per year, compounded continuously. Meanwhile, 18000 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?

Answer 1

 

 8000*exp(0.08t)  = 18000*(1.03)^t

 

exp(0.08t) = 18000(1.03)^t / 8000 <--- divides both sides by 8000

 

 0.08t = ln [18000(1.03)^t/8000]  <--- takes the log of both sides

 

 

 0 = ln [ 18000(1.03)^t/ 8000] - 0.08t = f(t)

 

The zero of this function must be found numerically

By examining the graph or using the bisection method or Newton's method

to approximate the zero.

 

the approximate zero is 16.077 , as I examined the graph on desmos.com

 

At 16 years:  8000*exp(16.077*0.08) =  28950.91

                   18000((1.03)^16.077 = 28950.53

 

 

My apologies, but I do not have the time to chase the 38 cents

round off error.