Question

2000 dollars is invested in a bank account at an interest rate of 8 percent per year, compounded continuously. Meanwhile, 11000 dollars is invested in a bank account at an interest rate of 4 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

Answer:

The two accounts will have the same balance after 41.8 years

Explanation:

Hi, first, let´s intruduce the mathematical expression for the future value of each investment.

$2,000 compounded continously

$FV=PV*e^{rt}$

$FV=2,000*e^{0.08t}$

$11,000 at 4% compounded annually (equivalent to effective annual)

$FV=PV(1+r)^{t}$

$FV=11,000(1+0.04)^{t}$

Since the problem is asking when the future value of both investment will reach an equal amount of money, we solve for "t" the resulting expression:

$2,000*e^{0.08t} =11,000(1+0.04)^{t}$

$\frac{e^{0.08t}}{(1+0.04)^{t}}=\frac{11,000}{2,000}$

$Ln(\frac{e^{0.08t}}{(1+0.04)^{t}} ) =Ln(\frac{11,000}{2,000} )$

$Ln(e^{0.08t})-Ln(1.04^{t} ) =Ln(\frac{11,000}{2,000} )$

$0.08*t-t*Ln(1.04) =Ln(\frac{11,000}{2,000} )$

$t(0.08-Ln(1.04) )=Ln(\frac{11,000}{2,000} )$

$t =\frac{Ln(\frac{11,000}{2,000} )}{(0.08-Ln(1.04)} =41.804264=41.8$

So, this 2 accounts will need 41.8 years to equal their balance. You can check your result by substituting "t" in both equations, they must have the same future value.

$FV=2,000*e^{0.08*41.8}=56,683.79$

$FV=11,000(1+0.04)^{41.8}=56,683.79$

Best of luck.