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
[tex]FV=PV*e^{rt}[/tex]
[tex]FV=2,000*e^{0.08t}[/tex]
$11,000 at 4% compounded annually (equivalent to effective annual)
[tex]FV=PV(1+r)^{t}[/tex]
[tex]FV=11,000(1+0.04)^{t}[/tex]
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:
[tex]2,000*e^{0.08t} =11,000(1+0.04)^{t}[/tex]
[tex]\frac{e^{0.08t}}{(1+0.04)^{t}}=\frac{11,000}{2,000}[/tex]
[tex]Ln(\frac{e^{0.08t}}{(1+0.04)^{t}} ) =Ln(\frac{11,000}{2,000} )[/tex]
[tex]Ln(e^{0.08t})-Ln(1.04^{t} ) =Ln(\frac{11,000}{2,000} )[/tex]
[tex]0.08*t-t*Ln(1.04) =Ln(\frac{11,000}{2,000} )[/tex]
[tex]t(0.08-Ln(1.04) )=Ln(\frac{11,000}{2,000} )[/tex]
[tex]t =\frac{Ln(\frac{11,000}{2,000} )}{(0.08-Ln(1.04)} =41.804264=41.8[/tex]
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.
[tex]FV=2,000*e^{0.08*41.8}=56,683.79[/tex]
[tex]FV=11,000(1+0.04)^{41.8}=56,683.79[/tex]
Best of luck.