|
Question 996347: Alice invests $5000 at Bob's bank and $8000 at Charlie's bank. Bob compounds interest continuously at a nominal rate of 10%. Charlie compounds interest continuously at a nominal rate of 7%.
A). In how many years will the two investments be worth the same amount?
B). When both investments are worth the same amount, how much will each be worth?
My work so far: 8000e^(.07t)=5000e^(.1t)
Thats all I've been able to figure out. I just do not know where to go from there.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Alice invests $5000 at Bob's bank and $8000 at Charlie's bank. Bob compounds interest continuously at a nominal rate of 10%. Charlie compounds interest continuously at a nominal rate of 7%.
A). In how many years will the two investments be worth the same amount?
Bob's interest:: 5000*e^(0.10t)
Charlie's interest:: 8000*e^(0.07t)
Solve:: 5000e^(0.10t) = 8000e^(0.07t)
e^(0.03t) = 8/5
0.03t = ln(8/5)
-----
t = (100/3)*0.47
t = 47/3
t = 12 1/3 years
=======================
B). When both investments are worth the same amount,
how much will each be worth?
5000e^(0.1*(47/3)) = $21293.21
-----
Cheers,
Stan H.
---------
|
|
|
| |
