If an investor places $100,000into an account which pays 3% interest compounded continously , and $110,000 into another account paying 2.5% interest compounded continuously, after how much time will the two accounts have the same balance..
can someone go through step by step?
2007-06-12 14:37:58 · 3 answers · asked by Joan L 1 in Science & Mathematics ➔ Mathematics
The equation for continuously compounded interest is A= Pe^(rt) where P is the principal amount, A is the amount after time t, r is the interest rate in decimal. So you are solving the following equation:
100,000e^(.03t) = 110,000e^(.025t)
Divide by 100,000:
e^.03t = 1.1e^.025t
Divide by e^.025t
e^.005t = 1.1
Ln 1.1 = .005t
t = Ln 1.1 / .005
t = 19.06 years
2007-06-12 14:45:32 · answer #1 · answered by Kathleen K 7 · 0⤊ 0⤋
Amt = prin e^(it)
Since the final amount is the same for both scenarios:
100,000 e^(0.03t) = 110,000 e^(0.025t)
divide both sides by 100,000
e^(0.03t) = 1.1 e^(0.025t)
take Ln of both sides
0.03t = ln 1.1 + 0.025t
0.005t = ln 1.1
t = 19.1 years
.
2007-06-12 21:50:16 · answer #2 · answered by Robert L 7 · 0⤊ 0⤋
110k 2.5% 1 year 112.750
100k 3.0% 1 year 103.000
i dont thank you gave enough info
below is a link to figgure it out yerself
2007-06-12 21:54:06 · answer #3 · answered by Anonymous · 0⤊ 1⤋