How long will it take for $1,000 to be worth $3,000 at 7% interest, compounded continuously?
part 1: How long will it take for $1,200 to double if it is invested at 8% annual interest compounded 4 times a year? Enter in exact calculations or round to 3 decimal places.
part 2: how long will it take if the interest is compounded continuously?
thank you!
Bullwinkle
interest compounded n times per annum:
A = P * [1 + (r/n)]^(nt)
interest compounded continuously:
A = P * e^(rt)
what more do you need ??
You're welcome !
Ω
llaffer
The equation you need is:
A = Pe^(rt)
where P is the principal amount, r is the interest rate, t is the number of years, and A is the amount in "t" years.
Plug in what we know and solve for t:
3000 = 1000e^(0.07t)
3 = e^(0.07t)
Natural Log of both sides:
ln(3) = 0.07t
ln(3) / 0.07 = t
Using a calculator:
1.09861 / 0.07 = t
t = 15.6944
It will take 15.6944 years
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For the part 1 of the other question, you need a different equation:
A = P(1 + r/n)^(nt)
Part 1 and 2 asks to find out how long it takes to double at 8% interest when compounded quarterly and continuously. Part 1 first:
2400 = 1200(1 + 0.08/4)^(4t)
2 = (1 + 0.08/4)^(4t)
2 = (1 + 0.02)^(4t)
2 = (1.02)^(4t)
Log of base 1.02 on each side:
log-1.02(2) = 4t
t = log-1.02(2) / 4
t = log(2) / (4 log(1.02))
using a calculator and rounding to 3 decimal places:
8.751 years
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Part 2, back to the continuously compounded equation from above:
A = Pe^(rt)
2400 = 1200e^(0.08t)
2 = e^(0.08t)
ln(2) = 0.08t
t = ln(2) / 0.08
t = 8.664
8.664 years (slightly less time than if compounded quarterly)