Compound Interest
See how to find the final value/time of the investment in compound interest
(yearly, monthly/daily, continuously).
4 examples and their solutions.
Compound Interest: Yearly
Formula
A = A0(1 + r)t
A: final value
A0: initial value
r: rate of change
t: time
A compounded interest meansA: final value
A0: initial value
r: rate of change
t: time
add the principle and the interest,
calculate the next interest,
and repeat this process.
The amount of money shows exponential growth.
Example
$1,000 investment is at a rate of 6% per year, compounded yearly. Find the estimated value of the investment 5 years later.
(Assume 1.065 = 1.338.)
Solution (Assume 1.065 = 1.338.)
A0 = 1000
r = 0.06 /year
t = 5 years
A = 1000⋅(1 + 0.06)5
= 1000⋅1.065
= 1000⋅1.338
= 1338
$1,338
r = 0.06 /year
t = 5 years
A = 1000⋅(1 + 0.06)5
= 1000⋅1.065
= 1000⋅1.338
= 1338
$1,338
Close
Example
$1,000 investment is at a rate of 6% per year, compounded yearly. After how many years will the investment worth more than $1,800?
(Assume log 1.8 = 0.255, log 1.06 = 0.025.)
Solution (Assume log 1.8 = 0.255, log 1.06 = 0.025.)
A0 = 1000
A = 1800
r = 0.06 /year
1000⋅(1 + 0.06)t = 1800
(1 + 0.06)t = 1.8
1.06t = 1.8
log 1.06t = log 1.8 - [1]
t log 1.06 = log 1.8
t⋅0.025 = 0.255
t = 0.2550.025
= 25525
= 10.xx
→ 11 years - [2]
A = 1800
r = 0.06 /year
1000⋅(1 + 0.06)t = 1800
(1 + 0.06)t = 1.8
1.06t = 1.8
log 1.06t = log 1.8 - [1]
t log 1.06 = log 1.8
t⋅0.025 = 0.255
t = 0.2550.025
= 25525
= 10.xx
→ 11 years - [2]
[1]
[2]
t = 10.xx
→ After 10.xx years,
the investment worth $1,800.
→ After 11 years,
the investment worth more than $1,800.
→ After 10.xx years,
the investment worth $1,800.
→ After 11 years,
the investment worth more than $1,800.
Close
Compound Interest: Monthly, Daily
Formula
A = A0 (1 + rn)tn
n: 12 (monthly), 365 (daily)
If the investment is compounded monthly or daily,n: 12 (monthly), 365 (daily)
and if the unit of the rate r is '/year',
then use this formula.
The rate for each period is decreased.
r → r/n
But the number of the period is incrased.
t → tn
Then the final value A increases.
Example
$1,000 investment is at a rate of 6% per year, compounded monthly. Find the estimated value of the investment 5 years later.
(Assume 1.00560 = 1.349.)
Solution (Assume 1.00560 = 1.349.)
A0 = $1000
r = 0.06 /year
n = 12 months/year
t = 5 years
A = 1000⋅(1 + 0.0612)5⋅12
= 1000⋅(1 + 0.005)60
= 1000⋅1.00560
= 1000⋅1.349
= 1349
$1,349
r = 0.06 /year
n = 12 months/year
t = 5 years
A = 1000⋅(1 + 0.0612)5⋅12
= 1000⋅(1 + 0.005)60
= 1000⋅1.00560
= 1000⋅1.349
= 1349
$1,349
Close
Compound Interest: Continuously
Formula
A = A0ert
A: final value
A0: initial value
e: constant number (= 2.71828...)
r: rate of change
t: time
A: final value
A0: initial value
e: constant number (= 2.71828...)
r: rate of change
t: time
Example
$1,000 investment is at a rate of 6% per year, compounded continuously. Find the estimated value of the investment 5 years later.
(Assume e0.3 = 1.350.)
Solution (Assume e0.3 = 1.350.)
A0 = $1000
r = 0.06 /year
t = 5 years
A = 1000⋅e0.06⋅5
= 1000⋅e0.3
= 1000⋅1.350
= 1350
$1,350
r = 0.06 /year
t = 5 years
A = 1000⋅e0.06⋅5
= 1000⋅e0.3
= 1000⋅1.350
= 1350
$1,350
Close