# 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 = A

A: final value

A

r: rate of change

t: time

A compounded interest means_{0}(1 + r)^{t}A: final value

A

_{0}: initial valuer: 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 1.06

Solution (Assume 1.06

^{5}= 1.338.) A

r = 0.06 /year

t = 5 years

A = 1000⋅(1 + 0.06)

= 1000⋅1.06

= 1000⋅1.338

= 1338

$1,338

_{0}= 1000r = 0.06 /year

t = 5 years

A = 1000⋅(1 + 0.06)

^{5}= 1000⋅1.06

^{5}= 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.)

A

A = 1800

r = 0.06 /year

1000⋅(1 + 0.06)

(1 + 0.06)

1.06

log 1.06

t log 1.06 = log 1.8

t⋅0.025 = 0.255

t = 0.2550.025

= 25525

= 10.xx

→ 11 years - [2]

_{0}= 1000A = 1800

r = 0.06 /year

1000⋅(1 + 0.06)

^{t}= 1800(1 + 0.06)

^{t}= 1.81.06

^{t}= 1.8log 1.06

^{t}= 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 = A

n: 12 (monthly), 365 (daily)

If the investment is compounded monthly or daily,_{0}(1 + rn)^{tn}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 increased.

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 1.005

Solution (Assume 1.005

^{60}= 1.349.) A

r = 0.06 /year

n = 12 months/year

t = 5 years

A = 1000⋅(1 + 0.0612)

= 1000⋅(1 + 0.005)

= 1000⋅1.005

= 1000⋅1.349

= 1349

$1,349

_{0}= $1000r = 0.06 /year

n = 12 months/year

t = 5 years

A = 1000⋅(1 + 0.0612)

^{5⋅12}= 1000⋅(1 + 0.005)

^{60}= 1000⋅1.005

^{60}= 1000⋅1.349

= 1349

$1,349

Close

## Compound Interest: Continuously

### Formula

A = A

A: final value

A

e: constant number (= 2.71828...)

r: rate of change

t: time

Constant e _{0}e^{rt}A: final value

A

_{0}: initial valuee: 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 e

Solution (Assume e

^{0.3}= 1.350.) A

r = 0.06 /year

t = 5 years

A = 1000⋅e

= 1000⋅e

= 1000⋅1.350

= 1350

$1,350

_{0}= $1000r = 0.06 /year

t = 5 years

A = 1000⋅e

^{0.06⋅5}= 1000⋅e

^{0.3}= 1000⋅1.350

= 1350

$1,350

Close