Question

Compound interest In Exercise,$3000 is invested in an account at interest rate r,compounded continuously.Find the time required for the amount to (a) double and (b) triple.

r = 0.085

Answer 1

(a) 8.15

(b) 12.92

Explanation:

Given: P = $3000, r = 0.085

`A = Pe^(rt)`

Where

A is the Amount

P is the Principal

r is the rate

t is the time

(a) For the amount to double, A = 2 × P

A = 2 × $3000

A = $6000

`6000 = 3000e^(0.085t)`

`(6000)/(3000) = e^(0.085t)`

`2 = e^(0.085t)`

Take
`log_(e)` of both sides

`log_(e)2 = log_(e)e^(0.085t)`

But
`log_(e)e = 1`

∴
`ln2 = 0.085t`

`t = (ln2)/(0.085)`

`t = (0.693)/(0.085)`

t = 8.15

(b) For the amount to double, A = 3 × P

A = 3 × $3000

A = $9000

`9000 = 3000e^(0.085t)`

`(9000)/(3000) = e^(0.085t)`

`3 = e^(0.085t)`

Take
`log_(e)` of both sides

`log_(e)3 = log_(e)e^(0.085t)`

But
`log_(e)e = 1`

∴
`ln3 = 0.085t`

`t = (ln3)/(0.085)`

`t = (1.0986)/(0.085)`

t = 12.92