1000 is deposited in fund A which earns interest of 12 annua

$1000 is deposited in fund A which earns interest of 12% annually. At the same time

$2000 is deposited in fund B which earns interest of 6% annually. In how many years

will the balance in fund A equal twice the balance in fund B?

Solution

Let t = the time to equal the funds

As

A = P(1 + r)^t

Then

1000(1 + 0.12)^t = 2[2000(1+0.06)^t]

1000(1.12)^t = 4000(1.06)^t

Rearranging,

(1.12/1.06)^t = 4

(1.0566)^t = 4

Taking the ln of both sides,

t ln (1.0566) = ln 4

t = ln(4)/ln(1.0566)

t = 25.178 years [ANSWER]