The function v(t)=6000(1.08)^t gives the value, V, of an investment t years after the initial investment. How many years until the value of the investment doubles?

Question

im having a real hard time since its not compound continuously, im not sure where to start.

Answer 1

if V = 6000(1.08)^t and the investment is doubled, then

12,000 = 6,000(1.08)^t

12,000/6,000 =(1.08)^t = 2

Then t(Log(1.08)) = log(2)

t = log(2)/Log(1.08) = 9 years.