the amount A in an account after t years from an initial principle P invested at an annual rate r compounded?

Question

continuously is given by A=Pe^(rt) where r is expressed as a decimal. How many years will it take an initial investment of $1000 to grow to $1700 at the rate of 4.42% compounded continuiously.

Answer 1

(1.0442) to the power n(number of years)=1.7, the ratio of the final to the original amounts.
From logarithms, n log (1.0442) = log 1.7. considering the same base (let us say 10).
log 1.7= 0.2304489214
log 1.0442= 0.017867719
hence, log1.7/log1.0442= 12.2685657176, say approximately in 12 years.

Answer 2

12.26857 years if interest rate of 4.42% is annual and compounded yearly

Answer 3

1700=1000*(e^.0442t)

1.7=e^.0442t

take the natural log of each side (log base e)

ln(1.7) = ln(e^.0442t) reduces to ln(1.7) = .0442t

t ~ 12 years

Answer 4

So the equation (from the problem) is
1700 = 1000e^(1.0442*t)
1.7 = e^(1.0442*t)
ln(1.7)/1.0442) = t
t = .508 periods


Doug