Question

Find the time required for a $1000 investment to (a) double at the interest intreats rate of 0.085, compounded continuously, and (b) triple at interest rate 0.085, compounded continuously.

Answer 1

ln(2) and ln(3) are in your results

Answer 2

time double = ln(2)/r

time triple = ln(3)/r

Answer 3

\[3P = Pe^{rt}\]
\[\frac{3P}{P} = e^{rt}\]
\[3 = e^{rt}\]
\[ln(3) = ln(e^{rt})\]
\[ln(3) = rt\]
\[\frac{ln(3)}{r} = t\]

Answer 4

same set up for the doubler

Answer 5

in case of doubling it is 8.47 years

Answer 6

for tripling it is 13.41 years