Question

Estimate how long it will take for $1000 to double if invested at 6% compounded monthly

Answer 1

\[Amount(t ( years)) = 1.06^t \times 1000\]
We'll need to solve the equation:
\[2000 = 1.06^t \times 1000\]
\[1.06^t = { 2000 \over 1000 }\]
\[1.06^t={2}\]
\[^{1.06}\log ( { 2 } ) = t\]
\[{ \ln(2) \over \ln(1.06) } \approx 11.9 \approx 12 \space years\]

Answer 2

Here compound amount is S = 2000.
principle amount is P=1000
rate of interest is r=6/100
We have to find out t
The formula for S is
\[\S=P(1+r)^{t}\]
\[2000=1000(1+6/100)^{t}\]
\[2=1.06^{t}\]
log2=t log1.06
t=log2/log1.06
t=0.3010/0.0253=11.9 months = 1year (Since interest was given in months)

Answer 3

ah yeah, per month. Sorry :)