Do anyone know the annual rate and time to double equation for an investment
72/rate
you need one or the other. rate or time. what exactly is the question? oh right, rule of 72 is a good approximation.
\[Number of years 2 double = \frac{72}{Interest Rate} years\]
Initial Investment $600 Amount after 10 years 19,205.00 What is the annual % rate? What is the time to double?
here rate is equal to 41.425 %
and time to double is 1.73 years
just solve 2=(1+rate)^x
solve \[600(1+r)^{10}=19,205\] for r \[(1+r)^{10}=\frac{3841}{120}=32\](rounded) \[1+r=\sqrt[10]{32}=\sqrt{2}\]
that is a hell of a rate of interest
\[r=\sqrt{2}-1=.4142\] as sheg said. but rounded it is \[.412421\] not \[.412425\]
time to double solve \[1.41421^x=2\] \[x=\frac{\ln(2)}{\ln(1.4121)}\]
I think sheg's interest rate assumes continuous compound interest and satellites is compounded yearly. don't understand sheg's doubling time. It should be very close to 2 years.
@PHI dear friend in case of compound interest u earn interest on principal as well as interest earned formula for compounding is \[Amount = Principal \times (1 + r )^n\] now do this problem Principal,P = $ 1000 rate, r = 41.425 % Number of years, n = 2 years and see how much amount u have after 2 years
actually if u will make aproximation then it will double in 1.73 yrs because i was using the rule of 72
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