# Calculate the time after which Money doubles

I while solving a problem in banking just thought to form a formula for the time period after which money deposited in bank at a compounded interest rate @$\alpha$ % p.a. .

Amount for compounded annually is :$$A=P[1+r]^t$$ Where t is time period in years and r is rate of interest. Now how to calculate it?

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$$A=P[1+r]^t$$ ? If r is positive, A > P. – oks Apr 1 '13 at 17:08
@oks Typo ! edited :) – ABC Apr 1 '13 at 17:10
Thanks, also $r$ is $\alpha$ – oks Apr 1 '13 at 17:13

## 1 Answer

$$A=P[1+r]^t$$ So to find when $P$ has doubled, solve

$$2P=P[1+r]^t \\ \Rightarrow t = \frac{\ln 2}{\ln(1 + r)}.$$

$t$ (for doubling) is often approximated by the "rule of 72" i.e. $$t \approx \frac{0.72}{r}.$$

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You need parentheses around $\ln (1+r)$. Approximating this with $r$ is reasonable. 0.72 is an approximation for $\ln 2$ which is easy to divide into. – Ross Millikan Apr 1 '13 at 17:12
rule of 72? and ln2=0.693... – ABC Apr 1 '13 at 17:13
The Rule of 72 contains two approximations: replacing ln(2) with 0.72. and replacing ln(1+r) with r. The value 72 cancels these two errors for r around 7.5% – DJohnM Apr 1 '13 at 17:40