In any artificial and natural growth at some rate, the quantity will double at regular intervals of time- doubling time. In any decline or decay process at a fixed rate, in the quantity will halve itself at constant intervals of time-half life.
For example: at a particular rate of interest, the amount of money will double in constant period of time.There is a very simple formula to calculate doubling time or half life.
Doubling time or half life = natural log [2] / rate of decay or growth.
simply
t = 0.6931/r
Note: Rate should be expressed in decimal numbers.
example: 5% = 5/100 = 0.05
the answer will be in the same unit as r
considering real life problems:
1. If the inflation rate is 6% per annum, when the prices will double?
t = 0.6931/0.06 = 11.5 years
Every 11.5 years, the prices will be double.
2. If a financial institution offers 14% per year as a rate of interest[compound interest] on your money, when it will double.
t = 0.6931/0.14 = 4.95 years
In nearly 5 years, it will double or every 5 years the money will double as given in the following table.
YEARS MONEY
0 1000
5 2000
10 4000
15 8000
20 16000
AND SO ON
one day you will become millionaire
3. A capacitor discharges at the rate of 15% per second. The initial current is 1 ampere or 1000 milliampere. When the current will nearly vanish?
half-life = 0.6931/0.15 = 4.6 seconds
TIME CURRENT
0 second 1000 Milli amperes
4.6 500
9.2 250
13.8 125
18.4 62.5
23 31.25
27.6 15.625
32.2 7.81
. .
. .
. .
The current becomes negligible after 30 seconds. but it will take infinity of time to become zero.
This simple and great formula solves many real life problems and amazes.
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For the serious readers,
we know , formula for growth and decay is
N = Ni*e^rt
e = 2.71.... = constant
r = rate of growth
+ve for growth
-ve for decay
Ni = initial quantity
N = quantity after time t
for doubling
Ni = 2Ni
hence
2
2 = e^rt
Taking log to the base e on both sides
ln 2 = ln e^rt
ln 2 = rt*ln e
ln 2 = rt*1
since ln e to the base e is 1
rt = ln 2
t = ln2/r
t = 0.6931/r