8
Exponential
Astonishment
Copyright © 2015, 2011, 2008 Pearson Education, Inc.
Chapter 8, Unit B, Slide 1
Unit 8B
Doubling Time and
Half-Life
Copyright © 2015, 2011, 2008 Pearson Education, Inc.
Chapter 8, Unit B, Slide 2
Doubling and Halving Times
The time required for each doubling in
exponential growth is called doubling time.
The time required for each halving in exponential
decay is called halving time.
Copyright © 2015, 2011, 2008 Pearson Education, Inc.
Chapter 8, Unit B, Slide 3
Doubling Time
After a time t, an exponentially growing quantity with
a doubling time of Tdouble increases in size by a factor
t Tdouble
of 2
. The new value of the growing quantity is
related to its initial value (at t = 0) by
New value = initial value × 2t/Tdouble
Copyright © 2015, 2011, 2008 Pearson Education, Inc.
Chapter 8, Unit B, Slide 4
Example
Compound interest (Unit 4B) produces exponential
growth because an interest-bearing account grows by
the same percentage each year. Suppose your bank
account has a doubling time of 13 years. By what factor
does your balance increase in 50 years?
Solution
The doubling time is Tdouble = 13 years, so after t = 50
years your balance increases by a factor of
2t/Tdouble = 250 yr/13 yr = 23.8462 ≈ 14.382
For example, if you start with a balance of $1000, in 50
years it will grow to $1000 × 14.382 = $14,382.
Copyright © 2015, 2011, 2008 Pearson Education, Inc.
Chapter 8, Unit B, Slide 5
Approximate Double Time Formula
(The Rule of 70)
For a quantity growing exponentially at a rate of
P% per time period, the doubling time is
approximately
70
Tdouble 
P
This approximation works best for small
growth rates and breaks down for growth rates
over about 15%.
Copyright © 2015, 2011, 2008 Pearson Education, Inc.
Chapter 8, Unit B, Slide 6
Example
World population doubled in the 40 years from 1960 to
2000. What was the average percentage growth rate during
this period? Contrast this growth rate with the 2012 growth
rate of 1.1% per year.
Solution
We answer the question by solving the approximate
doubling time formula for P. Multiplying both sides of the
formula by P and dividing both sides by Tdouble, we have
P
70
Tdouble
Copyright © 2015, 2011, 2008 Pearson Education, Inc.
Chapter 8, Unit B, Slide 7
Example (cont)
Substituting Tdouble = 40 years, we find
P
70
Tdouble
70

 1.75 / yr
40 yr
The average population growth rate between 1960
and 2000 was about P% = 1.75% per year. This is
significantly higher (by 0.65 percentage point) than
the 2012 growth rate of 1.1% per year.
Copyright © 2015, 2011, 2008 Pearson Education, Inc.
Chapter 8, Unit B, Slide 8
Exponential Decay and Half-Life
After a time t, an exponentially decaying quantity
with a half-life time of Thalf decreases in size by a
t Thalf
factor of (1 2)
. The new value of the decaying
quantity is related to its initial value (at t = 0) by
New value = initial value x (1/2)t/Thalf
Copyright © 2015, 2011, 2008 Pearson Education, Inc.
Chapter 8, Unit B, Slide 9
Example
Radioactive carbon-14 has a half-life of about
5700 years. It collects in organisms only
while they are alive. Once they are dead, it
only decays. What fraction of the carbon-14
in an animal bone still remains 1000 years
after the animal has died?
Copyright © 2015, 2011, 2008 Pearson Education, Inc.
Chapter 8, Unit B, Slide 10
Example (cont)
Solution
The half-life is Thalf = 5700 years, so the fraction of the
initial amount remaining after t = 1000 years is
1
 
2
t /Thalf
1000 yr/5700 yr
1
 
2
 0.885
For example, if the bone originally contained 1 kilogram
of carbon-14, the amount remaining after 1000 years is
approximately 0.885 kilogram.
Copyright © 2015, 2011, 2008 Pearson Education, Inc.
Chapter 8, Unit B, Slide 11
Example
Suppose that 100 pounds of Pu-239 is deposited at a
nuclear waste site. How much of it will still be present in
100,000 years?
Solution
The half-life of Pu-239 is Thalf = 24,000 years. Given an initial
amount of 100 pounds, the amount remaining after t =
100,000 years is
1
new value = initial value   
2
t /Thalf
100,000 yr/24,000 yr
1
 100 lb   
2
 5.6 lb
About 5.6 pounds of the original 100 pounds of Pu-239 will
still be present in 100,000 years.
Copyright © 2015, 2011, 2008 Pearson Education, Inc.
Chapter 8, Unit B, Slide 12
The Approximate Half-Life Formula
For a quantity decaying exponentially at a rate of P% per
time period, the half-life is approximately
Thalf
70

P
This approximation works best for small decay rates
and breaks down for decay rates over about 15%.
Copyright © 2015, 2011, 2008 Pearson Education, Inc.
Chapter 8, Unit B, Slide 13
Example
Suppose that inflation causes the value of the Russian
ruble to fall at a rate of 12% per year (relative to the
dollar). At this rate, approximately how long does it take
for the ruble to lose half its value?
Solution
We can use the approximate half-life formula because
the decay rate is less than 15%. The 12% decay rate
means we set P = 12/yr.
Thalf
70
70


 5.8 yr
P 12/ yr
The ruble loses half its value (against the dollar) in 6 yr.
Copyright © 2015, 2011, 2008 Pearson Education, Inc.
Chapter 8, Unit B, Slide 14
Exact Doubling Time and
Half-Life Formulas
For more precise work, use the exact formulas. These use the
fractional growth rate, r = P/100.
For an exponentially growing quantity with a fractional grow
rate r, the doubling time is
Tdouble 
log10 2
log10(1 + r)
For a exponentially decaying quantity, in which the fractional
decay rate r is negative (r < 0), the half-life is
Thalf  
log10 2
log10(1 + r)
Copyright © 2015, 2011, 2008 Pearson Education, Inc.
Chapter 8, Unit B, Slide 15
Example
A population of rats is growing at a rate of 80% per month.
Find the exact doubling time for this growth rate and
compare it to the doubling time found with the approximate
doubling time formula.
Solution
The growth rate of 80% per month means P = 80/mo or r =
0.8/mo. The doubling time is
Tdouble
log10 2
0.301030 0.301030



 1.18 mo
log10 1  0.8  log10 1.8  0.255273
or about 1.2 mo.
Copyright © 2015, 2011, 2008 Pearson Education, Inc.
Chapter 8, Unit B, Slide 16