Logarithms and Population Growth
Population growth problems resemble interest rate problems. The basic equation for solving population growth problems is :
n(t) = n(0) * e ^ (rt) where
n(0) is the initial population
n(t) is the population at some later time t
t is the length of time
r is the population growth rate, similar to the interest rate used in interest rate problems
By way of reminder the interest rate equation for continuously compounded interest rate was A = Pe^rt. Look familiar?
Logarithms are used to compute population growth rates because populations normally increase rapidly until environmental conditions, such as available space, food, and water quickly start to retard growth. Populations do not grow in a linear fashion. If has been found from studies of actual population growth rates that logarithms can be used to model, predict, and understand how populations grow.
This section will present simple problems for solving common questions about using logarithms to compute aspects of population growth.
Typical problems are:
- Compute time to double a population
- What is the rate of growth
- Set up the equation to model population growth rate
- What will the population be at some time in the future
- What was the original population
Regardless of the specific questions there are several things to do with population growth questions:
- Start with the equation to compute population growth which is n(t) = n(0) * e ^(rt)
- The basic equation has four variables. Enumerate known values from the problem that is to be solved. Identify the unknown variables. Determine a method of isolating the unknown values on one side of the equation.
- Compute time to double a population. This calculation has the advantage of rapidly eliminating variables from the equation making it easier to solve problems.
There are other ways to model population growth rates besides what is presented in this section. The examples and solutions are relatively simple compared to additional methods and uses of logarithms to solve population problems.
Example: World Population
World population in 2000 was 6.1 billion. Assuming growth rates of 1% and 1.4% per year compute the projected population in 2050
To answer this question start with the standard expression for computing population growth:
n(t) = n(0) * e^(rt) where
n(t) is the population in 2050
n(0) is the population at time t = 0 (or 2000) in this case
relative population growth rate r is either 1% (0.01) or 1.4% (0.014)
t is the number of years between the date of n(0) and n(t) which in this case is the difference between the yeat 2000 and 2050, or 50
Filling in the variables, to answer the first question, the expression is
n(2050) = 6.1 * e ^(0.01 * 50)
Computing this on a calculator or math program such as R the result is 9.9. This means if the world population increases at one percent a year then in 50 years the population will increase from 6.1 to 9.9 billion.
Assuming a 1.4% growth rate the expression is
n(2050) = 6.1 * e^(0.014 * 50)
Computing this on a calculator or math program such as R gives a result of 12.3 billion.
Example: Bacteria Culture
If the initial population of a bacteria culture is 10 and the population doubles every 1.5 hours then write an expression that can be used to compute the population for any value of time t.
This problem can be solved by computing the relative rate of growth using an expression with time to double the population.
Starting with the general expression for computing population growth:
n(t) = n(0) * e ^ (rt) where
n(t) = 2 * n(0) = 20 because the objective is to compute an expression using time to double
n(0) is 10
rate r is not known
time t is 1.5 hours
Substituting in the known values:
20 = 10 * e ^ (r * 1.5)
Divide both sides by 10
2 = e ^ (r * 1.5)
Take the natural log of both sides. This makes e go away and the exponent comes down
ln(2) = ln e ^ (r * 1.5) natural log of e^x is x
ln(2) = r * 1.5
r = ln(2) / 1.5 Note that growth rate is the natural log of two divided by time = ln(2)/t which is the same expression for rate used with radioactive decay and interest rates.
Now that a growth rate is computed a general expression for computing population at any time can be written:
n(t) = 10 * e ^((ln(2)/1.5) * t) 10 is the initial population and 1.5 is the time to double, measured in hours. So units for this expression are population and time in hours.
Next, compute the population after 35 hours. This requires using the known values in the standard expression for computing population growth and solving for n(t)
n(35) = n(0) * e ^(rt) where
n(0) is 10
rate r is ln(2)/1.5
time t is 35 hours
n(35) is population at 35 hours and is not known. The objective is to solve for n(35)
Substituting in the known values the expression is
n(35) = 10 * e ^ ((ln(2)/1.5) * 1.5)
Using a calculator the answer is about 106 million
How long does it take for the population to reach 10,000.
In this case the expression with values filled in is
10,000 = 10 * e^( (ln(2)/1.5) * t)
All the values are known except for time t and the objective is to solve for t.
The first step is to divide both sides by 10
1,000 = e^( (ln(2)/1.5) * t)
Taking the natural log of both sides makes e go away and the exponent comes down
ln(1000) = ln (e^( (ln(2)/1.5) * t))
ln(1000) = ln(2)/1.5 * t
To isolate time t multiply both sides by 1.5/ln(2)
(1.5 * ln(1000)) / ln(2)
Using a calculator or math program such as R gives an answer of 14.9 hours which seems reasonable given 35 hours to reach 106 million.
Example: Bacteria Population Growth
Assume a culture with an initial population of 25 doubles in five hours. Compute the relative rate of increase.
This computation can be done by resolving an equation representing time to double. Starting with the standard expression for population growth:
N(t) = n(0) * e ^(rt)
n(0) is given as 25
In this case, n(t) = 2x n(0)
time t is five
Substituting in these known values the expression now is
50 = 25 * e ^ (r * 5)
Divide both sides by n(0), which is 25, to simplify the expression
2 = e ^ (r * 5)
take the natural log of both sides to eliminate e and bring down the exponent:
ln(2) = ln(e^(r * 5))
ln(2) = r * 5
r = ln(2)/5
Next, write a general expression for growth of this population given the initial population and growth rate:
n(t) = 25 * e ^ ((ln(2)/5) * t
Compute time to reach population of one million
The only change from the general expression is to substitute in 1,000,000 for n(t) and then solve for time t.
1000000 = 25e^(rt) where rate r = ln(2)/5
Dividing both sides by 25 makes the 25 go away from the right side of the equation
It is a little easier to leave rate r as a variable symbol rather than use ln(2)/5. When it is needed the value can be put in.
40000 = e^rt
Take the natural log of both sides. This will make the exponent and e go away
ln(40000) = ln(e^rt) = r*t
dividing both sides by r isolates t and the objective is to solve for t
ln(40000)/r = t Now is a good time to substitute in the value for r which is ln(2)/5
ln(40000) / ln(2)/5 = t
Multiply the top and bottom of the left side by one, in the form of 5/ln(2) / 5/ln(2) and the fraction in the denominator goes away and the expression is simplified
ln(40000) * 5 / ln(2) = t
Using a calculator or math program such as R the result is about 76.4 hours.
This means that at the present relative rate of growth that the culture will grow from a population of 25 to one million in 76.4 hours.
Example: Bacteria growth rate
Initial bacterium count is 500 and the relative growth rate is 40 percent per hour.
Write an equation that models population after t number of hours.
The basic equation for population growth is
n(t) = n(0) * e^(rt)
Substituting in the initial population value n(0) and relative growth rate r of 40% gives:
n(t) = 500 * e ^(0.40 * t)
Compute population after ten hours.
This is done using the preceding equation, substituting in 10 for t
n(t) = 500 * e ^(0.40 * 10)
Using a calculator the population after ten hours is about 27,300
Compute when the bacteria count will reach 80,000
In this case n(t) is known and time t is the unknown and the objective is to solve for time t. The standard equation is:
n(t) = n(0) * e ^(rt)
Substituting in the known values the expression is:
80000 = 500 * e ^(0.40 * t).
The objective is to solve for t.
The equation should be first simplified by dividing by 500. This gives
160 = e ^ (0.40 * t)
Take the natural log of both sides to make e go away and bring down the exponents. The result is:
ln(160) = ln(e^(0.40 * t))
ln(160) = 0.40t
t = ln(160)/.40
On a calculator this is about 12.7 hours until the count reaches 80,000
Compute time to double the population.
In this case n(t) is two times n(0). The equation looks like this:
2*n(0) = n(0) * e^(rt)
The value of r is known and the objective is solve for t. When both sides are divided by n(0) that value goes away and 2 remains on the left side of the expression.
2 = e^(rt)
Taking the natural log of both sides takes e out of the equation and brings down the exponent
ln(2) = ln(e^(rt))
ln(2) = rt
Dividing both sides by rate r isolates the time variable
ln(2)/r = t
Rate r was previously given as 0.4
ln(2)/0.4 = t
On a calculator this is approximately 1.73 meaning it will take about 1.73 hours for the population to double given a relative growth rate of 40 percent, or 0.4, per hour.
Example: Fox Population
Given a relative annual growth rate of eight percent and a population in 2005 of 18,000.
Write an expression to model the population at any given number of years following 2005
The basic expression is
n(t) = n(0) * e^(rt)
n(0) is 18000
r is eight percent per year
With known values the expression now is
n(t) = 18000 * e ^(.08*t)
Estimate the population in 2013
time t is eight because 2013 is eight years after 2005
n(0) is 18000. Rate r was previously computed to be 0.08. n(0) is the population in 2005 and is given as 18000
n(2013) = 18000 * e^(0.08 * 8)
On a calculator or computer math program such as R the result is approximately 34,137
Example: Population of a Country
Given a relative growth rate of three percent a year and a population in 1995 of 110 million.
Write the expression that models population growth and compute population in 2020 assuming a two and then three percent growth rate.
Start with the standard expression for computing population growth.
n(t) = n(0) * e^(rt)
In this case n(0) is 1995 population of 110 million
rate r is 2% and 3%, or 0.02 and 0.03.
time t is the number of years from 1995 which is 25.
The model is
n(t) = 110 * e^(rt) where r is either 0.02 or 0.03 and t us the number of years since 1995
For a 2% growth rate the expression is
n(2020) = 110 * e ^(0.02 * 25) which on a calculator is just over 181 million
For a 3% relative growth rate the expression is
n(2020) = 110 * e ^(0.03 * 25) which on a calculator is almost 233 million.
Compute time to double the population for 2% and 3% relative growth rates
For 2% relative growth the expression is
2(110) = 110 * e^(rt) and the objective is to isolate and solve for time t. It is easier to work the equation if rate is substituted in when it is needed.
Divide both sides by 110 giving
2 = e^(rt)
Take the natural log of both sides to make e go away and remove the exponent. The result is
ln(2) = ln(e^*rt))
ln(2) = rt
Divide both sides by rate r and this isolates time t
ln(2)/r = t
If rate r is 2% the expression is
ln(2)/0.02 = t
t ~= 34.7 years
If rate r is 3% the expression is:
ln(2)/0.03 = t
t~= 23.1 years
This means with a two percent annual growth rate population doubles in 34.7 years, and doubles in 23.1 years if the growth rate is three percent per year.
Example: Population of a City
The population is 112,000 in 2006. Time to double is 18 years.
Write an exponential expression n(t) = n(0) e^(rt) that models this growth rate.
The standard expression for population growth rate is
n(t) = n(0) * e^(rt) The objective is to solve for rate r and then plug in the known value for n(0). Time to double is known to be 18 years. A model using double the population would use n(t) = 2*n(0) so the expression fould be
2*n(0) = n(0) * e^(rt) where time t is 18 years. Substituting in time and dividing both sides by n(0) gives
2 = e^(18r)
Take the natural log of both sides to eliminate both e and the exponent
ln(2) = ln(e^(18r))
ln(2) = 18r
ln(2)/18 = r
This means the annual rate of growth is ln(2)/18 which is approximately 0.0385
All the information needed to write a model is now available. The expression is
n(t) = 112,000 * e ((ln(2)/18) * t)
This expression can be used to compute the population for any number of years after 2006.
In how many years will the population be 500,000?
The expression for population is
n(t) = n(0) * e^(rt)
Final population n(t) is 500,000
Initial population n(0) is 112,000
Relative growth rate r is ln(2)/18
Using these values the expression is
500,000 = 112,000* e((ln(2)/18) * t) and the objective is to solve for t.
First, divide both sides by 112,000
500,000/112,000 = e^((ln(2)/18)*t)
The next step is to take the natural log of both sides in order to make e disappear and to bring down the exponent.
ln(500000/112000) = ln(2^((ln(2)/18) *t)
ln(500000/112000) = ln(2)/18*t multiplying both sides by 18/ln(2) isolates t
t = 18* ln(500000/112000)/ln(2)
Solving the expression on a calculator or using a math program such as R gives:
t ~= 38.9 years.
This means that 38.9 years after 2006 the population will be half a million.
Example: Bat Population
In 2009 the bat population in a French cave was 350,000. Time to double is 25 years.
Write an equation that models population growth.
The standard expression for population growth is
n(t) = n(0) * e^(rt)
To write an expression that models this population growth the relative growth rate must be computed. Using time to double in the expression can be used to compute relative growth rate which is needed in the expression
Time to double is 25 years so the expression is
2*n(0) = n(0) * e^(rt) Time t is known and the objective is to isolate rate r and then solve for it
Divide both sides by n(0). The result is
2 = e^(rt)
Take the natural log of both sides to make the exponent and e go away
ln(2) = ln(e^(rt))
ln(2) = rt
Divide both sides by time t to isolate rate r
ln(2)/t = r
Time t is known to be 25 years so the expression is
ln(2)/25 = r
Using a calculator or math program to solve this shows that r is approximately equal to 0.028. Leaving it in the form of ln(2)/25 is preferable because no precision is lost. 0.028 is a rounded number.
Knowing rate r allows writing a full expression that models population growth if time to double is 25 years and the initial population is 350,000
n(t) = 350000 * e^((ln(2)/25) * t)
With this information the time to reach any future population level can be computed. For example, the time to reach a population of 2 million is computed as follows:
n(t) = n(0) * e^(rt)
Substituting in the known values gives
2,000,000 = 350,000 * e ^((ln(2)/25) *t)
The next step is to divide both sides by 350000
2,000,000/350,000 = e * ^((ln(2)/25)*t)
Take the natural log of both sides to eliminate e and the exponent and the result is:
ln(2000000/350000) = ln(2)/25 * t
Multiply both sides by 25/ln(2) and t is now isolated
25 * ln(2000000/350000) / ln(2) = t
On a calculator this is about 62.9 years. This means it will take about 62.9 years for the population to grow fro 350,000 to 2,000,000.
Example: Deer Population
If initial population is 20,000 in 2003 and 31,000 in 2007 then write the expression for this population growth.
First, start with the standard expression for population growth:
n(t) = n(0) * e^(rt)
The objective is to compute rate r. That will give enough information to write an equation that models this population growth.
Known values are:
n(t) is 31,000
n(0) is the starting population which is 20,000
time t is 4 because the difference between 2003 and 2007 is four
Using these values the expression now is
31,000 = 20,000 * e^(r*4)
The objective is to solve for rate r. The first step is to divide both sides by 20,000
31000/20000 = e^(r4)
The next step is to take the natural log of both sides, which will eliminate the exponent and e
ln(31000/20000) = ln(e^(r4))
ln(31000/20000) = r4
It is possible and reasonable to simplify 31000/20000 to 31/20. It is not required because the natural log of 31000/20000 is equal to the natural log of 31/20.
If both sides are divided by 4 then rate r is isolated as
ln(31/20)/4 = r
On a calculator this is approximately 0.0196. This means the general expression for modeling this population growth is
n(t) = 20000 * e^(ln(31/20)/4*t)
Compute the population in 2011
This would mean that time t is 2011 - 2003 = 8
The starting expression is
n(t) = n(0) * e^(rt)
The initial population n(0), rate r, and time t are known and the objective is solve for time t. Substituting in the known values the expression is
n(2011) = 20000 * e^(ln(31/20)/4*8)
On a calculator the result is 48,050
In how many years will the population reach 100,000?
The standard expression is
n(t) = 20000 * e^(ln(31/20)/4 * t)
n(t) is 100,000 and the objective is to solve for t.
An alternative approach is to leave the expression with symbolic variables, isolate time t, and then fill in the known values. The advantage of this approach is that it reduces opportunity for error.
The original expression is:
n(t) = n(0) * e^(rt)
The first step is to divide both sides by n(0)
n(t)/n(0) = e^(rt)
The next step is to take the natural log of both sides. This eliminates the value e and the exponent:
ln(n(t)/n(0)) = ln(e^(rt))
ln(n(t)/n(0)) = rt
To isolate time t divide both sides by rate r
ln(n(t)/n(0))/r = t
Now the known values can be substituted in
n(t) is 100000
rate r is ln(31/20)/4
So the expression is
ln(100000/20000)/(ln(31/20)/4) = t
Note that 100000/20000 is 5 so the expression can be simplified to
ln(5)/(ln(31/20)/4)
The fraction can be simplified by multiplying numerator and denominator by 4/(ln(31/20)), giving
4 * ln(5) / ln(31/20)
On a calculator this is approximately 14.7
This means that 14.7 years after 2003 the population will reach 100,000.
Example: Setting up the standard equations
The general expression for computing population is
n(t) = n(0) * e^(rt)
To isolate starting population n(0) divide both sides of the original e^(rt)
n(0) = n(t)/(e^(rt))
To isolate rate first divide both sides of the original equation by n(0) giving
n(t) / n(0) = e^(rt)
Next take the natural log of both sides to eliminate the exponent and value e
ln(n(t)/n(0)) = ln(e^(rt))
ln(n(t)/n(0)) = rt
Divide both sides by time t to isolate rate r
ln(n(t)/n(0))/t = r
Multiply both sides of the preceding equation by t/r to isolate time t:
ln(n(t)/n(0))/r = t
Write a general expression for computing time to double:
This can be written from the preceding expression that has already isolated time r. If the objective is to compute time to double hten n(t) must equal 2x n(0). This means that ln((n(t)/n(0)) is equal to ln(2). The simplified expression would be
ln(2)/r = t
Now there are standard equations for computing time, rate, finial population, initial population, and time to double:
n(t) = n(0) * e^(rt) compute final population
n(0) = n(t) / (e^(rt)) compute initial population
ln(n(t)/n(0))/t = r compute rate r
ln(n(t)/n(0))/r = t compute time t
ln(2)/r = t compute time to double population
Example: Frog Population
The initial population is 100, increased to 225 two years later.
Write a general expression for computing population growth:
n(t) = n(0) * e^(rt)
n(0) is known and is 100 so the expression is
n(t) = 100 * e^(rt)
Rate can be computed by adding time and final population to the expression and isolating rate r
ln(n(t)/n(0))/t = r
Substituting in the known values the expression is
ln(225/100)/2 = r
Using a calculator
r ~= 0.405
How long will it take the population to reach 75000?
This requires isolating time t. The initial and final populations are known and rate r has been computed. The expression is
t = ln(n(t)/n(0))/r
t = (ln(75000/100)/(ln(225/100)/2)
t = 2 *ln(75000/100)/ln(225/100)
r ~= 16.3 years.
Compute the population in 15 years
The expression for this is:
n(t) = n(0) * e^(rt)
n(15) = 100 * e^(ln(225/100)/2*15)
n(15) ~= 43,789
Compute time to double:
t = ln(2)/r
t = ln(2)/(ln(225/100)/2)
t = 2 * ln(2) / ln(225/100)
t ~= 1.71 years. This is reasonable given two years to grow from 100 to 225.