Applications Using Logarithmic Functions
This section will discuss applications using logarithmic functions. While there are
many applications that use logarithmic functions, we will examine three. These
applications involve formulas. Thus, understanding each formula and how to use
them is important.
Earthquakes – Richter Scale
Earthquakes are complicated events, and so the intent here is not to provide a
complete discussion of the science involved in them. Rather, we will look at a
simplified version of the Richter scale. The Richter scale measures the magnitude
of an earthquake by comparing the amplitude of the seismic waves of the given
earthquake to those of a “magnitude 0 event”, which was chosen to be a seismograph
reading of 0.001 millimeters recorded on a seismometer 100 kilometers from the
earthquake’s epicenter. Specifically, the magnitude of an earthquake is given by:
M(x) = log (
x
)
0.001
where x is the seismograph reading in millimeters of the earthquake recorded 100
kilometers from the epicenter.
Example 1:
following:
Use the formula for the magnitude of an earthquake and do the
a. Show that M(0.001) = 0.
b. Compute M(80,000).
c. Show that an earthquake which registered 6.7 on the Richter scale had a
seismograph reading ten times larger than one which measured 5.7.
Modified from College Algebra, by Carl Stitz, PhD and Jeff Zeager, PhD, CC-BY 2013; Intermediate
Algebra, by Andrew Gloag, Anne Gloag, and Mara Landers, CK-12 Foundation, CC-BY 2013; and
Precalculus (College Algebra), by David Lippman and Melanie Rasmussen, CC-BY 2015. Licensed under a
Creative Commons Attribution 3.0 Unported License (http://creativecommons.org/licenses/by/3.0)
a. M(0.001) = (
0.001
)
Substitute 0.001 in for x
0.001
= log (1)
=0
b. M(80,000) = log (
Reduce the fraction (Divide)
80,000
0.001
)
= log (80,000,000)
≈ 7.9
x
c. log (
Reduce the fraction (Divide)
) = 6.7
0.001
x
0.001
x
0.001
= 106.7
Convert to exponential form
(Hint: log is base 10)
= 5011872.3
106.7 ≈ 5011872.3
x = 5011872.3(0.001)
x ≈ 5011.87
log (
Substitute 80,000 in for x
x
0.001
x
Clear the fraction
Round to the hundredth
) = 5.7
0.001
x
0.001
= 105.7
Convert to exponential form
(Hint: log is base 10)
= 501187.23
105.7 ≈ 501187.23
x = 501187.23(0.001)
x ≈ 501.187
Clear the fraction
Round to the thousandth
501.187 × 10 = 5011.87, so 6.7 is ten times larger than 5.7
Modified from College Algebra, by Carl Stitz, PhD and Jeff Zeager, PhD, CC-BY 2013; Intermediate
Algebra, by Andrew Gloag, Anne Gloag, and Mara Landers, CK-12 Foundation, CC-BY 2013; and
Precalculus (College Algebra), by David Lippman and Melanie Rasmussen, CC-BY 2015. Licensed under a
Creative Commons Attribution 3.0 Unported License (http://creativecommons.org/licenses/by/3.0)
Population Growth
Exponential growth has a general growth pattern of y = a (bx). Previously, we looked
at the formula for compound interest, A = P(1 + r)t, which is an example of
exponential growth where b is (1 + r). Population growth is also an example of
exponential growth but is often modeled with a function in which b is e.
Specifically, population growth is given by:
P(t) = P0 e rt
where P0 represents the initial population, r is the growth rate or growth constant,
and t is time in years.
Example 2: Use the formula for population growth to find the following:
a. A population grows from 100 to 130 in 2 years. Find the growth rate.
b. A population grows from 20,000 to 35,000 in 10 years. Find the growth
rate.
c. Using the growth rate found in b, when will the population reach
1,000,000?
a.
t=2
P(t) = P(2) = 130
P0 = 100
130 = 100 e r2
130
100
= e r2
ln(1.3) = ln(e r2)
ln(1.3) = r2
ln(1.3)
2
0.26236
2
2 years
Population in 2 years
Initial population
Substitute values into the formula
Divide by 100
Take the natural logarithm of both sides
Recall: ln of e to any power is that exponent
=r
Divide by 2
=r
ln(1.3) ≈ 0.26236
r ≈ 0.1312
The population is growing at a rate of approximately 13.12% per year.
Modified from College Algebra, by Carl Stitz, PhD and Jeff Zeager, PhD, CC-BY 2013; Intermediate
Algebra, by Andrew Gloag, Anne Gloag, and Mara Landers, CK-12 Foundation, CC-BY 2013; and
Precalculus (College Algebra), by David Lippman and Melanie Rasmussen, CC-BY 2015. Licensed under a
Creative Commons Attribution 3.0 Unported License (http://creativecommons.org/licenses/by/3.0)
b.
t = 10
P(t) = P(10) = 35,000
P0 = 20,000
10 years
Population in 10 years
Initial population
35,000 = 20,000 e r10
Substitute values into the formula
35,000
= e r10
20,000
Divide by 20,000
ln(1.75) = ln(e r10)
ln(1.75) = r10
ln(1.75)
10
0.559616
10
Take the natural logarithm of both sides
Recall: ln of e to any power is that exponent
=r
Divide by 10
=r
ln(1.75) ≈ 0.559616
r ≈ 0.056
The population is growing at a rate of approximately 5.6% per year.
c.
r ≈ 0.056
P(t) = 1,000,000
P0 = 20,000
Rate from b above
Population of 1,000,000
Initial population
1,000,000 = 20,000 e0.056t
1,000,000
20,000
= e0.056t
ln(50) = ln(e0.056t)
ln(50) = 0.056t
ln(50)
0.056
3.912023
0.056
Substitute values into the formula
Divide by 20,000
Take the natural logarithm of both sides
Recall: ln of e to any power is that exponent
=t
Divide by 0.056
=t
ln(50) ≈ 3.912023
t ≈ 70
In approximately 70 years the population will reach 1,000,000.
Modified from College Algebra, by Carl Stitz, PhD and Jeff Zeager, PhD, CC-BY 2013; Intermediate
Algebra, by Andrew Gloag, Anne Gloag, and Mara Landers, CK-12 Foundation, CC-BY 2013; and
Precalculus (College Algebra), by David Lippman and Melanie Rasmussen, CC-BY 2015. Licensed under a
Creative Commons Attribution 3.0 Unported License (http://creativecommons.org/licenses/by/3.0)
Radioactive Decay
Certain naturally occurring radioactive isotopes are unstable and change over time.
This change is called radioactive decay. One of the common terms associated with
radioactive decay is half-life which is the time it takes for half the substance to decay.
The general growth equation, y = a (bx), is also used when growth goes backward –
decays. Half-life is found by solving for half of the remaining original amount, i.e.,
1
1
when y = a. So the half-life decay equation becomes a = a(bx). However, in
2
2
scientific texts b is more commonly replaced with e. Specifically, radioactive decay
is given by:
1
2
a = a (e rt)
where r is rate of decay and t is time in years.
Example 3: Carbon-14 is a radioactive isotope that is present in organic materials
and is commonly used for dating historical artifacts. Carbon-14 has a half-life of
5,730 years. Use the formula for radioactive decay to find the following:
a. Find the rate of decay for carbon-14.
b. If a bone fragment is found that contains 20% of its original carbon-14,
how old is the bone?
a.
1
2
a = a (e 5730r)
0.5 = e 5730r
5730r
ln(0.5) = ln(e
)
ln(0.5) = 5730r
ln(0.5)
5730
−0.69315
5730
Substitute 5730 into the formula for t
1
Divide by a, and write as 0.5
2
Take the natural logarithm of both sides
Recall: ln of e to any power is that exponent
=r
Divide by 5730
=r
ln(0.5) ≈ -0.69315
r ≈ -0.000121
Modified from College Algebra, by Carl Stitz, PhD and Jeff Zeager, PhD, CC-BY 2013; Intermediate
Algebra, by Andrew Gloag, Anne Gloag, and Mara Landers, CK-12 Foundation, CC-BY 2013; and
Precalculus (College Algebra), by David Lippman and Melanie Rasmussen, CC-BY 2015. Licensed under a
Creative Commons Attribution 3.0 Unported License (http://creativecommons.org/licenses/by/3.0)
b. 0.20a = a (e -0.000121t)
0.20 = e -0.000121t
ln(0.20) = ln(e -0.000121t)
ln(0.20) = -0.000121t
ln(0.20)
−0.000121
−1.609438
−0.000121
20% of the original amount is 0.20a
Divide by a
Take the natural logarithm of both sides
Recall: ln of e to any power is that exponent
=t
Divide by -0.000121
=t
ln(0.20) ≈ -1.609438
t ≈ 13301
The bone fragment is approximately 13,300 years old.
Modified from College Algebra, by Carl Stitz, PhD and Jeff Zeager, PhD, CC-BY 2013; Intermediate
Algebra, by Andrew Gloag, Anne Gloag, and Mara Landers, CK-12 Foundation, CC-BY 2013; and
Precalculus (College Algebra), by David Lippman and Melanie Rasmussen, CC-BY 2015. Licensed under a
Creative Commons Attribution 3.0 Unported License (http://creativecommons.org/licenses/by/3.0)