Opening the book…
- Note to the Instructor (click to expand)
- This is the culmination of the exponential and logarithmic function material—skipping this would be criminal. If any area in Mathematics deserves extra time for applications, it's this chapter.
- (click to expand)
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- Sets and Numbers
The following is a list of learning objectives for this section.
- Learning Objectives (click to expand)
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- Model exponential growth and decay.
- Use Newton's Law of Cooling.
- Use logistic-growth models.
- Express an exponential model in base \(e\).
- Use logarithmic models.
We have already explored some basic applications of exponential and logarithmic functions. In this section, we explore some important applications in more depth, including radioactive isotopes and Newton's Law of Cooling.
Modeling Exponential Growth and Decay
In real-world applications, we need to model the behavior of a function. In mathematical modeling, we choose a familiar general function with properties that suggest that it will model the real-world phenomenon we wish to analyze. In the case of rapid growth, we may choose the exponential growth function \(P(t) = P_0 e^{kt}\), where \(P_0\) is equal to the value at time zero, \(e\) is Euler's constant, and \(k\) is a positive constant that determines the rate (percentage) of growth. We may use the exponential growth function in applications involving doubling time, the time it takes for a quantity to double.
Such phenomena as wildlife populations, financial investments, biological samples, and natural resources may exhibit growth based on a doubling time.
On the other hand, if a quantity is falling rapidly toward zero, without ever reaching zero, then we should probably choose the exponential decay model. Again, we have the form \(P(t) = P_0 e^{kt}\) where \(P_0\) is the starting value, and \(e\) is Euler's constant. Now \(k\) is a negative constant that determines the rate of decay. We may use the exponential decay model when we are calculating half-life, or the time it takes for a substance to exponentially decay to half of its original quantity. We use half-life in applications involving radioactive isotopes.
In our choice of a function to serve as a mathematical model, we often use data points gathered by careful observation and measurement to construct points on a graph and hope we can recognize the shape of the graph. Exponential growth and decay graphs have a distinctive shape, as we can see in Figure \(\PageIndex{2}\) and Figure \(\PageIndex{3}\). It is important to remember that, although parts of each of the two graphs seem to lie on the \(x\)-axis, they are really a tiny distance above the \(x\)-axis.
Exponential growth and decay often involve very large or very small numbers. To describe these numbers, we often use orders of magnitude.
For example, the distance to the nearest star, Proxima Centauri, measured in kilometers, is \(40,113,497,200,000\) kilometers. Expressed in scientific notation, this is \(4.01134972 \times 10^{13}\). So, we could describe this number as having order of magnitude 13.
A population of bacteria doubles every hour. If the culture started with \(10\) bacteria, graph the population as a function of time.
- Solution
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When an amount grows at a fixed percent per unit time, the growth is exponential. To find \(P_0\) we use the fact that \(P_0\) is the amount at time zero, so \(P_0 = 10\). To find \(k\), use the fact that after one hour (\(t=1\)) the population doubles from \(10\) to \(20\). The formula is derived as follows\[\begin{array}{rrclcl} & 20 & = & 10 e^{k \cdot 1} & \quad & \left( \text{substituting} \right) \\[6pt] \implies & 2 & = & e^k & \quad & \left( \text{dividing both sides by }10 \right) \\[6pt] \implies & \ln 2 & = & k & \quad & \left( \text{applying the natural logarithm to both sides} \right) \\[6pt] \end{array}\nonumber\]So \(k = \ln (2)\). Thus, the equation we want to graph is\[y = 10 e^{(\ln 2)t} = 10 \left( e^{\ln2} \right)^t = 10 \cdot 2^t.\nonumber\]The graph is shown in Figure \(\PageIndex{4}\).
Figure \(\PageIndex{4}\): The graph of \(P(t) = 10 e^{(\ln 2)t}\) The population of bacteria after ten hours is \(10,240\). We could describe this amount as being an order of magnitude 4 because it can be written as \(1.024 \times 10^4\). The population of bacteria after twenty hours is \(10,485,760 \approx 1.049 \times 10^7\), which is an order of magnitude of 7, so we could say that the population has increased by three orders of magnitude in ten hours.
Half-Life
In previous sections, we learned the properties and rules for both exponential and logarithmic functions. We have seen that any exponential function can be written as a logarithmic function and vice versa. We have used exponents to solve logarithmic equations and logarithms to solve exponential equations. We are now ready to combine our skills to solve equations that model real-world situations, whether the unknown is in an exponent or in the argument of a logarithm.
One such application is in science, in calculating the time it takes for half of the unstable material in a sample of a radioactive substance to decay, called its half-life. Table \(\PageIndex{1}\) lists the half-life for several of the more common radioactive substances.
| Substance | Use | Half-life |
|---|---|---|
| gallium-67 | nuclear medicine | 80 hours |
| cobalt-60 | manufacturing | 5.3 years |
| technetium-99m | nuclear medicine | 6 hours |
| americium-241 | construction | 432 years |
| carbon-14 | archeological dating | 5,715 years |
| uranium-235 | atomic power | 703,800,000 years |
To find the half-life of a function describing exponential decay, we need to know how long it takes for the initial population, \(P_0\), to decay by half to \(\frac{1}{2} P_0\). This requires us to solve the following equation:\[\dfrac{1}{2} P_0 = P_0 e^{kt}.\nonumber\]We find that the half-life depends only on the constant \(k\) and not on the starting quantity \(P_0\).
The formula is derived as follows:\[\begin{array}{rrclcl}
& \dfrac{1}{2} P_0 & = & P_0 e^{kt} & & \\[6pt]
\implies & \dfrac{1}{2} & = & e^{kt} & \quad & \left( \text{dividing both sides by }P_0 \right) \\[6pt]
\implies & \ln \left(\dfrac{1}{2}\right) & = & k t & \quad & \left( \text{taking the natural logarithm of both sides} \right) \\[6pt]
\implies & \dfrac{\ln \left( 1/2 \right)}{k} & = & t & \quad & \left( \text{dividing both sides by }k \right) \\[6pt]
\implies & \dfrac{\ln \left( 2^{-1} \right)}{k} & = & t & \quad & \left( \text{Laws of Exponents} \right) \\[6pt]
\implies & - \dfrac{\ln \left( 2 \right)}{k} & = & t & \quad & \left( \text{Laws of Logarithms} \right) \\[6pt]
\end{array}\nonumber\]Since \(t\), the time, is positive, \(k\) must, as expected, be negative. This gives us the half-life formula.
While most students tend to memorize this formula, you are expected to understand it and derive it, if needed.
The half-life of carbon-14 is 5,730 years. Express the amount of carbon-14 remaining as a function of time, \(t\).
- Solution
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This formula is derived as follows:\[\begin{array}{rrclcl} & P(t) & = & P_0 e^{kt} & & \\[6pt] \implies & \dfrac{1}{2} P_0 & = & P_0 e^{k \cdot 5730} & \quad & \left( \text{substituting} \right) \\[6pt] \implies & \dfrac{1}{2} & = & e^{k \cdot 5730} & \quad & \left( \text{dividing both sides by }P_0 \right) \\[6pt] \implies & \ln \left(\dfrac{1}{2}\right) & = & 5730 k & \quad & \left( \text{taking the natural logarithm of both sides} \right) \\[6pt] \implies & \dfrac{\ln \left(1/2\right)}{5730} & = & k & \quad & \left( \text{dividing both sides by }5730 \right) \\[6pt] \end{array}\nonumber\]Substituting this back into the original function, we get\[P(t) = P_0 e^{\left( \frac{\ln(1/2)}{5730} \right)t}.\nonumber\]We observe that the coefficient of \(t\), \(\frac{\ln(0.5)}{5730} \approx -1.2097 \times 10^{-4}\), is negative, as expected in the case of exponential decay.
The half-life of plutonium-244 is 80,000,000 years. Find a function that gives the amount of plutonium-244 remaining as a function of time, measured in years.
Radiocarbon Dating
The formula for exponential decay is important in radiocarbon dating, which is used to calculate the approximate date a plant or animal died. Radiocarbon dating was discovered in 1949 by Willard Libby, who won a Nobel Prize for his discovery. It compares the difference between the ratio of two isotopes of carbon in an organic artifact or fossil to the ratio of those two isotopes in the air. It is believed to be accurate to within about \(1\%\) error for plants or animals that died within the last 60,000 years.
Carbon-14 is a radioactive isotope of carbon that has a half-life of 5,730 years. It occurs in small quantities in the carbon dioxide in the air we breathe. Most of the carbon on Earth is carbon-12, which has an atomic weight of 12 and is not radioactive. Scientists have determined the ratio of carbon-14 to carbon-12 in the air for the last 60,000 years, using tree rings and other organic samples of known dates—although the ratio has changed slightly over the centuries.
As long as a plant or animal is alive, the ratio of the two isotopes of carbon in its body is close to the ratio in the atmosphere. When it dies, the carbon-14 in its body decays and is not replaced. By comparing the ratio of carbon-14 to carbon-12 in a decaying sample to the known ratio in the atmosphere, the date the plant or animal died can be approximated.
Since the half-life of carbon-14 is 5,730 years, we derive a formula for the amount present at any time \(t\) as follows:\[\begin{array}{rrclcl} & P(t) & = & P_0 e^{kt} & & \\[6pt] \implies & \dfrac{1}{2} P_0 & = & P_0 e^{k \cdot 5730} & \quad & \left( \text{after 5,730 years, half the initial amount remains} \right) \\[6pt] \implies & \dfrac{1}{2} & = & e^{5730 k} & \quad & \left( \text{dividing both sides by }P_0 \right) \\[6pt] \implies & \ln \left(\frac{1}{2}\right) & = & 5730 k & \quad & \left( \text{taking the natural logarithm of both sides} \right) \\[6pt] \implies & \dfrac{\ln (0.5)}{5730} & = & k & \quad & \left( \text{dividing both sides by }5730 \right) \\[6pt] \end{array}\nonumber\]Thus, the formula for the amount of carbon-14 remaining after \(t\) years is\[P(t) \approx P_0 e^{ \left( \frac{\ln(0.5)}{5730} \right)t},\nonumber\]where
- \(P(t)\) is the amount of carbon-14 remaining at time \(t\), and
- \(P_0\) is the amount of carbon-14 when the plant or animal began decaying.
Again, I highly discourage you from memorizing this formula—it's fast to derive, and the half-life changes based on substance given. It is much preferred to derive the formula each time.
A bone fragment is found that contains 20% of its original carbon-14. To the nearest year, how old is the bone?
- Solution
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We are told that 20% of an initial sample remains. That is, the current amount of carbon-14 in the sample is \(0.20 P_0\). Therefore, we want to solve \(0.2 P_0 = P_0 e^{kt}\). This gives\[0.2 = e^{kt} \implies \ln 0.2 = k t \implies t = \dfrac{\ln 0.2}{k}.\nonumber\]Since we already have computed \(k\) for carbon-14 earlier, we substitute that value into this formula:\[t = \dfrac{\ln 0.2}{k} = \dfrac{\ln 0.2}{\frac{\ln 0.5}{5730}} = \dfrac{5730 \ln(0.2)}{\ln{0.5}} \approx 13305\nonumber\]The bone fragment is about 13,305 years old.
In reality, the instruments that measure the percentage of carbon-14 are extremely sensitive and a scientist would need to do much more work than we did in order to be satisfied. Even so, carbon dating is only accurate to about 1%, so the age of the sample from Example \(\PageIndex{3}\) should be given as\[13,301 \text{ years } \pm \text{ 1% of } 13,301 \text{ years } = 13,301 \pm 133 \text{ years}.\nonumber\]
Cesium-137 has a half-life of about 30 years. If we begin with 200 mg of cesium-137, will it take more or less than 230 years until only 1 milligram remains?
Calculating Doubling Time
For decaying quantities, we determined how long it took for half of a substance to decay. For growing quantities, we might want to find out how long it takes for a quantity to double. As we mentioned above, the time it takes for a quantity to double is called the doubling time.
Given the basic exponential growth equation \(P(t) = P_0 e^{kt}\), doubling time can be found by solving for when the original quantity has doubled—that is, by solving \(2 P_0 = P_0 e^{kt}\). Rather than attracting you to memorize a formula, let's show how to derive it.
According to Moore's Law, the doubling time for the number of transistors that can be put on a computer chip is approximately two years. Give a function that describes this behavior.
- Solution
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The formula is derived as follows:\[\begin{array}{rrclcl} & P(t) & = & P_0 e^{kt} & & \\[6pt] \implies & 2 P_0 & = & P_0 e^{k \cdot 2} & \quad & \left( \text{substituting }t = 2 \text{ and }P(2) = 2P_0 \right) \\[6pt] \implies & 2 & = & e^{2k} & \quad & \left( \text{dividing both sides by }P_0 \right) \\[6pt] \implies & \ln 2 & = & 2k & \quad & \left( \text{taking the natural log of both sides} \right) \\[6pt] \implies & \dfrac{\ln 2}{2} & = & k & \quad & \left( \text{dividing both sides by }2 \right) \\[6pt] \end{array}\nonumber\]Thus, the function is\[P(t) = P_0 e^{(\ln (2)/2) t}.\nonumber\]
Recent data suggests that, as of 2013, the rate of growth predicted by Moore's Law no longer holds. Growth has slowed to a doubling time of approximately three years. Find the new function that takes that longer doubling time into account.
Newton's Law of Cooling
Exponential decay can also be applied to temperature. When a hot object is left in surrounding air that is at a lower temperature, the object's temperature will decrease exponentially, leveling off as it approaches the surrounding air temperature. On a graph of the temperature function, the leveling off will correspond to a horizontal asymptote at the temperature of the surrounding air. Unless the room temperature is zero, this will correspond to a vertical shift of the generic exponential decay function. This translation leads to Newton's Law of Cooling, the scientific formula for temperature as a function of time as an object's temperature is equalized with the ambient temperature.
A cheesecake is taken out of the oven with an ideal internal temperature of \(165^{\circ} \text{F}\) and is placed into a \(35^{\circ} \text{F}\) refrigerator. After 10 minutes, the cheesecake has cooled to \(150^{\circ} \text{F}\). If we must wait until the cheesecake has cooled to \(70^{\circ} \text{F}\) before we eat it, how long will we have to wait?
- Solution
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Because the surrounding air temperature in the refrigerator is 35 degrees, the cheesecake's temperature will decay exponentially toward 35, following the equation\[T(t) = 35 + (T_0 - 35) e^{kt}.\nonumber\]We know the initial temperature was 165, so \(T(0) = T_0 =165\). Therefore,\[T(t) = 35 + (165 - 35) e^{k \cdot t} = 35 + 130 e^{kt}.\nonumber\]We were given another data point, \(T(10)=150\), which we can use to solve for \(k\).\[\begin{array}{rrclcl}
& 150 & = & 35 + 130 e^{k \cdot 10} & \quad & \left( \text{subsituting the point }\left( 10, 150 \right) \right) \\[6pt]
\implies & 115 & = & 130 e^{10k} & \quad & \left( \text{subtracting }35\text{ from both sides} \right) \\[6pt]
\implies & \dfrac{115}{130} & = & e^{10k} & \quad & \left( \text{dividing both sides by }130 \right) \\[6pt]
\implies & \ln\left(\frac{115}{130}\right) & = & 10k & \quad & \left( \text{taking the natural log of both sides} \right) \\[6pt]
\implies & \dfrac{\ln\left(\frac{115}{130}\right)}{10} & = & k & \quad & \left( \text{dividing both sides by }10 \right) \\[6pt]
\end{array}\nonumber\]Hence, \(k \approx -0.0123\). This gives us the equation for the cooling of the cheesecake:\[T(t) \approx 35 + 130 e^{-0.0123t}.\nonumber\]Now we can solve for the time it will take for the temperature to cool to 70 degrees. To keep numbers as exact as possible, we wait until the end to replace \(k\) with our found value.\[\begin{array}{rrclcl} & 70 & = & 35 + 130 e^{kt} & \quad & \left( \text{substituting }70\text{ for }T(t) \right) \\[6pt] \implies & 35 & = & 130 e^{kt} & \quad & \left( \text{subtracting }35\text{ from both sides} \right) \\[6pt] \implies & \dfrac{35}{130} & = & e^{kt} & \quad & \left( \text{dividing both sides by }130 \right) \\[6pt] \implies & \ln\left(\frac{35}{130}\right) & = & kt & \quad & \left( \text{taking the natural log of both sides} \right) \\[6pt] \implies & \dfrac{\ln\left(35/130\right)}{k} & = & t & \quad & \left( \text{dividing both sides by }k \right) \\[6pt] \implies & \dfrac{\ln\left(35/130\right)}{\frac{\ln\left(115/130\right)}{10}} & = & t & \quad & \left( \text{substituting }k = \dfrac{\ln\left(\frac{115}{130}\right)}{10} \right) \\[6pt] \end{array}\nonumber\]Computing this value, we get \(t \approx 106.68\). Thus, it will take about 107 minutes, or one hour and 47 minutes, for the cheesecake to cool to \(70^{\circ} \text{F}\).
A pitcher of water at 40 degrees Fahrenheit is placed into a 70 degree room. One hour later, the temperature has risen to 45 degrees. How long will it take for the temperature to rise to 60 degrees?
Using Logistic Growth Models
Exponential growth cannot continue forever. Exponential models, while they may be useful in the short term, tend to fall apart the longer they continue. Consider an aspiring writer who writes a single line on day one and plans to double the number of lines she writes each day for a month. By the end of the month, she must write over 17 billion lines, or one-half-billion pages. It is impractical, if not impossible, for anyone to write that much in such a short period of time. Eventually, an exponential model must begin to approach some limiting value, and then the growth is forced to slow. For this reason, it is often better to use a model with an upper bound instead of an exponential growth model, though the exponential growth model is still useful over a short term, before approaching the limiting value.
The logistic growth model is approximately exponential at first, but it has a reduced rate of growth as the output approaches the model's upper bound, or carrying capacity. The graph in Figure \(\PageIndex{5}\) shows how the growth rate changes over time. The graph increases from left to right, but the growth rate only increases until it reaches its point of maximum growth rate, at which point the rate of increase decreases.
An influenza epidemic spreads through a population rapidly, at a rate that depends on two factors: The more people who have the flu, the more rapidly it spreads, and also the more uninfected people there are, the more rapidly it spreads. These two factors make the logistic model a good one to study the spread of communicable diseases. And, clearly, there is a maximum value for the number of people infected: the entire population.
For example, at time \(t=0\) there is one person in a community of 1,000 people who has the flu. So, in that community, at most 1,000 people can have the flu. Researchers find that for this particular strain of the flu, the logistic growth constant is \(b = 0.6030\). Estimate the number of people in this community who will have had this flu after ten days. Predict how many people in this community will have had this flu after a long period of time has passed.
- Solution
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We substitute the given data into the logistic growth model\[f(t) = \dfrac{c}{1 + a e^{-bt}}.\nonumber\]Because at most 1,000 people, the entire population of the community, can get the flu, we know the limiting value is \(c=1000\). To find \(a\), we use the formula that the number of cases at time \(t=0\) is \(\frac{c}{1+a} = 1\), from which it follows that \(a=999\). This model predicts that, after ten days, the number of people who have had the flu is \(f(10)= \frac{1000}{1 + 999 e^{-0.6030(10)}} \approx 293.9\). Because the actual number must be a whole number (a person has either had the flu or not) we round down to 293. In the long term, the number of people who will contract the flu is the limiting value, \(c=1000\).
Remember that, because we are dealing with a virus, we cannot predict with certainty the number of people infected. The model only approximates the number of people infected and will not give us exact or actual values.
The graph in Figure \(\PageIndex{6}\) gives a good picture of how this model fits the data.
Figure \(\PageIndex{6}\): The graph of \(f(t)= \frac{1000}{1 + 999 e^{-0.6030t}}\).
Using the model in Example \(\PageIndex{6}\), estimate the number of cases of flu on day 15.
Expressing an Exponential Model in Base \(e\)
While powers and logarithms of any base can be used in modeling, the two most common bases are \(10\) and \(e\). In Science and Mathematics, the base \(e\) is often preferred. We can use Laws of Exponents and Laws of Logarithms to change any base to base \(e\).
Given a model with the form \(y = a b^x\), we change it to the form \(y = A_0 e^{kx}\) using the following steps.
- Rewrite \(y = a b^x\) as \(y = a e^{\ln\left( b^x \right)}\).
- Use the Power Rule of Logarithms to rewrite \(y\) as \(y = a e^{x \ln( b )} = a e^{\ln( b )x}\).
- Note that \(a = A_0\) and \(k = \ln( b )\) in the equation \(y = A_0 e^{kx}\).
Change the function \(y = 2.5 (3.1)^x\) so that this same function is written in the form \(y = A_0 e^{kx}\).
- Solution
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The formula is derived as follows:\[\begin{array}{rclcl} y & = & 2.5(3.1)^x & & \\[6pt] & = & 2.5 e^{\ln\left( 3.1^x \right)} & \quad & \left( \text{Inverse Property of Exponentials and Logarithms} \right) \\[6pt] & = & 2.5 e^{x\ln\left( 3.1 \right)} & \quad & \left( \text{Laws of Logarithms} \right) \\[6pt] & = & 2.5 e^{\ln\left( 3.1 \right) x} & \quad & \left( \text{Commutative Property of Multiplication} \right) \\[6pt] \end{array}\nonumber\]
Change the function \(y = 3(0.5)^x\) to one having \(e\) as the base.
Logarithmic Scales
For quantities that vary greatly in magnitude, a standard scale of measurement is not always effective, and utilizing logarithms can make the values more manageable. For example, if the average distances from the sun to the major bodies in our solar system are listed, you see they vary greatly.
| Planet | Distance (millions of km) |
| Mercury | 58 |
| Venus | 108 |
| Earth | 150 |
| Mars | 228 |
| Jupiter | 779 |
| Saturn | 1430 |
| Uranus | 2880 |
| Neptune | 4500 |
Placed on a linear scale—one with equally spaced values—these values get bunched up (see Figure \(\PageIndex{7}\)).
However, computing the logarithm of each value and plotting these new values on a number line results in a more manageable graph, and makes the relative distances more apparent, as seen in Figure \(\PageIndex{8}\) below Table \(\PageIndex{3}\). (It is interesting to note the large gap between Mars and Jupiter on the log number line. The asteroid belt is located there, which scientists believe is a planet that never formed because of the effects of the gravity of Jupiter.)
| Planet | Distance (millions of km) | log(distance) |
| Mercury | 58 | 1.76 |
| Venus | 108 | 2.03 |
| Earth | 150 | 2.18 |
| Mars | 228 | 2.36 |
| Jupiter | 779 | 2.89 |
| Saturn | 1430 | 3.16 |
| Uranus | 2880 | 3.46 |
| Neptune | 4500 | 3.65 |
Sometimes, as shown above in Figure \(\PageIndex{8}\), the scale on a logarithmic number line will show the log values, but more commonly the original values are listed as powers of 10, as shown below in Figure \(\PageIndex{9}\).
Estimate the value of point \(P\) on the log scale above
- Solution
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The point \(P\) appears to be half way between \(-2\) and \(-1\) in log value. So if \(V\) is the value of this point,\[\log (V) \approx -1.5.\nonumber\]Rewriting in exponential form,\[V \approx 10^{-1.5} \approx 0.0316.\nonumber\]
Place the number 6000 on a logarithmic scale.
- Solution
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Since \(\log (6000) \approx 3.8\), this point would belong on the log scale about here:
Figure \(\PageIndex{10}\)
Plot the data in the table below on a logarithmic scale (From http://www.epd.gov.hk/epd/noise_educ...1/intro_5.html, retrieved Oct 2, 2010).
| Source of Sound/Noise | Approximate Sound Pressure in \(\mu\) Pa (micro Pascals) |
| Launching of the Space Shuttle | 2000,000,000 |
| Full Symphony Orchestra | 2000,000 |
| Diesel Freight Train at High Speed at 25 m | 200,000 |
| Normal Conversation | 20,000 |
| Soft Whispering at 2 m in Library | 2,000 |
| Unoccupied Broadcast Studio | 200 |
| Softest Sound a human can hear | 20 |
- Answer
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Figure \(\PageIndex{11}\)
Notice that on the log scale above Figure \(\PageIndex{9}\), the visual distance on the scale between points \(A\) and \(B\) and between \(C\) and \(D\) is the same. When looking at the values these points correspond to, notice \(B\) is ten times the value of \(A\), and \(D\) is ten times the value of \(C\). A visual (linear) difference between points corresponds to a relative (ratio) change between the corresponding values.
Logarithms are useful for showing these relative changes. For example, comparing $1,000,000 to $10,000, the first is 100 times larger than the second.\[\dfrac{1,000,000}{10,000} = 100 = 10^2.\nonumber\]Likewise, comparing $1000 to $10, the first is 100 times larger than the second.\[\dfrac{1,000}{10} = 100 = 10^2.\nonumber\]When one quantity is roughly ten times larger than another, its exponent when written in scientific notation increases by 1. Thus, we say it is one order of magnitude larger. In both cases described above, the first number was two orders of magnitude larger than the second.
Notice that the order of magnitude can be found as the common logarithm of the ratio of the quantities. On the log scale above, \(B\) is one order of magnitude larger than \(A\), and \(D\) is one order of magnitude larger than \(C\).
On the log scale in Figure \(\PageIndex{9}\), how many orders of magnitude larger is \(C\) than \(B\)?
- Solution
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The value \(B\) corresponds to \(10^2 = 100\) and the value \(C\) corresponds to \(10^5 = 100,000\). The relative change is \(\frac{100,000}{100} = 1000 = \frac{10^5}{10^2} = 10^3\). The log of this value is \(3\). Therefore, \(C\) is three orders of magnitude greater than \(B\), which can be seen on the log scale by the visual difference between the points on the scale.
Using the table from Checkpoint \(\PageIndex{9}\), what is the difference of order of magnitude between the softest sound a human can hear and the launching of the space shuttle?
- Answer
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\(\frac{2 \times 10^9}{2 \times 10^1} = 10^8\). The sound pressure in \(\mu\) Pa created by launching the space shuttle is 8 orders of magnitude greater than the sound pressure in \(\mu\) Pa created by the softest sound a human ear can hear.
Earthquakes
An example of a logarithmic scale is the Richter Scale used for earthquakes. While not the only scale for measuring earthquakes, it is the most commonly known. The Richter scale compares the amplitude, \(A\), of the earthquake's seismographic trace with the amplitude, \(A_0\), of the smallest detectable earthquake.
If one earthquake has a Richter value of 6.0, and another has a magnitude of 8.0, how much more powerful (in terms of earth movement) is the second earthquake?
- Solution
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Since the first earthquake has Richter value of 6.0, we can find the amount of earth movement for that quake, which we'll denote \(A_1\). The value of \(A_0\) is not particularly relevant, so we will not replace it with its value.\[\begin{array}{rrclcl} & 6.0 & = & \log\left( \dfrac{A_1}{A_0} \right) & & \\[6pt] \implies & 10^{6.0} & = & \dfrac{A_1}{A_0} & \quad & \left( \text{rewriting the logarithm in exponential form} \right) \\[6pt] \implies & 10^{6.0} A_0 & = & A_1 & \quad & \left( \text{multiplying both sides by }A_0 \right) \\[6pt] \end{array}\nonumber\]This tells us the first earthquake has about \(10^6\) times more earth movement than the baseline measure.
Doing the same with the second earthquake, \(A_2\), with a magnitude of 8.0,\[\begin{array}{rrclcl} & 8.0 & = & \log\left( \dfrac{A_2}{A_0} \right) & & \\[6pt] \implies & 10^{8.0} & = & \dfrac{A_2}{A_0} & \quad & \left( \text{rewriting the logarithm in exponential form} \right) \\[6pt] \implies & 10^{8.0} A_0 & = & A_2 & \quad & \left( \text{multiplying both sides by }A_0 \right) \\[6pt] \end{array}\nonumber\]Comparing the earth movement of the second earthquake to the first,\[\dfrac{A_2}{A_1} = \dfrac{10^{8} A_0} {10^6 A_0} = 10^2 = 100.\nonumber\]The second value's earth movement is 100 times as large as the first earthquake.
One earthquake has magnitude of 3.0. If a second earthquake has twice as much earth movement as the first earthquake, find the magnitude of the second quake.
- Solution
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Since the first quake has magnitude 3.0,\[3.0 = \log \left( \dfrac{A}{A_0} \right).\nonumber\]Solving for \(A\),\[\begin{array}{rrclcl} & 3.0 & = & \log\left( \dfrac{A}{A_0} \right) & & \\[6pt] \implies & 10^{3.0} & = & \dfrac{A}{A_0} & \quad & \left( \text{rewriting in exponential form} \right) \\[6pt] \implies & 10^{3.0} A_0 & = & A & \quad & \left( \text{multiplying both sides by }A_0 \right) \\[6pt] \end{array}\nonumber\]Since the second earthquake has twice as much earth movement, for the second quake we get\[A = 2 \cdot 10^{3.0} A_0.\nonumber\]Finding the magnitude,\[\begin{array}{rcl} M & = & \log\left( \dfrac{2 \cdot 10^{3.0} A_0}{A_0} \right) \\[6pt] & = & \log\left( 2 \cdot 10^{3.0} \right) \\[6pt] & \approx & 3.3010 \\[6pt] \end{array}\nonumber\]The second earthquake with twice as much earth movement will have a magnitude of about 3.3.
In fact, using the Laws of Logarithms, we could show that whenever the earth movement doubles, the Richter magnitude will increase by about 0.301:\[\begin{array}{rcl} M & = & \log\left( \dfrac{2A}{A_0} \right) \\[6pt] & = & \log\left( 2 \cdot \dfrac{A}{A_0} \right) \\[6pt] & = & \log\left( 2 \right) + \log \left( \dfrac{A}{A_0} \right) \\[6pt] & \approx & 0.3010 + \log \left( \dfrac{A}{A_0} \right) \\[6pt] \end{array}\nonumber\]This illustrates the most important feature of a log scale: that multiplying the quantity being considered will add to the scale value, and vice versa.
Logarithms are useful in measuring the intensities of many quantities based on senses. Earthquakes (the Richter scale), sound (decibels) and acids and bases (pH) are common examples. We now present yet a different use of the basic logarithm function, password strength.
The information entropy \(H\), in bits, of a randomly generated password consisting of \(L\) characters is given by\[H = L \log_{2}(N),\nonumber\]where \(N\) is the number of possible symbols for each character in the password. In general, the higher the entropy, the stronger the password.
- If a \(7\) character case-sensitive password is comprised of letters and numbers only, find the associated information entropy.
- How many possible symbol options per character are required to produce a \(7\) character password with an information entropy of \(50\) bits?
- Solutions
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- There are \(26\) letters in the alphabet, \(52\) if upper and lower case letters are counted as different. There are \(10\) digits (\(0\) through \(9\)) for a total of \(N=62\) symbols. Since the password is to be \(7\) characters long, \(L = 7\). Thus, \(H = 7 \log_{2}(62) = \frac{7 \ln(62)}{\ln(2)} \approx 41.68\).
- We have \(L = 7\) and \(H=50\) and we need to find \(N\). Solving the equation \(50 = 7 \log_{2}(N)\) gives \(N = 2^{50/7} \approx 141.323\), so we would need \(142\) different symbols to choose from.
Chemical systems known as buffer solutions can adjust to small changes in acidity to maintain a range of pH values. Buffer solutions have a wide variety of applications, from maintaining a healthy fish tank to regulating the pH levels in blood.
Blood is a buffer solution. When carbon dioxide is absorbed into the bloodstream, it produces carbonic acid and lowers the pH. The body compensates by producing bicarbonate, a weak base, to partially neutralize the acid. The equation which models blood pH in this situation is\[\text{pH} = 6.1 + \log\left(\dfrac{800}{x} \right),\nonumber\]where \(x\) is the partial pressure of carbon dioxide in arterial blood, measured in torr. Find the partial pressure of carbon dioxide in arterial blood if the pH is \(7.4\).
- Solution
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We set \(\text{pH} = 7.4\) and get \(7.4 = 6.1 + \log\left(\frac{800}{x} \right)\), or \(\log\left(\frac{800}{x} \right) = 1.3\). Solving, we find \(x = \frac{800}{10^{1.3}} \approx 40.09\). Hence, the partial pressure of carbon dioxide in the blood is about \(40\) torr.
This page titled 13.7: Exponential and Logarithmic Models was last modified on Tue, 08 Jul 2025 17:47:34 GMT and is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Roy Simpson, Cosumnes River College (OpenStax) via source content that was edited to the style and standards of the LibreTexts platform.
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