Logarithms - Practice Problems
Solving Earth science problems with Logarithms
This module is undergoing classroom implementation with the Math Your Earth Science Majors Need project. The module is available for public use, but it will likely be revised after classroom testing.
Water Chemistry - determining pH
Salt marsh and ghost forest in a coastal environment.
Provenance: Alex Manda, East Carolina University
Reuse: This item is offered under a Creative Commons Attribution-NonCommercial-ShareAlike license http://creativecommons.org/licenses/by-nc-sa/3.0/ You may reuse this item for non-commercial purposes as long as you provide attribution and offer any derivative works under a similar license.
pH is the measure of the level of alkalinity or acidity of a substance. It is important to know the pH of water in the environment because pH can influence the biological and chemical processes that occur in natural water bodies.
Problem 1. As a member of a water quality team, you have been tasked with determining the pH of a sample of water from a local wetland. If the pH of a water sample is 7.5, what is the hydrogen molar ion concentration ([H+]) of the water sample?
`pH=-log([H^+])`
Step 1. Identify the base of the logarithm
The base of the logarithm is 10. Because there is no explicit base, it is implied that the base is 10.
Step 2. Identify the known and unknown variables in the equation
There are two variables in this problem: pH and [H+]. pH is known (=7.5) and [H+] is unknown.
Step 3. Rearrange and/or rewrite the equation to isolate the unknown variable as a log term
To isolate the unknown variable [
H+], you must transform the logarithmic equation into an exponential equation:
Start with `-log([H^+])=pH`
Then, multiply both sides by -1
`log([H^+])=-pH`
Then, rewrite the logarithmic relationship in exponential form to isolate the unknown variable. Use the relation between logarithms and exponential expressions: `log_(b)(a)=c leftrightarrow b^c=a`
The expression in exponential form is therefore:
`[H^+]=10^(-pH)`
Step 4. Plug the known values into the equation and simplify
The equation to use is:
`[H^+]=10^(-pH)`
Plugging in the known pH value:
`[H^+]=10^-7.5M` - M is the units for molarity
Step 5. Solve for the unknown variable
Technically, you can stop at 10-7.5 M because that is a valid value. However, you may want to convert to the simplified form:
`[H^+]=10^-7.5M=3.2x10^(-8) M`
Scientific calculators will have a
xy or 10
x option that you will use to evaluate the equation:
`[H^+]=10^-7.5=`0.000000032, which simplifies to 3.2x10-8 M
Click on any box (called a cell) in the spreadsheet.
Type the following expression in the cell to solve the equation:
"=10^(-7.5)"
Press Enter to reveal your answer in the cell.
The answer is 3.2E-08, which is the engineering notation for 3.2x10-8 M.
Acoustics - measuring the intensity of sound
Illustration of the Krakatoa eruption in 1883.
Provenance: Krakatoa Committee, Royal Society Great Britain. Provided by Houghton Library, Harvard University. Accessed at https://picryl.com/media/houghton-71-1250-krakatoa-1883-eruption-bdd31f
Reuse: This item is in the public domain and maybe reused freely without restriction.
The study of sound and its effects on the environment is a topic of increasing interest in the Earth sciences as the effects of noise on the environment and organisms become more apparent. Some of the loudest sounds ever recorded include volcanic eruptions that can be heard hundreds of kilometers away.
Problem 2: The eruption of Krakatoa in 1883 was one of the loudest sounds ever recorded at an estimated 235 decibels (dB) at the volcano. The sound of the eruption was heard about 3,000 miles away and at ~180 dB about 100 miles away. Calculate the intensity in watts per square meter of the sound at the volcano during the eruption given the following relationship between loudness level (D) and intensity (I):
`D=10log(I/(10^(-12)))`
Step 1 Identify the base of the logarithm
The base of the logarithm is 10. Because there is no explicit base, it is implied that the base is 10.
Step 2 Identify the known and unknown variables in the equation
The known variable is `D` and the unknown variable is `I`
Step 3. Rearrange and/or rewrite the equation to isolate the unknown variable as a log term
Start with the loudness equation:
`10log(I/(10^(-12)))=D`
Divide by 10 to isolate the log term:
`log(I/(10^(-12)))=D/(10)`
Use the Quotient Rule to separate the complex log argument:
`log(I)-log(10^(-12))=D/(10)`
Isolate the log term containing the unknown variable, `I`
`log(I)=D/(10)+log(10^(-12))`
Step 4. Plug the known values into the equation and simplify
`log(I)=D/(10)+log(10^(-12))`
Plug in the known variable, `D`
`log(I)=(235 "dB")/10 +log(10^(-12))`
Simplify. Remember that `log(10^x)=x`
`log(I)=23.5+ -12=11.5`
Step 5. Solve for the unknown variable
`log(I)=11.5`
Convert to exponent form using the identity: `log_b(a)=c leftrightarrow b^c=a`
`log(I)=11.5 leftrightarrow I=10^(11.5) (watts)/m^(2)`
Structural Geology - Power-law relationships of fractures
Fault offset of road during the Ridgecrest 2019 earthquake.
Provenance: USGS via https://commons.wikimedia.org/wiki/File:July_5,_2019,_Ridgecrest_earthquake_road_offset.jpg
Reuse: This item is in the public domain and maybe reused freely without restriction.
In structural geology, a brittle fault is defined as a discontinuity in rocks that has an appreciable displacement. Researchers have shown that the displacement of a fault is related to the length of the fault through the following 'power-law' relationship:
`log (D) = log (a) + k log (L)`
Where `D` = displacement
`a` = constant representing the width of a scaling relationship
`L` = Length of a fracture
`k` = constant for the power-law exponent
Problem 3: While mapping fractures in the field, you find that the displacement across a fault is 1.73 m. Assuming `a` = 0.0058 and `k` =1.19, use the logarithmic form of the power-law relationship to estimate the length of the fault.
Step 1. Identify the base of the logarithm
The base of the logarithm is 10. Because there is no explicit base, it is implied that the base is 10.
Step 2. Identify the known and unknown variables in the equation
The known variable is `D` and the unknown variable is `L`
Step 3. Rearrange and/or rewrite the equation to isolate the unknown variable as a log term.
Start with a logarithmic form of the power-law relationship
`log (a) + k log (L) = log (D) `
Subtract log (a) from both sides of the equation
`k log (L) = log (D) - log (a)`
Divide both sides by k
`log (L) = 1/k(log (D) - log (a)) `
Step 4. Plug the known values into the equation and simplify
Replace `k`, `D` and `a` with the appropriate values in the logarithmic equation:
`log (L) = 1/k(log (D) - log (a)) `
`log (L) = 1/1.19(log (1.73) - log (0.0058))`
Simplify the right side of the expression:`log (L) = 1/1.19(log (1.73) - log (0.0058))`
`log (L) = 0.840 (0.24 - (-2.24))`
`log (L) = 0.840 (2.48)`
`log (L) = 2.08`
Step 5. Solve for the unknown variable.
First, convert
`log (L) = 2.08` to exponential form
`L = 10^2.08`
The answer is 120 m
Power law relationships are typically derived from plotting graphs of two variables (e.g., displacement vs trace length) in logarithmic scales. A straight line can be fit through these data points to show the relationship.
Plot of maximum displacement versus trace length of faults.
Provenance: Alex Manda, East Carolina University
Reuse: This item is offered under a Creative Commons Attribution-NonCommercial-ShareAlike license http://creativecommons.org/licenses/by-nc-sa/3.0/ You may reuse this item for non-commercial purposes as long as you provide attribution and offer any derivative works under a similar license.
Geochemistry - Calculating equilibrium constants
Aridisol soils with halite precipitated at the surface in central Nevada
Provenance: Soil Science Society of America Marbut Memorial Slide Set https://www.soils.org/media-gallery/
Reuse: This item is offered under a Creative Commons Attribution-NonCommercial-ShareAlike license http://creativecommons.org/licenses/by-nc-sa/3.0/ You may reuse this item for non-commercial purposes as long as you provide attribution and offer any derivative works under a similar license.
Equilibrium constants,
K, are used in geochemistry to represent the equilibrium conditions of a reaction of interest and are compared to current conditions to understand how the reaction will proceed. In mineralogy, soils, and aquatic geochemistry, they are essential to understanding whether minerals will dissolve or precipitate or the acidity of a solution.
K values are derived from thermodynamic constants according to the reaction:
`ΔG_r=-RTln(K)`
where `ΔG_r` is the Gibbs free energy of the reaction, `R` is the gas constant, and `T` is the temperature in Kelvin.
Problem 4: Desert soils, called aridisols, accumulate salts including halite due to the arid environments they develop in. Halite easily dissolves when exposed to fresh water but will precipitate halite again when enough water evaporates, often at the surface. The reaction of interest in this case is the dissolution of halite according to the reaction `NaCl leftrightarrow Na^++Cl^-`.
Calculate the value of K for the halite dissolution reaction given that `ΔG_r` = -9.06 (kJ/mol), `R` = 8.31x10-3 kJ/mol-K, and `T` = 298K (25oC).
Including the gas constant and temperature into the equation above simplifies it to:
`ΔG_r=-2.48ln(K)`
Step 1. Identify the base of the logarithm
This equation uses the natural log (ln), which is log base e
Step 2. Identify the known and unknown variables in the equation
In the simplified equation, `ΔG_r` is known (-9.1 kJ/mol) and K is the unknown variable.
Step 3. Rearrange and/or rewrite the equation to isolate the unknown variable as a log term
Start with the simplified equation for calculating K:
`-2.48ln(K)=ΔG_r`
Isolate the term with K by dividing by -2.48:
`ln(K)=ΔG_r/(-2.48)`
Step 4. Plug the known values into the equation and simplify
`ln(K)=ΔG_r/(-2.48)`
Replace `ΔG_r` with -9.06
`ln(K)=(-9.06)/(-2.48)=3.65`
Step 5. Solve for the unknown variable
To solve for K, you must rewrite the logarithmic relationship in exponential form to isolate the unknown variable. Use the relation between logarithms and exponential expressions: `log_(b)(a)=c leftrightarrow b^c=a`
`ln(K)=3.6 leftrightarrow K=e^(3.65)`
And simplify the exponential expression:
`K=e^(3.65)=38.47`
Remember that K values do not have units. This positive number indicates that salt dissolves readily in water!
Scientific calculators will have an
ex or EXP option that you will use to evaluate the equation. Enter the following into your calculator:
e^(3.65)
The answer is 38.47.
Click on any box (called a cell) in the spreadsheet.
Type the following expression in the cell to solve the equation:
=EXP(3.65)
Press Enter to reveal your answer in the cell.
The answer is 38.47.
Sedimentology - Using the phi scale to calculate the size of sediments
Grain size frequency plot on phi size scale.
Provenance: Alex Manda, East Carolina University
Reuse: This item is offered under a Creative Commons Attribution-NonCommercial-ShareAlike license http://creativecommons.org/licenses/by-nc-sa/3.0/ You may reuse this item for non-commercial purposes as long as you provide attribution and offer any derivative works under a similar license.
Sedimentologists use the phi (φ) scale to describe the distribution of sediment sizes. The phi scale is logarithmic and thus a way to look at a wide range of particle sizes in a manageable form.
Problem 5: A sedimentologist went to a sandy beach to collect a sediment sample. After collecting a sediment sample, the sedimentologist created a size frequency plot on the phi scale. The plot revealed that most of the sediments had a phi value of 1. If the phi scale is related to the diameter of sediments by the following relationship, `φ=-log_(2)(d)`, where d is the diameter of the sediment in millimeters, use the phi value to calculate the size of sediment in millimeters.
Step 1. Identify the base of the logarithm
`φ=-log_(2)(d)` uses log base 2.
Step 2. Identify the known and unknown variables in the equation
The value of φ is 1, the unknown variable is the diameter of sediments (d).
Step 3. Rearrange and/or rewrite the equation to isolate the unknown variable as a log term
You cannot solve for a log base 2 equation directly on most scientific calculators. Instead, you must convert this expression to a common log base using the association:
`log_(b)(a)=((log_(10)(a))/(log_(10)(b)))` `φ=-log_(2)(d)`
`-φ=log_(10)(d)/log_(10)(2)`
`-φlog_(10)(2)=log_(10)(d)`
Step 4. Plug the known values into the equation and simplify.
In this case, φ = 1
So, from
`-φlog_(10)(2)=log_(10)(d)`
we get,
`-log_(10)(2)=log_(10)(d)`
After simplifying,
`-0.30103=log_(10)(d)`
Step 5. Solve for the unknown variable
From `-0.30103=log_(10)(d)`
`10^(-0.30103)=d`
The answer is 0.5 mm
Scientific calculators will have a 10
x option that you will use to evaluate the equation. Enter the following into your calculator:
10(-0.30103)
The answer is 0.4999 which rounds to 0.5 (one significant figure).
Click on any box (called a cell) in the spreadsheet.
Type the following expression in the cell to solve the equation:
=10^(-0.30103)
Press Enter to reveal your answer in the cell.
The answer is 0.4999 which rounds to 0.5 (one significant figure).
Seismology - The moment magnitude scale
Example earthquake magnitudes and energy release equivalents
Provenance: EarthScope Consortium adapted from original by Gavin Hayes.
Reuse: This item is offered under a Creative Commons Attribution-NonCommercial-ShareAlike license http://creativecommons.org/licenses/by-nc-sa/3.0/ You may reuse this item for non-commercial purposes as long as you provide attribution and offer any derivative works under a similar license.
The energy released by an earthquake can vary tremendously from one event to the next. Seismologists developed the logarithmic Moment Magnitude scale to make it easier to compare this energy (seismic moment) between different earthquakes. Understanding these relationships and the probability of a certain magnitude earthquake occurring is important to policy- and decision-making, particularly in areas more directly affected by larger earthquakes.
Problem 6: On April 27, 2024, an earthquake with a seismic moment of 1.6 x 1025 joules occurred beneath the Indian Ocean ~100km south of Banjar, Indonesia. Calculate the magnitude of this earthquake given the Moment Magnitude formula:
`M_w=2/3 * log(M_o) - 10.7`
where `M_w` is the moment magnitude, and `M_o` is the seismic moment (with units of joules) which is based on the rigidity, slip, and area of the rupture zone.
Step 1. Identify the base of the logarithm
The equation to use is `M_w=2/3 * log(M_o) - 10.7`. The base of the logarithm in this equation is 10. Because there is no explicit base, it is implied that the base is 10.
Step 2. Identify the known and unknown variables in the equation
There are two variables in this problem: `M_w` and `M_o`. `M_o` is known (=1.6 x 1025 joules) and `M_w` is unknown.
Step 3. Rearrange and/or rewrite the equation to isolate the unknown variable as a log term
The equation is already in a form where the unknown variable (`M_w`) is isolated. Because this is the value we are solving for, it does not need to be transformed into a log.
Step 4. Plug the known values into the equation and simplify
The equation to use is:
`M_w=2/3 * log(M_o)-10.7`
Plugging in the known `M_o` value:
`M_w=2/3 * log(1.6 xx 10^25)-10.7`
Simplify further by solving `log(1.6 xx 10^25)=25.2` and plugging back into the equation and remember for order of operations, multiplication is done before subtraction
`M_w=2/3 * 25.2 - 10.7=16.8 - 10.7=6.1` (to two significant figures)
Step 5. Solve for the unknown variable
You already solved for the unknown in the previous step: `M_w=6.1`.
Next steps
TAKE THE QUIZ!!
I think I'm competent with logarithms and I am ready to take the quiz! This link takes you to WAMAP. If your instructor has not given you instructions about WAMAP, you may not have to take the quiz.Or you can go back to the Logarithms explanation page.