Engaging students: Vectors in two dimensions

In my capstone class for future secondary math teachers, I ask my students to come up with ideas for engaging their students with different topics in the secondary mathematics curriculum. In other words, the point of the assignment was not to devise a full-blown lesson plan on this topic. Instead, I asked my students to think about three different ways of getting their students interested in the topic in the first place.

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission comes from my former student Sarah McCall. Her topic, from Precalculus: vectors in two dimensions.

What interesting (i.e., uncontrived) word problems using this topic can your students do now?

For such an applicable topic, I believe that it is beneficial to have students see how this might apply to their lives and to real world problems. I selected the following word problems because they are challenging, but I think it is necessary for students to be a little frustrated initially so that they are able to learn well and remember what they’ve learned.

1. A DC-10 jumbo jet maintains an airspeed of 550 mph in a southwesterly direction. The velocity of the jet stream is a constant 80 mph from the west. Find the actual speed and direction of the aircraft.

2. The pilot of an aircraft wishes to head directly east, but is faced with a wind speed of 40 mph from the northwest. If the pilot maintains an airspeed of 250 mph, what compass heading should be maintained? What is the actual speed of the aircraft?

3. A river has a constant current of 3 kph. At what angle to a boat dock should a motorboat, capable of maintaining a constant speed of 20 kph, be headed in order to reach a point directly opposite the dock? If the river is ½ a kilometer wide, how long will it take to cross?

Because these problems are difficult, students would be instructed to work together to complete them. This would alleviate some frustrations and “stuck” feelings by allowing them to ask for help. Ultimately, talking through what they are doing and successfully completing challenging problems will take students to a deeper level of involvement with their own learning.

How could you as a teacher create an activity or project that involves your topic?

I believe vectors are fairly easy to teach because there are so many real life applications of vectors. However, it can be difficult to get students initially engaged. For this activity, I would have students work in groups to complete a project inspired by Khan Academy’s videos on vector word problems. Students would split off into groups and watch each of the three videos on Khan Academy that have to do with applications of vectors in two dimensions. Using these videos as an example, students will be instructed to come up with a short presentation or video that teaches other students about vectors in two dimensions using real world applications and examples.

How has this topic appeared in pop culture (movies, TV, current music, video games, etc.)?

Immediately when I see vectors, I think of one specific movie quote from my late childhood that I’ll always remember. The villain named Vector from Despicable Me who “commits crimes with both direction AND magnitude” is a fellow math nerd and is therefore one of my favorite Disney villains of all time. So of course, I had to find the clip (linked below) because I think it is absolutely perfect for engaging students in a lesson about vectors as soon as they walk in the door, and it is memorable and educational. I would refer back to this video several times throughout the lesson and in future lessons because it is a catchy way to remember the two components to vectors. This would also be great to kick off a unit on scalars and vectors, because it would get kids laughing and therefore engaged, plus they will always remember the difference between a scalar and a vector (direction AND magnitude!).

References:

Engaging students: Graphing Sine and Cosine Functions

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission comes from my former student Saundra Francis. Her topic, from Precalculus: graphing sine and cosine functions.

How has this topic appeared in pop culture (movies, TV, current music, video games, etc.)?

To engage students you can have them record a song using the WavePad app. Have students choose a clip of their favorite song to record. Once they record the song they choose, the app will display the sound waves compiled that are used to create the sounds in the song the song. Students will realize that sound waves are in the form of the sine function. This will engage students since you would have related the topic of graphing sine and cosine functions to their favorite song. You could also have students create their own sounds and record them with the app to see the graph associated with the sound they made. Students can look at their sound and other classmates sounds are recognize differences in the waves, you can relate this to the equation f(x)=asin(bx+c)+d. You can them work with students to discover what the constants terms mean in relation to the parent function of sine.

How has this topic appeared in high culture (art, classical music, theater, etc.)?

Sine waves are the basis of sound. Have a piece by Beethoven playing while students are entering the classroom. Tell students that Beethoven was able to create music while he was in the process of becoming deaf. Ask students how they think Beethoven was able to create music in spite of that set back. After you have students share some answers show them the video above which explains how Beethoven’s music (all music) is related to sine waves. The music of his “Moonlight Sonata” is explained using math in the TED-Ed video. While Beethoven did not use this method to create his music he said that he knew what the music looked like. This will show students an example of how sine graphs are used in real life and get them interested in graphing sine and cosine functions.

How can technology (YouTube, Khan Academy [khanacademy.org], Vi Hart, Geometers Sketchpad, graphing calculators, etc.) be used to effectively engage students with this topic? Note: It’s not enough to say “such-and-such is a great website”; you need to explain in some detail why it’s a great website.

Students will be given a TI-Nspire calculator in order to discover how changing the amplitude, period, horizontal shift, or vertical shift changes the equation of sine. Students will start with the graph of f(x)=sinx. They will then manipulate the graph on the calculator to change the function. Have them move the function up and down, right and left, and work with the slope of sinx and the slope of the x. Have students write dawn some of their new functions and sketch the graph. They will then compare how changing the graph effects the equation of f(x)=sinx. Introduce f(x)=asin(bx+c)+d . Give students some time to compare the functions that they created to the formula and describe how each constant changes the graph. Students will hopefully discover how the function f(x)=asin(bx+c)+d relates to amplitude, period, horizontal shift, and vertical shift.

References
1. https://education.ti.com/en/timathnspired/us/detail?id=4E9BA7808CA74F6599BD5EA2037C088A&t=C52AEC55A39243D182772F76318B901C
2. https://www.smore.com/gy9h4-sine-waves-and-music
3. https://www.youtube.com/watch?time_continue=16&v=zAxT0mRGuoY

Engaging students: Dot product

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission comes from my former student Trent Pope. His topic, from Precalculus: computing a dot product.

What interesting (i.e., uncontrived) word problems using this topic can your students do now? (You may find resources such as http://www.spacemath.nasa.gov to be very helpful in this regard; feel free to suggest others.)

This website gives an example of a word problem that students could solve, and it has real-world applications. It is not a complete worksheet for students to work on. The teacher would have to create more word problems incorporating the idea of this website. The example on this web page is that you are a local store owner and are selling beef, chicken, and vegetable pies 3 days a week. The owner has a list of how many pies he sells a day and how much they cost. The cost of beef pies are $3, chicken pies are $4, and vegetable pies are $2. On Monday he sells 13 beef, 8 chicken, and 6 vegetable pies. Tuesday he sells 9 beef, 7 chicken, and 4 vegetable pies. Finally, on Thursday the owner sold 15 beef, 6 chicken, and 3 vegetable pies. Now, let’s think about how we can solve for the total number of sales for Monday. First, we would solve for the sales of the beef pies by multiplying the price of the pie and the number we sold. Then we would do the same for chicken and vegetable pies. After finding the sales of the three pies, we would add up sales to get the total amount for the day. In this case, we would get $83 of sales on Monday. The students would do the same thing for the other days the store is open. This is an example of the dot product of matrices in a word problem.

https://www.mathsisfun.com/algebra/matrix-multiplying.html

How could you as a teacher create an activity or project that involves your topic?

An idea I was able to see in an actual classroom during observation this week was the use of Fantasy Football in matrices. A teacher at Lake Dallas High School has her classes in a Football Fantasy League competing against each other. The way they started this activity is that the students have to keep up with the points that their teams are earning. They are doing this by the information the teacher gives them about how to score their players. Each class chooses one quarterback, running back, wide receiver, kicker, and defense to represent their team. The point system is the same as in the online fantasy. For instance, Aaron Rodgers, quarterback for the Green Bay Packers, throws for 300 yards, two touchdowns, and one interception. The points Rodgers earns you for the week comes from taking the several yards and multiplying by the points earned for each yard. Then, do the same for touchdowns and interceptions. After computing this, you will then add the numbers up to get the total points you receive from Aaron for the week. This is using dot product because we have two matrices, which are the stats that the player receives in the game, and the points you get for those same stats. By doing this activity, the students would be working on this aspect of pre-calculus for the entire football season.

Graphing calculators would be a great way to use technology to teach this topic. When computing the dot product of two matrices, there are two ways to do it. One is by hand and the other is a calculator. As the teacher, it would be more efficient for you to see how students are learning the material by having them compute it by hand, but no student wants to do that with every problem. A way the teacher could incorporate solving for the dot product using a calculator in an engaging way would be to have students complete a scavenger hunt. In the scavenger hunt, students will have to solve problems of the dot product to get the next clue and move on to the next. The idea of this would be for the students to show that they can work the calculator and actually get answers. You could have anywhere from five to ten questions for them to solve and decoy answers throughout the room with little mishaps. This would get the students up and moving for this activity

Decimal Approximations of Logarithms (Part 5)

While some common (i.e., base-10) logarithms work out evenly, like $\log_{10} 10,000$ , most do not. Here is the typical output when a scientific calculator computes a logarithm:

In today’s post, I’ll summarize the past few posts to describe how talented Algebra II students, who have just been introduced to logarithms, can develop proficiency with the Laws of Logarithms while also understanding that the above answer is not just a meaningless jumble of digits. The only tools students will need are

The Laws of Logarithms,
A hand-held scientific calculator, and
Access to the Wolfram Alpha website or a Microsoft Excel spreadsheet that I’ve written.

To estimate $\log_{10} 5.1264$ , Algebra II students can try to find a power of 5.1264 that is close to a power of 10. In principle, this can be done by just multiplying by $5.1264$ until an answer decently close to $5.1264$ arises. For the teacher who’s guiding students through this exploration, it might be helpful to know the answer ahead of time.

One way to do this is to use Wolfram Alpha to find the convergents of $\log_{10} 5.1264$ . If you click this link, you’ll see that I entered

Convergents[Log[10,5.1264],15]

A little explanation is in order:

Convergents, predictably, is the Mathematica command for finding the convergents of a given number.
Log[10,5.1264] is the base-10 logarithm of 5.1264. By contrast, Log[5.1264] is the natural logarithm of 5.1264. Mathematica employs the convention of that $\log$ should be used for natural logarithms instead of $\ln$ , as base-10 logarithms are next to useless for mathematical researchers. That said, I freely concede that this convention is confusing to students who grew up thinking that $\log$ should be used for base-10 logarithms and $\ln$ for natural logarithms. (See also my standard joke about using natural logarithms.) Naturally, the $5.1264$ can be changed for other logarithms.
The 15 means that I want Wolfram Alpha to give me the first 15 convergents of $\log_{10} 5.1264$ . In most cases, that’s enough terms to provide a convergent whose denominator is at least six digits long. In the rare instance when this doesn’t happen, a number larger than 15 can be entered.

From Wolfram Alpha, I see that $\displaystyle \frac{22}{31}$ is the last convergent with a numerator less than 100. For the purposes of this exploration, I interpret these fractions as follows:

The best suitable power of $5.1264$ for an easy approximation on a scientific calculation will be $(5.1264)^{31}$ . In this context, “best” means something that’s close to a power of 10 but less than $10^{100}$ . Students entering $(5.1264)^{31}$ into a calculator will find

$(5.1264)^{31} \approx 1.009687994 \times 10^{22}$

$(5.1264)^{31} \approx 10^{22}$

In other words, the denominator of the convergent $\displaystyle \frac{22}{31}$ gives the exponent for $5.1264$ , while the numerator gives the exponent for the approximated power of 10. Continuing with the Laws of Logarithms,

$\log_{10} (5.1264)^{31} \approx \log_{10} 10^{22}$

$31 \log_{10} 5.1264 \approx 22$

$\log_{10} 5.1264 \approx \displaystyle \frac{22}{31}$

$\log_{10} 5.1264 \approx 0.709677\dots$

A quick check with a calculator shows that this approximation is accurate to three decimal places. This alone should convince many students that the above apparently random jumble of digits is not so random after all.

While the above discussion should be enough for many students, some students may want to know how to find the rest of the decimal places with this technique. To answer this question, we again turn to the convergents of $\log_{10} 5.1264$ from Wolfram Alpha. From this list, we see that $\displaystyle \frac{89,337}{125,860}$ is the first convergent with a denominator at least six digits long. The student therefore has two options:

Option #1. Ask the student to use Wolfram Alpha to raise $5.1264$ to the denominator of this convergent. Surprisingly to the student, but not surprisingly to the teacher who knows about this convergent, the answer is very close to a power of 10: $10^{89,337}$ . The student can then use the Laws of Logarithms as before:

$\log_{10} (5.1264)^{125,860} \approx \log_{10} 10^{89,337}$

$125,860 \log_{10} 5.1264 \approx 89,337$

$\log_{10} 5.1264 \approx \displaystyle \frac{89,337}{125,860}$

$\log_{10} 5.1264 \approx 0.70981249006\dots$ ,

which matches the output of the calculator.

Option #2. Ask the student to “trick” a hand-held calculator into finding $(5.1264)^{125,860}$ . This option requires the use of the convergent with the largest numerator less than 100, which was $\displaystyle \frac{22}{31}$ .

Option #2A: Use the Microsoft Excel spreadsheet that I’ve written to perform the calculations that follow.
Option #2B: The student divides the smaller denominators into the larger denominator and finds the quotient and remainder. It turns out that $125,860 = 31 \times 4060 + 0$ . (This is a rare case where there happens to be no remainder.) Next, the student uses a hand-held calculate to compute

$\displaystyle \left( \frac{(5.1264)^{31}}{10^{22}} \right)^{4060} \times (5.1264)^0$

In this example, the $\times (5.1264)^0$ is of course superfluous, but I include it here to show where the remainder should be placed. Entering this in a calculator yields a result that is close to $10^{17}$ . (The teacher should be aware that some of the last few digits may differ from the more precise result given by Wolfram Alpha due to round-off error, but this discrepancy won’t matter for the purposes of the student’s explorations.) In other words,

$\displaystyle \left( \frac{(5.1264)^{31}}{10^{22}} \right)^{4060} \times (5.1264)^0 \approx 10^{17}$ ,

which may be rearranged as

$(5.1264)^{125,860} \approx 10^{89,337}$

after using the Laws of Exponents. From this point, the derivation follows the steps in Option #1.

Decimal Approximations of Logarithms (Part 4)

While some common (i.e., base-10) logarithms work out evenly, like $\log_{10} 10,000$ , most do not. Here is the typical output when a scientific calculator computes a logarithm:

To a student first learning logarithms, the answer is just an apparently random jumble of digits; indeed, it can proven that the answer is irrational. With a little prompting, a teacher can get his/her students wondering about how people 50 years ago could have figured this out without a calculator. This leads to a natural pedagogical question:

Can good Algebra II students, using only the tools at their disposal, understand how decimal expansions of base-10 logarithms could have been found before computers were invented?

Here’s a trial-and-error technique — an exploration activity — that is within the grasp of Algebra II students. It’s simple to understand; it’s just a lot of work. The only tools that are needed are

The Laws of Logarithms
A hand-held scientific calculator
Access to the Wolfram Alpha website (optional)
A lot of patience multiplying $x$ by itself repeatedly in a quest to find integer powers of $x$ that are close to powers of $10$ .

In the previous post in this series, we found that

$3^{153} \approx 10^{73}$

and

$3^{323,641} \approx 10^{154,416}$ .

Using the Laws of Logarithms on the latter provides an approximation of $\log_{10} 3$ that is accurate to an astounding ten decimal places:

$\log_{10} 3^{323,641} \approx \log_{10} 10^{154,416}$

$323,641 \log_{10} 3 \approx 154,416$

$\log_{10} 3 \approx \displaystyle \frac{154,416}{323,641} \approx 0.477121254723598\dots$ .

Compare with:

$\log_{10} 3 \approx 0.47712125471966\dots$

Since hand-held calculators will generate identical outputs for these two expressions (up to the display capabilities of the calculator), this may lead to the misconception that the irrational number $\log_{10} 3$ is actually equal to the rational number $\displaystyle \frac{154,416}{323,641}$ , so I’ll emphasize again that these two numbers are not equal but are instead really, really close to each other.

We now turn to a question that was deferred in the previous post.

Student: How did you know to raise 3 to the 323,641st power?

Teacher: I just multiplied 3 by itself a few hundred thousand times.

Student: C’mon, really. How did you know?

While I don’t doubt that some of our ancestors used this technique to find logarithms — at least before the discovery of calculus — today’s students are not going to be that patient. Instead, to find suitable powers quickly, we will use ideas from the mathematical theory of continued fractions: see Wikipedia, Mathworld, or this excellent self-contained book for more details.

To approximate $\log_{10} x$ , the technique outlined in this series suggests finding integers $m$ and $n$ so that

$x^n \approx 10^m$ ,

or, equivalently,

$\log_{10} x^n \approx \log_{10} 10^m$

$n \log_{10} x \approx m$

$\log_{10} x \approx \displaystyle \frac{m}{n}$ .

In other words, we’re looking for rational numbers that are reasonable close to $\log_{10} x$ . Terrific candidates for such rational numbers are the convergents to the continued fraction expansion of $\log_{10} x$ . I’ll defer to the references above for how these convergents can be computed, so let me cut to the chase. One way these can be quickly obtained is the free website Wolfram Alpha. For example, the first few convergents of $\log_{10} 3$ are

$\displaystyle \frac{1}{2}, \frac{10}{21}, \frac{21}{44}, \frac{52}{109}, and \frac{73}{153}$ .

A larger convergent is $\frac{154,416}{323,641}$ , our familiar friend from the previous post in this series.

As more terms are taken, these convergents get closer and closer to $\log_{10} 3$ . In fact:

Each convergent is the best possible rational approximation to $\log_{10} 3$ using a denominator that’s less than the denominator of the next convergent. For example, the second convergent $\displaystyle \frac{10}{21}$ is the closest rational number to $\log_{10} 3$ that has a denominator less than $44$ , the denominator of the third convergent.
The convergents alternate between slightly greater than $\log_{10} 3$ and slightly less than $\log_{10} 3$ .
Each convergent $\displaystyle \frac{m}{n}$ is guaranteed to be within $\displaystyle \frac{1}{n^2}$ of $\log_{10} 3$ . (In fact, if $\displaystyle \frac{m}{n}$ and $\displaystyle \frac{p}{q}$ are consecutive convergents, then $\displaystyle \frac{m}{n}$ is guaranteed to be within $\displaystyle \frac{1}{nq}$ of $\log_{10} 3$ .)
As a practical upshot of the previous point: if the denominator of the convergent $\displaystyle \frac{m}{n}$ is at least six digits long (that is, greater than $10^5$ ), then $\displaystyle \frac{m}{n}$ must be within $\displaystyle \frac{1}{(10^5)^2} = 10^{-10}$ of $\log_{10} 3$ … and it’ll probably be significantly closer than that.

So convergents provide a way for teachers to maintain the illusion that they found a power like $3^{323,641}$ by laborious calculation, when in fact they were quickly found through modern computing.

Decimal Approximations of Logarithms (Part 3)

While some common (i.e., base-10) logarithms work out evenly, like $\log_{10} 10,000$ , most do not. Here is the typical output when a scientific calculator computes a logarithm:

Can good Algebra II students, using only the tools at their disposal, understand how decimal expansions of base-10 logarithms could have been found before computers were invented?

Here’s a trial-and-error technique — an exploration activity — that is within the grasp of Algebra II students. It’s simple to understand; it’s just a lot of work.

To approximate $\log_{10} x$ , look for integer powers of $x$ that are close to powers of 10.

In the previous post in this series, we essentially used trial and error to find such powers of 3. We found

$3^{153} \approx 9.989689095 \times 10^{72} \approx 10^{73}$ ,

from which we can conclude

$\log_{10} 3^{153} \approx \log_{10} 10^{73}$

$153 \log_{10} 3 \approx 73$

$\log_{10} 3 \approx \displaystyle \frac{73}{153} \approx 0.477124$ .

This approximation is accurate to five decimal places.

By now, I’d imagine that our student would be convinced that logarithms aren’t just a random jumble of digits… there’s a process (albeit a complicated process) for obtaining these decimal expansions. Of course, this process isn’t the best process, but it works and it only uses techniques at the level of an Algebra II student who’s learning about logarithms for the first time.

If nothing else, hopefully this lesson will give students a little more appreciation for their ancestors who had to perform these kinds of calculations without the benefit of modern computing.

We also saw in the previous post that larger powers can result in better and better approximation. Finding suitable powers gets harder and harder as the exponent gets larger. However, when a better approximation is found, the improvement can be dramatic. Indeed, the decimal expansion of a logarithm can be obtained up to the accuracy of a hand-held calculator with a little patience. For example, let’s compute

$3^{323,641}$

Predictably, the complaint will arise: “How did you know to try $323,641$ ?” The flippant and awe-inspiring answer is, “I just kept multiplying by 3.”

I’ll give the real answer that question later in this series.

Postponing the answer to that question for now, there are a couple ways for students to compute this using readily available technology. Perhaps the most user-friendly is the free resource Wolfram Alpha:

$3^{323,641} \approx 9.999970671 \times 10^{154,415} \approx 10^{154,416}$ .

That said, students can also perform this computation by creatively using their handheld calculators. Most calculators will return an overflow error if a direct computation of $3^{323,641}$ is attempted; the number is simply too big. A way around this is by using the above approximation $3^{153} \approx 10^{73}$ , so that $3^{153}/10^{73} \approx 1$ . Therefore, we can take large powers of $3^{153}/10^{73}$ without worrying about an overflow error.

In particular, let’s divide $323,641$ by $153$. A little work shows that

$\displaystyle \frac{323,641}{153} = \displaystyle 2115 \frac{46}{153}$ ,

$323,641 = 153 \times 2115 + 46$ .

This suggests that we try to compute

$\displaystyle \left( \frac{3^{153}}{10^{73}} \right)^{2115} \times 3^{46}$ ,

and a hand-held calculator can be used to show that this expression is approximately equal to $10^{21}$ . Some of the last few digits will be incorrect because of unavoidable round-off errors, but the approximation of $10^{21}$ — all that’s needed for the present exercise — will still be evident.

By the Laws of Exponents, we see that

$\displaystyle \left( \frac{3^{153}}{10^{73}} \right)^{2115} \times 3^{46} \approx 10^{21}$

$\displaystyle \frac{3^{153 \times 2115 + 46}}{10^{73 \times 2115}} \approx 10^{21}$

$\displaystyle \frac{3^{323,641}}{10^{154,395}} \approx 10^{21}$

$3^{323,641} \approx 10^{154,395} \times 10^{21}$

$3^{323,641} \approx 10^{154,395+21}$

$3^{323,641} \approx 10^{154,416}$ .

Whichever technique is used, we can now use the Laws of Logarithms to approximate $\log_{10} 3$ :

$\log_{10} 3^{323,641} \approx \log_{10} 10^{154,416}$

$323,641 \log_{10} 3 \approx 154,416$

$\log_{10} 3 \approx \displaystyle \frac{154,416}{323,641} \approx 0.477121254723598\dots$ .

This approximation matches the decimal expansion of $\log_{10} 3$ to an astounding ten decimal places:

$\log_{10} 3 \approx 0.47712125471966\dots$

Summarizing, Algebra II students can find the decimal expansion of $\log_{10} x$ can be found up to the accuracy of a hand-held scientific calculator. The only tools that are needed are

The Laws of Logarithms
A hand-held scientific calculator
Access to the Wolfram Alpha website (optional)
A lot of patience multiply $x$ by itself repeatedly in a quest to find integer powers of $x$ that are close to powers of $10$ .

While I don’t have a specific reference, I’d be stunned if none of our ancestors tried something along these lines in the years between the discovery of logarithms (1614) and calculus (1666 or 1684).

Decimal Approximations of Logarithms (Part 2)

While some common (i.e., base-10) logarithms work out evenly, like $\log_{10} 10,000$ , most do not. Here is the typical output when a scientific calculator computes a logarithm:

Can good Algebra II students, using only the tools at their disposal, understand how decimal expansions of base-10 logarithms could have been found before computers were invented?

Here’s a trial-and-error technique — an exploration activity — that is within the grasp of Algebra II students. It’s simple to understand; it’s just a lot of work. While I don’t have a specific reference, I’d be stunned if none of our ancestors tried something along these lines in the years between the discovery of logarithms (1614) and calculus (1666 or 1684).

To approximate $\log_{10} x$ , look for integer powers of $x$ that are close to powers of 10.

I’ll illustrate this idea with $\log_{10} 3$ .

$3^1 = 3$

$3^2 = 9$

Not bad… already, we’ve come across a power of 3 that’s decently close to a power of 10. We see that

$3^2 = 9 < 10^1$

and therefore

$\log_{10} 3^2 < 1$

$2 \log_{10} 3< 1$

$\log_{10} 3< \displaystyle \frac{1}{2} = 0.5$

Let’s keep going. We just keep multiplying by 3 until we find something close to a power of 10. In principle, these calculations could be done by hand, but Algebra II students can speed things up a bit by using their scientific calculators.

$3^3 = 27$

$3^4 = 81$

$3^5 = 243$

$3^6 = 729$

$3^7 = 2,187$

$3^8 = 6,561$

$3^9 = 19,683$

$3^{10} = 59,049$

$3^{11} = 177,147$

$3^{12} = 531,441$

$3^{13} = 1,594,323$

$3^{14} = 4,782,969$

$3^{15} = 14,348,907$

$3^{16} = 43,046,721$

$3^{17} = 129,140,163$

$3^{18} = 387,420,489$

$3^{19} = 1,162,261,467$

$3^{20} = 3,486,784,401$

$3^{21} = 10,460,353,203$

This looks pretty good too. (Students using a standard ten-digit scientific calculator, of course, won’t be able to see all 11 digits.) We see that

$3^{21} > 10^{10}$

and therefore

$\log_{10} 3^{21} > \log_{10} 10^{10}$

$21 \log_{10} 3 > 10$

$\log_{10} 3 > \displaystyle \frac{10}{21} = 0.476190\dots$

Summarizing our work so far, we have

$0.476190\dots < \log_{10} 3 < 0.5$ .

We also note that this latest approximation actually gives the first two digits in the decimal expansion of $\log_{10} 3$ .

To get a better approximation of $\log_{10} 3$ , we keep going. I wouldn’t blame Algebra II students a bit if they use their scientific calculators for these computations — but, ideally, they should realize that these calculations could be done by hand by someone very persistent.

$3^{22} = 31,381,059,609$

$3^{23} = 94,143,178,827$

$3^{24} = 282,429,536,481$

$3^{25} = 847,288,609,443$

$3^{26} = 2,541,865,828,329$

$3^{27} = 7,625,597,484,987$

$3^{28} = 22,876,792,454,961$

$3^{29} = 68,630,377,364,883$

$3^{30} = 205,891,132,094,649$

$3^{31} = 617,673,396,283,947$

$3^{32} = 1,853,020,188,851,841$

$3^{33} = 5,559,060,566,555,523$

$3^{34} = 16,677,181,699,666,569$

$3^{35} = 50,031,545,098,999,707$

$3^{36} = 150,094,635,296,999,121$

$3^{37} = 450,283,905,890,997,363$

$3^{38} = 1,350,851,717,672,992,089$

$3^{39} = 4,052,555,153,018,976,267$

$3^{40} = 12,157,665,459,056,928,801$

$3^{41} = 36,472,996,377,170,786,403$

$3^{42} = 109,418,989,131,512,359,209$

$3^{43} = 328,256,967,394,537,077,627$

$3^{44} = 984,770,902,183,611,232,881$

Using this last line, we obtain

$3^{44} < 10^{21}$

and therefore

$\log_{10} 3^{44} < \log_{10} 10^{21}$

$44 \log_{10} 3 < 21$

$\log_{10} 3 < \displaystyle \frac{21}{44} = 0.477273\dots$

Summarizing our work so far, we have

$0.476190\dots < \log_{10} 3 < 0.477273\dots$ .

A quick check with a calculator shows that $\log_{10} 3 = 0.477121\dots$ . In other words,

This technique actually works!
This last approximation of $0.477273\dots$ actually produced the first three decimal places of the correct answer!

With a little more work, the approximations

$3^{109} \approx 1.014417574 \times 10^{52} > 10^{52}$

$3^{153} \approx 9.989689095 \times 10^{72} < 10^{73}$

can be found, yielding the tighter inequalities

$\displaystyle \frac{52}{109} < \log_{10} 3 < \displaystyle \frac{73}{153}$ ,

$0.477064\dots < \log_{10} 3 < 0.477124$ .

Now we’re really getting close… the last approximation is accurate to five decimal places.

Decimal Approximations of Logarithms (Part 1)

My latest article on mathematics education, titled “Developing Intuition for Logarithms,” was published this month in the “My Favorite Lesson” section of the September 2018 issue of the journal Mathematics Teacher. This is a lesson that I taught for years to my Precalculus students, and I teach it currently to math majors who are aspiring high school teachers. Per copyright law, I can’t reproduce the article here, though the gist of the article appeared in an earlier blog post from five years ago.

Rather than repeat the article here, I thought I would write about some extra thoughts on developing intuition for logarithms that, due to space limitations, I was not able to include in the published article.

While some common (i.e., base-10) logarithms work out evenly, like $\log_{10} 10,000$ , most do not. Here is the typical output when a scientific calculator computes a logarithm:

Can good Algebra II students, using only the tools at their disposal, understand how decimal expansions of base-10 logarithms could have been found before computers were invented?

Students who know calculus, of course, can do these computations since

$\log_{10} x = \displaystyle \frac{\ln x}{\ln 10}$ ,

and the Taylor series

$\ln (1+t) = t - \displaystyle \frac{t^2}{2} + \frac{t^3}{3} - \frac{t^4}{4} + \dots$ ,

a standard topic in second-semester calculus, can be used to calculate $\ln x$ for values of $x$ close to 1. However, a calculation using a power series is probably inaccessible to bright Algebra II students, no matter how precocious they are. (Besides, in real life, calculators don’t actually use Taylor series to perform these calculations; see the article CORDIC: How Hand Calculators Calculate, which appeared in College Mathematics Journal, for more details.)

In this series, I’ll discuss a technique that Algebra II students can use to find the decimal expansions of base-10 logarithms to surprisingly high precision using only tools that they’ve learned in Algebra II. This technique won’t be very efficient, but it should be completely accessible to students who are learning about base-10 logarithms for the first time. All that will be required are the Laws of Logarithms and a standard scientific calculator. A little bit of patience can yield the first few decimal places. And either a lot of patience, a teacher who knows how to use Wolfram Alpha appropriately, or a spreadsheet that I wrote can be used to obtain the decimal approximations of logarithms up to the digits displayed on a scientific calculator.

I’ll start this discussion in my next post.