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

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:

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.

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},

or

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

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.

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:

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. 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},

or

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:

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?

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.

log(an)

Source: https://www.facebook.com/MathAwesomeness/photos/a.342252885964433.1073741828.342251349297920/628770990645953/?type=3&theater

Engaging students: Computing logarithms with base 10

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 Katelyn Kutch. Her topic, from Precalculus: computing logarithms with base 10.

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How has this topic appeared in the news?

http://www.seeitmarket.com/the-log-blog-trading-with-music-and-logarithmic-scale-investing-14879/ . This website gives an insight into logarithms that many students would not know and I think that what is has to say is quite interesting. While this may not be a news article, it includes many methods in which logarithms can and are being used in the world. It also gives some insight into the history of logarithms. I feel like showing the students this website would get them interested in logarithms because they can see what logarithms can do, like tell us the magnitude of an earthquake on the Richter Scale. Students may not find logarithms interesting, but I feel like most would find this interesting.

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

http://mathequalslove.blogspot.com/2014/01/introducing-logarithms-with-foldables.html . This website gives multiples games that teachers can do with logarithms, not just base 10, but for all logarithms. The teacher had foldables that the students put their notes in for logarithms and personally, as a kinesthetic learner, that is something that I loved when teachers did it. It helped me visually put down the notes and it was something that I could keep to refer to. The teacher also had Log War, Log Bingo, and Log Speed Dating. Students always respond better when a sense of fun is involved in the lesson and this teacher proved that when one of her students asked about another game involving the subject. The games are ones that students interact with the teacher, with each other, and it enhances their own thinking because they are having to do calculations, correctly, in order to win the game. This seems like a wonderful website to pull from when wanting to do something fun with a lesson.

 

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What interesting things can you say about the people who contributed to the discovery and/or the development of this topic?

In 1614 a Scottish mathematician by the name of John Napier published his discovery for logarithms. Napier worked with an English mathematician by the name of Henry Briggs. The two of them adjusted Napier’s original logarithm to the form that we use today. After Napier passed away, Briggs continued their work alone and published, in 1624, a table of logarithms that calculated 14 decimal places for numbers between 1 and 20,000, and numbers between 90,000 and 100,000. In 1628 Adriaan Vlacq, a Dutch publisher, published a 10 decimal place table for values between 1 and 100,000, which included the values for 70,000 that were not previously published. Both men worked on setting up log trigonometric tables. Later, the notation Log(y) was adopted in 1675, by Leibniz, and soon after he connected Log(y) to the integral of dy/y.

 

 

Engaging students: Solving logarithmic equations

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 again comes from my former student Anna Park. Her topic: how to engage Algebra II or Precalculus students when solving logarithmic equations.

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Application:

 

The students will each be given a card with a) a logarithmic equation solution and b) a new logarithmic equation. The student that has a number one on the back of their card will begin the game. The student will stand up and tell the rest of the class what they have for b) the Log equation they have, then the student with the corresponding card will read their solution a) to the first students problem. If that student is correct they will read part b) the new log equation. Then another student that has the logarithmic solution will stand up and say their solution a) and then read their new log equation b). This will continue until the last student stands with their new equation and it loops back to student number one’s solution. This will end the game. This game requires students to solve logarithmic equations and recognize how to rewrite a logarithmic equation. There will be an appropriate amount of time before the game begins so the students can work backwards to find their logarithmic equation that matches their solution.

 

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History:

 

John Napier was the mathematician that introduced logarithms. The way he came up with logarithms is very fascinating, especially how long it took him to develop the logarithm table. He first published his work on logarithms in 1614. He published the findings under “A Description of the Wonderful Table of Logarithms.” He named them logarithms after two Greek words; logos, meaning proportion, and arithmos, meaning number. His discovery was based off of his imagination of two particles traveling along two parallel lines. One line had infinite length and the other had a finite length. He imagined both particles starting at the same horizontal positions with the same velocity. The first line’s velocity was proportional to the distance, which meant that the particle was covering equal distance in equal time. Whereas the second particle’s velocity was proportional with the distance remaining. His findings were that the distance not covered by the second line was the sine and the distance of the first line was the logarithm of the sine. This showed that the sines decreased and the logarithms increased. This also resulted in the sines decreasing in geometric proportion and the logarithms increasing in arithmetic proportion. He made his logarithm tables by taking increments of arc (theta) every minute, listing the sine of each minute by arc, and the corresponding logarithm. Completing his tables, Napier computed roughly ten million entries, and he selected the appropriate values. Napier said that his findings and completing this table took him about 20 years, which means he probably started his work in 1594.

Resource: http://www.maa.org/press/periodicals/convergence/logarithms-the-early-history-of-a-familiar-function-john-napier-introduces-logarithms

 

 

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Technology:

 

I have found that when it comes to remembering rules, sometime the cheesiest of songs help student’s to remember the rules. It is also a very good engage before the students start with the lesson. The chorus is typically the most important content for the student’s to remember. Here are two videos that would help the student’s to remember how to compute logarithms.

The first video is a song from Youtube set to the song Thriller by Michael Jackson. The song is produced very well and is very engaging throughout the whole song.

The Second video is of a student’s project  on Youtube of how to remember how to compute logarithms to the song Under the sea by the little mermaid. Though the production isn’t as good as the first video, the young girls do a good job at explaining how to solve logarithms.

 

 

My Favorite One-Liners: Part 80

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

Today’s awful pun comes courtesy of Math With Bad Drawings. Suppose we need to solve for x in the following equation:

2^{2x+1} = 3^{x}.

Naturally, the first step is taking the logarithm of both sides. But with which base? There are two reasonable options for most handheld scientific calculators: base-10 and base-e. So I’ll tell the class my preference:

I’m organic; I only use natural logs.

 

My Favorite One-Liners: Part 68

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

When discussing the Laws of Logarithms, I’ll make a big deal of the fact that one law converts a multiplication problem into a simpler addition problem, while another law converts exponentiation into a simpler multiplication problem.

After a few practice problems — and about 3 minutes before the end of class — I’ll inform my class that I’m about to tell the world’s worst math joke. Here it is:

After the flood, the ark landed, and Noah and the animals got out. And God said to Noah, “Go forth, be fruitful, and multiply.” So they disembarked.

Some time later, Noah went walking around and saw the two dogs with their baby puppies and the two cats with their baby kittens. However, he also came across two unhappy, frustrated, and disgruntled snakes. The snakes said to Noah, “We’re having some problems here; would you mind knocking down a tree for us?”

Noah says, “OK,” knocks down a tree, and goes off to continue his inspections.

Some time later, Noah returns, and sure enough, the two snakes are surrounding by baby snakes. Noah asked, “What happened?”

The snakes replied, “Well, you see, we’re adders. We need logs to multiply.”

After the laughter and groans subside, I then dismiss my class for the day:

Go forth, and multiply (pointing to the door of the classroom). For most of you, don’t be fruitful yet, but multiply. You’re dismissed.