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I'm going to teach exponentiation to my 7th grade math students, and I like the exponentiation rules to arise in a natural way (through real-world problems or meaningful math problems). Do you have any such problem in mind?

Thanks.

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    $\begingroup$ Something with orders of magnitude/scales/standard form? $\endgroup$ – nickjamesuk Jan 12 '15 at 15:43
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    $\begingroup$ What about simply pointing out how one multiplies $7000$ and $400$? -- Multiply $7$ and $4$ and then add $3+2=5$ zeros to the result. Similarly with any pair of (rounded off) large numbers, such as occurs in considerations of how far stars are, numbers of molecules in a cup of water, etc. At least for me (and most of my school classmates), this well known shortcut of adding zeros, and using scientific notation that one starts seeing in science classes and in everyday reading, seemed to do well in motivating general exponent behavior. $\endgroup$ – Dave L Renfro Jan 12 '15 at 17:08
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    $\begingroup$ Whatever problems you adopt, I'd suggest that you discuss and work them out with the class before introducing the exponential notation. Save the new notation for when they have a pain point that will be relieved by it, eg, when you ask them how many different 30 number combinations are possible with some 1-60 padlock. The prospect of writing 60*...*60 30 times should elicit some groans. The new notation will be a relief, and their adoption of it will be motivated by the same reasons as its original creation. $\endgroup$ – NiloCK Jan 12 '15 at 17:20
  • $\begingroup$ Did anyone tried the combinatorial interpretation in terms of functions, that is, $|A|^{|B|}$ is the number of functions from $B$ to $A$? If yes, what were the results? (Of course, I would not start like this, but rather like @Toscho suggested and get to functions/mappings much, much later.) $\endgroup$ – dtldarek Jan 12 '15 at 20:34
  • $\begingroup$ Consider if that's really a good use of time? Generally I find that the traditional format is best: definitions first, then theorems, proofs, and applications. Doing the discovery-based method can quickly suck up class time and leaves students muddled about exactly what has been shown and what they should know at a particular point. At the end, using the rules to deal with operations and problems in scientific notation is a fine capstone. $\endgroup$ – Daniel R. Collins Sep 21 '15 at 6:02
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might i suggest a different approach that I have found very helpful when teaching about exponentiation rather than real world examples? I have found that until students understand why a rule works (i.e. the derivation or something similar), they won't be able to understand how to use it.

Instead of teaching the rules to the students, have them expand all exponents out by hand and combine like factors just like they have (hopefully) been doing and this should lead them to discover the rules for themselves. If they ever forget a rule, they can just go back to how they discovered them, by expanding out exponents, and essentially "derive" the rule right there. so for example present them this problem:

$$4x^4y\cdot3x^5y^2$$ Which they can expand to $$4x^4y\cdot3x^5y^2 = 4 \cdot x \cdot x \cdot x \cdot x \cdot y \cdot 3 \cdot x \cdot x \cdot x \cdot x \cdot x \cdot y \cdot y $$ And simplifying yields

$$12x^9y^3$$

The same idea works exactly the same for each of the other exponent rules so for instance for the division/subtraction rule (whatever it is called:))

$$\frac{12x^5y^8}{2x^2y^3} = \frac{12 \cdot x \cdot x \cdot x \cdot x \cdot x \cdot y \cdot y \cdot y \cdot y \cdot y \cdot y \cdot y \cdot y}{2 \cdot x\cdot x \cdot y \cdot y \cdot y} = 6x^3y^5$$

By expanding out the exponents it is a lot easier to see what cancels and what can be combined. Doing this all by hand is a huge pain and should motivate your students to try and figure out the rules as fast as possible. Reinforcing what it actually means to raise something to an exponent by having them do this all manually, will certainly help down the line when the expressions/equations get more complicated.

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  • $\begingroup$ This is what I do, too. Huge lights went on behind my students eyes when I presented them with this explanation of why the exponent rules work the way they do. $\endgroup$ – David Steinberg Jan 15 '15 at 21:16
  • $\begingroup$ I agree with this 100%. I work as a math tutor and EVERY TIME exponents come up, you end up going through all the rules again... I tell them they need a way to "invent" the rules themselves, simple examples as indicated above, when do we add, when do we subtract, when do we multiply. $\endgroup$ – Tom De Vries Feb 3 '15 at 15:58
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Here's something I used to do in college algebra and precalculus classes, from the mid 1980s to the mid 2000s, which has the additional advantage of being an example in which estimation is used. I think you can adapt this to your case. (Use metric system units if appropriate.) Of course, you'll want to go a lot slower than I do below, which is written for the teacher, not the student.

Question: How many times would one theoretically have to successively fold a sheet of paper in order for the (theoretical) folded thickness to equal the distance from the Earth to the Moon?

The thickness of $300$ sheets is about an inch. You can hold up a book to show how to come up with this. The moon's distance is about $240000$ miles $= 24 \times 10^4$ miles $= 24 \times 10^4 \times 5280$ feet, which is about $25 \times 5 \times 10^4 \times 10^3$ feet $= 125 \times 10^7$ feet $= 125 \times 10^7 \times 12$ inches, which is about $100 \times 10^7 \times 10 = 10^{10}$ inches $= 300 \times 10^{10}$ sheet-thicknesses $= 3 \times 10^{12}$ sheet-thicknesses, so we need to know how many doublings gives this last number. $10$ doublings gives us almost exactly $1000 = 10^3$ (only a $2$% error), so $20$ doublings gives $10^{6},$ $30$ doublings gives $10^{9},$ $40$ doublings gives $10^{12},$ and two more doublings after this (for a total of $42$ doublings) gives us $4 \times 10^{12} > 3 \times 10^{12}.$ You can also use $2^{10}$ is approximately $10^3$ (i.e. $10$ doublings has the effect of approximately multiplying by $1000)$ to estimate other situations with geometric growth when the doubling time (or half-life) is known.

Also worth pursuing is how insensitive the final result is to the approximations used. For example, if the cummulative effect of all the estimates is a factor of $2,$ then the number of foldings is off by $1$ (i.e. it's between $41$ and $43).$ If the cummulative effect of all the estimates is a factor of $4,$ then the number of foldings is off by $2$ (i.e. it's between $40$ and $44).$

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  • $\begingroup$ 200 sheets / inch is much more common than 300 sheets per inch. One American paper-folder is famous for making extremely complicated origami models, out of small sheets of paper. Other people had trouble making the models so small. It turned out that he had a source of origami-quality paper that was 250 sheets / inch, instead of the more usual 200 sheets / inch. $\endgroup$ – Jasper Jan 12 '15 at 19:38
  • $\begingroup$ @Jasper: I just checked a random book on my shelf (only a handful where I'm at, over 99.7% to 99.8% of my math books are at home), Bennett/Nelson's Mathematics for Elementary Teachers (4th edition), and using a ruler I have with me, I'm getting 650 (maybe even up to 680 pages) for an inch, at least when I press the pages tightly together. The 300 figure came from an old post of mine (I've written about this example several times in the past 10 or so years), and I think it originally came from a guess students and I made in class one time (without a ruler, just estimating). $\endgroup$ – Dave L Renfro Jan 12 '15 at 19:49
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You could use combinatorics:

How many words with 4 letters of 26 letter alphabet do exist? If you have the answer: How many passphrases of two word with 4 letters of the same alphabet do exist? How many passphrases of a 4 letter word and a 6 letter word do exist?

The real-world problem arises from the security of passwords (for Facebook for example). Another example is the range of telephone numbers, license plates, IP-addresses that sometimes happen to be too short for their application.

Of course you could also go into probability theory and use Bernoulli processes.

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I let the students discover the rules for exponentiation as a consequence of finding prime factors. Students like the shorthand notation $24=2\times2\times2\times3=2^3\times3$ and related questions arise naturally. Given the prime factors of 24 what is the prime factor decomposition of 240? I like to juxtapose $ 2+2+2+2=4\times2$ with $2\times2\times2\times2=2^4$ Why do the rules work for one set and not the other? What's the same or different between the cases? Will the rules ever fail? I let the students apply the rules with algebra after developing their own abstract ideas.

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