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.
Mathematics Educators Stack Exchange is a question and answer site for those involved in the field of teaching mathematics. It only takes a minute to sign up.
Sign up to join this communityI'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.
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.
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 $44).$
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.
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.