SWBAT use logarithms to solve a simple exponential equation.

What turns multiplication into addition and division into subtraction? Logarithms, that what!

5 minutes

Today begins with a pair of sprints just like the previous lessons. Exponent Sprint 9 is normal, but Exponent Sprint 10 contains an obvious trick question: 4^x = 95.

I won't mention this to the students, of course, until they're finished. Since they've seen "trick" questions on these sprints before (such as 0^-1 = x), some of them may write something like "undefined" or "impossible" for *x*. That's fine for now.

5 minutes

When the sprint is over, it's time to introduce logarithms. I use the problem from the sprint as a concrete example to show that we need some way to "find *x* if b^*x* = a". In other words, we need to define the inverse of the exponentiation operation, just like division is the inverse of multiplication.

By experimenting with a calculator, we find that if 95 is regarded as a power of 4, then "the logarithm of 95 to the base 4 is approximately 3.2849278". I show the students how to write and say this. I also emphasize that as strange as the word "logarithm" seems, it can simply be thought of as a synonym for "exponent". The difference is that we tend to use the word logarithm when we're trying to solve for an unknown exponent. I explain it this way:

For good measure, I'll ask the students to try another one on their own: 4^x = 572. (This number will come up later in the lesson.) They should approximate it with their calculators, and I'll confirm the true value out to seven decimal places: 4.5799357.

(For these initial problems, I've intentionally used a base other than 10 or e because I want my students to first "find the unknown exponent" via approximation. By experimenting on their own, some may figure out what the log and ln keys are for on their calculators, and I applaud them for it. I also do not introduce the logarithm notation yet, because I want my students to get comfortable with the word first. Later they'll see the notation as an abbreviation of the verbal expression.)

15 minutes

So, where did logarithms come from?

It's time to launch into a little history lecture about Sir John Napier, Baron of Merchiston (1550 - 1617). (See my notes. For *much* more detail check out this essay. Also, I highly recommend Eli Maor's *e: The Story of a Number*, Chapters 1 & 2. This is probably the best place to begin learning more about logarithms.)

Napier was a bit of a renaissance man, and mathematics was one of his passions. He was particularly interested in developing more efficient methods of calculating. One of these inventions - the logarithm - was introduced and explained in his 1614 work *Mirifici Logarithmorum Canonis Constructio*, just three years before his death...

As I tell the story, I emphasize these points:

1. Napier *postulated* that every number may be regarded as a power of any other number. (Recall that a postulate is an unproved assumption. Is Napier's postulate reasonable?)

2. Napier's logarithms are *exactly the same as exponents*. Napier picked a single number to use consistently as the base (1 - 10^-7) and then found the exponent/logarithm for other numbers with reference to this base. Knowing the exponents made computations easier.

10 minutes

Now I say, "We've established what a logarithm is (a new name for something familiar) and who invented them. But *why*? Why were mathematicians and scientists so excited about them? What motivated them to immediately begin computing logarithms for all of the numbers from 1 - 100,000 ... to 10 decimal places ... *by hand*?! Why did mathematicians keep using these tables of logarithms for the next 350 years?"

With this question hanging in the air, I pass out these problems (each student gets just one), face down. It's going to be a race, so no peeking! The slips of paper look the same, and the students don't know that half the class gets a subtraction problem, while the other half gets a division problem. Remember, Napier didn't have a calculator, so you don't get one either.

Ready, set, go! (Check out this video to see how the race ends.)

The point is that addition/subtraction is much easier (and typically more error-free) than multiplication/division. After it's all over, I'll write the two problems on the board and ask the students to use a calculator to solve them. Then, I'll ask the students to all take the difference of the subtraction problem and raise 4 to that power. Lo, and behold, it's the quotient of the division problem!

10 minutes

To summarize, I'll say:

"The takeaway from the race is two-fold. First, subtraction is easier than division, so it would be really nice if we could find a way to turn division problems into subtraction problems. Second, the the numbers used in the subtraction problem were the logarithms of the numbers in the division problem. And, as we saw, the difference was the logarithm of the quotient."

"So..."

At this point, I'm going to pause expectantly. As they mull over what they've just seen, I expect the light will begin to dawn for a number of students. "Of course! If you divide two powers, then you find the difference of the exponents! 527 and 95 are both powers of 4, so if we find the difference of the exponents - I mean, logarithms - then that should give us the exponent (or logarithm) for the quotient. If we raise 4 to that power, we'll get the actual quotient."

I'll do my best to stay quiet now and let the students do the explaining. I might ask if they could illustrate with a simpler division problem, such as 32 / 8. Once it seems like everyone understands what we're seeing, I'll use this focus on division to introduce the meaning of the word logarithm - "ratio number" - as an aside.

Finally, I'll summarize by saying that if Napier wanted to divide two numbers, he'd look up their logarithms in his handy-dandy table and then find the difference of the logs. (Students are always shocked to hear about these tables: hundreds of thousands of logs calculated by hand to many decimal places!) Next, he'd find the number whose logarithm is equal to this difference, that number would be the quotient. Along the way, I'd be sure to illustrate how this is *identical* to the familiar property of exponents.

This is a great point in the lesson to reiterate how important logarithms were to scientists and mathematicians in Napier's day. His invention made their computations much easier and much less prone to error!

5 minutes

The students have seen and done a lot today, but I want to leave them with one last thing to think about. I'll write the following homework problems on the board:

1. Find an approximate value for x, correct to three decimal places: 3^x = 17

2. Find an approximate value for x, correct to three decimal places: 3^x = 23

3. How might Napier use your x-values to multiply 17 by 23?

A task like this is also a great one for a 3-2-1 Exit Ticket. It'll be useful to see the students' responses before the next lesson!