Introducing logarithms?

1. Sep 29, 2003

arcnets

Hi all,
I wonder what is the best way to introduce logarithms when you're teaching.

My "approach #1" is the one I consider the most natural:
You introduce exponential functions as f(x) = bx, and ask what is the derivative.
It turns out
df/dx = lim(h->0) (bh-1)/h bx.
Now, actually
ln(b) = lim(h->0) (bh-1)/h.
You could give this to the students as a definition of ln().
Now you define e by ln(e) = 1 and so on...
This is unsatisfactory since ln() should be defined as the inverse of exp(). And since it doesn't give a recipe to compute e.

My "approach #2" is that you don't tell the students anything about logarithms until you have defined e.
Works like this:
We want f'(x) = f(x), so we look for a number e which satisfies
lim(h->0) (eh-1)/h = 1.
Let's say h = 1/n, so
lim(n->[oo]) n(e1/n-1) = 1,
or
e = lim (n->[oo]) (1 + 1/n)
Then you introduce exp(x) = ex, and, as it's inverse, ln(x).
Now, let
f(x) = bx = exp(x ln(b)),
we find
df/dx = ln(b) * f(x),
meaning the limes appearing above equals ln(b).
This is also unsatisfactory since only the natural logarithm is introduced, not any other logarithms.

So, my "approach #3" is, you introduce all the logarithms (i.e. logb() as inverse of b()), and then introduce e (as in approach #2), and then ln() as inverse of exp(). This is also unsatisfactory, since you don't use d/dx bx = ln(b)* bx.

Which approach do you think is the best?

Last edited: Sep 29, 2003
2. Sep 30, 2003

HallsofIvy

Yet a fourth way, not as "natural" but has many points in its favor, is to introduce the natural logarithm as the anti-derivative of 1/x.

That's easy to motivate since "-1" is the one power for which the anti-derivative power law does not work. All of the properties of ln(x) follow easily from that definition. Having show that ln(x) is one-to-one, define exp(x) as the inverse function to ln(x).

One can THEN show that exp(x)= e^x where e is defined by ln(e)= 1.

3. Sep 30, 2003

arcnets

Good idea, HallsofIvy. Though it's sort of non-standard...

4. Sep 30, 2003

chroot

Staff Emeritus
I would encourage introducing logs well before calculus. It's easy to introduce them as simply the inverse operation to exponentiation.

- Warren

5. Sep 30, 2003

arcnets

Thanks chroot. But this means you don't tell how a logarithm is actually calculated - all you can tell the students is "to find logb(x), type ln(x)/ln(b)". They will always wonder what the ln() means...

6. Sep 30, 2003

Integral

Staff Emeritus
Off the wall idea, find some old slide rules, show how it works and what it can do, then ask, why? Seems to me this would be a very nice way to motivate logs.

7. Sep 30, 2003

chroot

Staff Emeritus
You could just tell them log10 x / log10 b instead.

Don't confuse the concepts of logs with the concepts of e, at least not at first. The log10 is a very simple concept, the inverse operation to exponentiation of 10, and is not related in any necessary way to e.

When they get to calculus, or series sums, you can introduce the natural log.

- Warren

8. Sep 30, 2003

Tom Mattson

Staff Emeritus
Yes, that's how I learned it--3 years before calculus. I'll never forget what the algebra teacher said, "A logarithm is an exponent." Once I had that grasped, the wierd addition and subtraction rules for logs (you know, ln(a)+ln(b)=ln(ab) and ln(a)-ln(b)=ln(a/b)) were not so bad.

9. Oct 1, 2003

arcnets

Thanks.

Integral: You mean "why can it multiply by just adding lengths" don't you? Yes, I guess students will be curious to find out.

chroot, Tom: Good idea to start with log10. Because it's also on most calculators.

10. Oct 6, 2003

phoenixthoth

in precalculus, i defined log base b as the inverse of b^x. it is the case with all logarithms, not just base e, that it is often difficult to calculate by hand. when it came to base e, i'd ask questions like:
what is ln(e^3) or ln(sqrt(e)) or ln(e^ln(e)).

when it came to base 8 i asked what is log(4), log(16), etc.

then i told them the calc could do other things like log(5.7) with some precision.

the advantage is that calculus isn't required. then you can prove that the definite integral from 1 to x of 1/t is also an inverse of e^x and then prove inverses are unique to arrive at the result that that definite integral is the natural log.

let's let that integral as a function of x be called I(x). it's a fun exercise to prove that e^I(x) = x and I(e^x) = x to show that I(x) is the inverse of e^x.

when it came to defining e, i would even tell my precalculus students about tangent lines and that e is defined to be the base for which the slope of the tangent line when x = 0 through b^x is 1. i couldn't really tell them that base e was more important than base ten because of the elegance of the derivate of e^x and ln(x), making e^x quite important for differential equations, which are used in a lot of science (as far as i know). then, oddly enough, a connection to compound interest came up later and that probably seemed curious to them. why that odd geometric definition should have anything to do with interest is kinda interesting.

i remember asking my algebra teacher what ln(-1) is and she had no idea. i was apalled. back then, i thought i had invented a new kind of imaginary number...

11. Oct 6, 2003

arcnets

phoenixthoth,
yes I understand your concept. Interesting.

12. Jul 23, 2004

mathwonk

I think no one has found a way to do logs that is really "best", and apparently we are all experimenting. I sometimes do it as follows, with a mix of all the ideas above.

First I review in detail how take exponents, i.e. what a^n means for integers. Then observe the law a^(n+m) = a^n.a^m holds. then use that to motivate why we define a^(1/n) as we do, and a^(-n) as we do. Then I ask how to define a^x for real x, and point out the difficulty in this.

Then I ask if the exponential function is continuous or even differentiable. We have essentially defined it to be continuous so i do not belabor this, but go on the more interesting question of diferentiability.

We start with a familiar case, f(x) = 10^x and try to take the limit defining the derivative. Although we see it reduces to the limit of (10^h - 1)/h as h approaches 0, we note that this is a very non trivial limit to compute. Even showing it exists is no snap.

Even if we assume it exists and equals C, we have a derivative formula that reads

f'(x) = Cf(x), where we do not know what C equals, a highly unsatisfactory situation.

So we observe that in order to show the exponential function is differentiable, it would suffice to work backwards, i.e. to search through all differentiable functions and try to find one which looked a lot like the exponential function. This seems crazy, but actually is what works.

To do it we have to know how to recognize the exponential function when we find it, so I point out that the properties discussed at the beginning, essentially prove that the exponential function 10^x is the only continuous function f(x) with f(x+y) = f(x).f(y) for all real x, f(0) = 1, and f(1) = 10.

So if we can locate a differentiable function with those properties, it must equal the function 10^x, which will prove 10^x is differentiable.

Now we do not know how to do this, so we take advantage of the inverse function rule. I.e. we ask what theorems we can use to guarantee a function is differentiable. Well, if a function is differentiable and has non zero derivative, then its inverse is also differentiable. So we use the inverse function rule to see what the derivative of an inverse function to 10^x would look like.

Of course it turns out the inverse function, log10(x) would have derivative equal to 1/(Cx), where C is the same mystery constant that showed up in the derivative problem for 10^x.

Now we are looking not for a function f with derivative equal to Cf, but for a function g with derivative equal to 1/(Cx). Now we have the FTC to tell us that certainly there is a differentiable function out there with derivative 1/(Cx), namely the indefinite Riemann integral of 1/(Cx), taken from 1 to x, for any positive x.

Then we ask whether this area function has any resemblance to a logarithm function; Well of course we recognize a logarithm function analogously to the rule for exponentials. I.e. log10(x) is the only continuous function g(x) defined for all positive reals, such that g(1) = 0, g(xy) = g(x) + g(y) for all real x,y, and also g(10) = 1. So we have to check those properties for our integral function g.

Now it is easy to show that any differentiable function g with derivative equal to 1/(Cx), does satisfy g(xy) = g(x) + g(y). And if we start our indefinite integral at x=1, we also get g(1) = 0, and we can show that there must be some point K where g(K) = 1. But what is C when K = 10?

It is easier to first study the case C = 1. Then the number K such that g(e) = 1, is christened e, for euler. We also use the notation ln(x) for this case of g(x) = loge(x).
Then it follows easily that when K = 10, then C = ln(10).

Now that we know that the inverse of the exponential function is differentiable, and know its derivative, we deduce the differentiability and the derivative formula for the exponential, our original goal.

I.e Finally we have shown, by a clever backdoor method, that 10^x is differentiable, and that its derivative is ln(10).(10^x).

This method seems completely natural to me, and explains every phenomenon that arises, i.e. in a way that a brilliant person might have thought of herself, but has the disadvantage of being very long.

There is unfortunately no other treatment I have seen that is intellectually natural and covers all the bases. I have thus concluded that this topic either deserves a lot of time, or should be left incompletely treated.