The function is the exponential function with *doubling time K* and *initial value A_0*. I’ve capitalized the constants (as I usually try to do); the units of *K* are (of course) those of *t* (which can be arbitrary; our assumption yesterday that *t* would be in years was for yesterday’s material only). Thus, for example, a population of one million (at time zero; look at what happens in the exponential function under consideration; *A_0* is a million) that doubles every 32 years would be represented by *f(x) = 10^6 * 2^(t/32)*. Now what?

Well, I suppose one should plot the co-ordinate pairs having *t* equal to (say) 0, 32, and 64 and sketch the curve; look at the calculator; that kind of thing. But we’ll assume everybody gets it. You probably oughta be looking at a source with graphs in it if you don’t already understand this theory pretty well; I’m pretty helpless with online graphics.

Now, one could define the *same* exponential function in terms of its “tripling time” instead of doubling time; indeed putting *f(x) = 3*10^6* in our example yields *3 = 2^(t/32)*; taking logs (base two) on both sides gives us *t/32 = log_2 (3)* and hence *t = 32 log_2 (3)*…about 50.72 years. So *g(x) = 10^6 * 3^(t/50.72)* is essentially the same function as *f* (to make *g exactly* equal to *f*, replace “50.72” with the *exact* value (the “formula”; for some reason students tend to believe that decimals are more “exact” than the algebra they approximate [and have a hard time letting go of this idea]).

What I have in mind is some alternative universe (or future class) where I would hold back on introducing the “exponential growth rate” for at least one section and have students work on exercises about doubling times (and half-lives; I intended to have mentioned by now that when *K* is negative one has an exponential *decay* rather than an exponential growth and that *-K* is then called the *half-life* of the quantity being measured [usually in this context, “number of molecules of a certain radioactive isotope”]). Things are already hard enough to assimilate without some weird number unknown to elementary mathematics being thrown into the mix.

Passing back and forth between the bases of 2 and 3 could, at least for some students, make it easier to see what’s going on in transforming to the base *e*… and this “interchangiblity” of the bases seems to be something that students typically seem not to quite “get”. Anyhow, the graphs I’m thinking of here—and’ll have on the board in a few hours—would be *w(x)= 2^x* and *h(x)= 3^x* (on the same co-ordinate axes). If we think of “beginning at” 2^x and “going to” 3^x, we’re *applying a transformation*. This one looks like a horizontal “shrinking” (note in particular that the fixed point is on the *y*-axis). And indeed it is. *Assuming* this for a moment, we can conclude that the functions *w* and *h* are related by *h(x) = w(Lx)* for some “scaling factor” *L*.

I’ll interupt myself to remark that we know *this* from work in the first week of the course… adds and subtracts are *shifts*, multiplies and divides are *scalings*, and sign changes are *reflections*; things that happen to *y* go up and down and things that happen to *x* go side-to-side; finally (and most trickily) such side-to-side moves occur “backward” in the sense that, for example, adding a positive number to the “argument” (replacing *f(x)* with *f(x+3)*, say) calls for *subtracting* from the *x* co-ordinates in computing points on the “new” graph (shifting *left* by three units in our example).

Returning to our exponential functions, we’ve used *geometry* to see that 3^x can be written as 2^(Lx) for some L (the scaling factor). Some “simple” algebra (perfectly natural and straightforward *once one has done the exercises*) tells us that *L = log_2 (3)* here; indeed . It would hardly be an exaggeration to say this is what *log_a* is *for*: determining the scaling factors for passing from one exponential function to another. Gotta go; see you soon I hope.

March 9, 2009 at 1:03 am

I’ve read all the recent blogs and missing class was not helpful!!

March 9, 2009 at 12:31 pm

missing class is… i hope… *seldom* helpful.

of course, when necessary, part of the remedy

is: even more exercises than usual!