Exponentials and Logs

I'm going to very very briefly go over stuff that was covered last year, and then explain the new stuff.

Differentiating exponential equations: Multiply by the derivative of the power of e, keep everything else the same. (e.g. e^(2x) becomes 2e^(2x))
Integrating exponential equations: Divide by the derivative of the power of e, keep everything else the same (e.g. e^(2x) becomes (1/2)e^(2x))

ln (natural log) is a log to base e
A log with no base specified is a log to base 10

Converting logs to exponentials:

Remember, the bottom number (the a in this case) is the base.
According to Wikipedia, the b is called the argument in the logarithmic function (top) and the answer in the exponential function (bottom).
Also according to Wikipedia, the c is called the answer in the logarithmic function and the power in the exponential function.
Though Wikipedia also suggests that the terms "argument" and "power" are interchangeable.

Adding and subtracting logs: You know how when you multiply numbers with powers, you add the powers, and when you divide numbers with powers, you subtract the powers? Well, logs are the opposite. You multiply when you need to add and divide when you need to subtract. I know, I'm terrible at explaining. And there's the power law too, which involves bringing the power around to the front- also a pretty terrible explanation. So here's some formulae, courtesy of Microsoft Word:

Differentiating logarithmic equations: This method ONLY works for natural logs- logs with base e, or ln.

Derivative of the function ln (f(x)) is given by (f '(x))/ (f(x)) - that is, the derivative of the argument divided by the argument (see "converting logs to exponentials," two sections above).

If you don't have a natural log, you need to convert whatever you have into a natural log. To convert an exponential into a natural log, take a natural log of both sides. For example, 2^x = y can be rearranged to ln (2^x) = ln y and then x ln 2 = ln y and finally x = (ln y) / (ln 2). From there you can find dx/dy and then dy/dx.

To convert a log of another base into a natural log, first convert to exponential form and then use the above method.

Integrating to give logarithmic equations: The new stuff! Finally!

Remember how back in 3AB we were told that you can't find the integral of x^(-1) using the old "raise the power by one and divide by the new power" rule because then you'd end up dividing by 0? Well, fear no more, because now we have a new secret weapon! Natural logs!

You see, x^(-1) = 1/x. And the derivative of ln x = 1/x! Therefore, the integral of x^(-1) must be ln x! (EDIT: Actually, it's ln x + c.)

Well, actually, it's slightly more complicated than that. You see, if you want to find the area under the curve of y = x^(-1) you'd want to use integration, which would lead you to using ln x. The thing is, though, the curve of y = x^(-1) exists for negative values as well as positive values, while ln x is undefined for negative values. To get around this, we actually have to write the integral of x^(-1) as ln |x|. No, ln |x| + c. Oops.

Now for some more fancy stuff! Clearly, there are other logarithmic functions out there than ln x, and therefore lots more equations with logs as their integrals. Let's see how we can work this one out.

First of all, put the bit at the bottom into a natural log (probably not really correct terminology, but whatever). That is to say, if you have 2x + 3 at the bottom, make it ln (2x + 3) and move it to the top. Now, hopefully you also have a multiple of the derivative- in this case 2- sitting at the top with it. (If not, you can't integrate.) Then divide this multiple by the derivative.

Wow, today must be a day for terrible explanations. So here's an example that might shed some light on the matter.

(integral sign) (2 + cos 2x)/(4x + sin 2x) dx

Step 1: Put the bit at the bottom into a natural log and move it on top.

(2 + cos 2x) ln (4x + sin 2x) dx

Step 2: Divide by the derivative of the stuff that was originally at the bottom.

((2 + cos 2x) ln (4x + sin 2x))/(4 + 2 cos 2x)
= ((2 + cos 2x) ln (4x + sin 2x))/(2 (2 + cos 2x))
= (1/2) ln(4x + sin 2x)

Step 3: DON'T FORGET THE PLUS C!!!!

= (1/2) ln(4x + sin 2x) + c

(Please note: I don't know for sure if the above method is mathematically correct or whatever... but hey, it works. Just make sure you've got the derivative of the stuff at the bottom at the top, and you should be fine. Otherwise, use the substitution method as I outlined here: http://year11misadventures.blogspot.com.au/2013/05/integrating-by-substitution.html.)