Logarithms

[Shwang]

Hello

I have a math final coming up and expect to do moderate to good. My one problem with the course this year has been logarithms. Can anyone offer a good source for logs, because the way my teacher teaches it, I just don't get it, so I might have better luck on my own.

It is just basic logarithms, but I really don't know it, so "Logarithms for Dummies" would be very helpful.

Thank you

[junglistic]

Basically the only way to 'learn' or understand logs, are to just think of them as another operation which caries its own rules.

You will have to memorize the rules, and practice practice practice. But in the end, its not really much differnt then mulitplication/division/addition/subtraction.

[pig]

Without knowing exactly what's giving the problems, the best I could come up with is links. junglistic pretty much nailed it, in my opinion - logs are another set of functions, kind of like sines and cosines, that map one set of numbers to another set, using a specific set of rules, with specific properties. I think that the most useful stuff in working with logs is learning the definition of the log and it's inverse relationship to the exponential function. Hope these help.


http://mathworld.wolfram.com/Logarithm.html
http://courses.open.ac.uk/t305/maths5.htm

[phukraut]

pigglet has it right. if you know the laws of exponents then you know the laws of logarithms. one is just the complement of the other. if you begin to look at it this way, and try to ignore the unfortunate notation, then your problems ought to clear up quickly.

one rule which confuses a lot of people seems to be:

log x^y = y log x; (moving the exponent). but if you recall in exponent laws,

(b^x)^y = b^(xy). see how that works?

start with log x^y = y log x, where the logarithm has base "b", then apply exponents to each side.

b^(log x^y) = b^(y log x). now simplify using the exponent rule above and the idea that exponents "cancel out" logs and vice versa:

on the left side you get x^y, and on the right you get (b^(log x))^y = (x)^y = x^y, which is the same as the left side. that's why that log rule works. the others are similar. good luck.

[Yakk]

start with
log (a*b) = log a + log b
log x^a = a * log x

and every other property of logs falls out, with a bit of cheating.

The different types of logs are tricker.

Start with the definition of a log to a fixed base.
log_a x = k
=>
a^k = x

Now, the goal is to find the log_b(x), the log of x to a different base.

Well,
a^k = x
=>
log_b a^k = log_b x
=>
k * log_b a = log_b x
but k = log_a x so
log_a x * log_b a = log_b x

which is how you convert between bases.

I find, personally, that knowing how to work out the important results makes it easier for me to work with something.

[Slims]

I just repeat to my self "a log is an exponent"