Sunday, May 18, 2008

Playing with Logarithms

Logarithms, or logs, are very useful and powerful tools in mathematics. They can be used for many different problems, and all their properties make them very easy to manipulate.

Today I have another question from Yahoo Answers, but I'll take only the good part:

loga x = c
logb x = d
Find logab x in terms of c and d

This question looks simple. After all, logs have so many properties, there must be one with multiplication of base.

Well, there isn't.

In second thought, there is, but it's not written in any textbook, and to get to it we need to go through many log properties. Shall we start?

First, we can use the log property of reversal of base and argument:
loga x = 1/logx a

Now we can write this for both definitions above:
logx a = 1/loga x = 1/c
logx b = 1/logb x = 1/d

Using the addition property of logs that says: logx ab = logx a + logx b, we can say:
logx ab = 1 / c + 1 / d = d/cd + c/cd = (c + d) / cd.

Now all we need is to reverse the log back to the one we want:
logab x = cd / (c + d)

As you can see, logs don't require too much complicated mathematics, but rather original thinking and creativity.

Do you have any log problems?

Yours,
Nadav

nadavs

3 comments:

Anonymous said...

Thanks so much, helped a lot. I wished you could of explained it better though...

Anonymous said...

but you said that 1/log(x) = 1/c and that 1/log(x) b = 1/d so how can log(x) a + log(x) b = 1/c + 1/d ? don't log(x) a and log(x) b also have to be in fractions as you said before?
get everything except that...please help!!!

Anonymous said...

ok no got it now after pondering a lot!!!!=)