Enough is enough, ever since i started doing some efforts in maths, i seem to be having some quite good enough success with it, but unfortunately as most of you all know i kind of always escaped classes when i was in sc or hsc, [specially when they were working on logs] so here after trying to learn a bit about it, here’s something that hopefully might help others.

so let’s get this done…

we shall start by taking an example…

**2 ^{3}=8**, (in words we say two to the power of 3 equals 8, 2x2x2=8).

in log terms we will say..

**log _{2}8=3 **, (in words we shall say the logarithm to the base 2 of 8 is 3)

that is, 2 to the power of WHAT will be 8? hence in this case it is 3. [You may use your calculator to verify that 2^{3 }is infact 8]

so the general form is

**a ^{b} = c** and

**log**

_{a}c=bnow let’s get some practice:

1.) Find the Value of log_{5}25 ?

Now, most people who [like me] never went to those log maths class or who have been sleeping like a log when they were in that class would try hard as hell to try to figure out how the heck to change the base of log in their calculator so as they can computer it directly on their calculators, but heh, don’t tire yourself, most probably there isn’t [atleast in the calculators that you are allowed to bring in exams]

there are only two things log related in the calculators, log, which is base 10, and ln which is base e, we’ll use log.

SO,

back to the question, basically we are asking 5 to the power of WHAT gives 25??

- [step1.] place it in the form
**a**^{b}= c

we are given log_{5}25, in the formlog, it will be log_{a}c=b_{5}25=b, hence the unknown here is b

a=5, c=25, b=?

5^{b}=25

- Solve the equation 5
^{b}=25

5^{b}=25

b log 5 = log 25

b=(log 25) / (log 5) [use calculator]

b=2

et voila! we got the answer.

Simple isn’t it? [i really wonder why i really hated that then] … probably coz i always found alternatives

With that you shall be able to solve any logarithm related equations, just find and replace. Till then, good luck and well, feel free to comment and add suffs, that’s how it works, coz am sure there;s a lot more to that. My next post will probably be on jacobi’s convergence or solving recurrence relations :p.

http://www.blogged4ever.com/2008/09/19/logarithm/

+$3|v3n

[note i still hate maths because i don't find it useful and soo illogical while programming is soo natural and soo logical]

ps. lemme know in case of errors.