Log Base Formula

The concept  of logarithms arrived  from John Napier's work in the early 17th century. Since then, logarithm tables are used. logarithms were an important key to simplifying scientific calculations. Today's there are numerous applications of logarithms.

The logarithms of a number 'N' to the base 'a' (a>0, a`!=` 1)  is defined as the exponent (or power) to which the base (a) is raised to produce the number (N).

i.e;      logarithmbase( Positive number) = any real number

Mathematically , if  ax = N  a>0, a`!=` ,  N>0

then  loga N = x

Conversely, loga N = x

=>  ax = N

In the above context, it is to be noted that logarithmic function can be defined in real number system only if following conditions are satisfied at a time.

Understanding Formula for Percent is always challenging for me but thanks to all math help websites to help me out.

base is positive.
base is not equal to 1.
number ( whose logarithm is to be found out ) is positive.

Important Log Base Formula:

1) loga(1) =0 ,

where  'a' is any base.   The log of any base is zero.

2) logN N =1

The log of a number to the same base is one.

3) loga N =   1/ logN a

Here base and number are interchanged.  =>  loga N *logN a =1

4) a loga N = N

5) loga b * logbc * logc a =1

Circle of log base

Here a,b and c are arranged in such a way that each number moves to the base once alternatively.

In such case, the continued product of the logarithm of the numbers is equal to 1. It is also true for more than 3 numbers.

log base formula

7 ):  ( a). loga N > 0, if a>1, N>1

or      a<1 n="" p="">
(b). loga N < 0, if a>1, N<1 p="">
or   a<1 n="">1.

8) (a). If  a (=base) > 1, and loga N1 > logaN2, then N1 > N2

(b)   If  a (=base) < 1, and loga N1 > logaN2, then  N1 Examples on Log Base:

Q:1 Evaluate log2log2 log2 16

Sol:  log2log2 log2 16

= log2log2 log2 24

= log2  log2 4              ( since log2 2=1 )

=log2 log222  =  log2 2  =1

Algebra is widely used in day to day activities watch out for my forthcoming posts on limit test for convergence and civil services syllabus 2013. I am sure they will be helpful.

Q:2 Evaluate '(logm  (logk m))/ logk (logm k)'

Sol:            '(logm  (logk m))/ logk (logm k)'

let  logk m =x    so we can write as logm k = 1/x

now we can write log mx  / logk  (1/x)  =  logm x/ -logk x

so our answer =   -logm k