Life Math for Golden Age

LOGARITHM

                    ( from Greek words : 'logos' -proportion  & 'arithmes' - number)
The way I like to think about logarithms is that they
are just another language for describing
exponentials.
In exponential language ,we emphasize the base;
logarithms emphasize the exponent. Each property
of logarithms is just a property of exponentials,
expressed from a new point of view in the
logarithmic language.
    Assuming that we know the fundamental
exponential    properties:
if a >0, then
1) a^{n *}a^m= a ( n + m )
2) aⁿ/a^m = a ^{(n-m )}
3) (a^m)ⁿ = a^(m*n)
If b^x= y ( x is the exponent, b is the base
and y is the base raised to that exponent),
then we can rephrase this as
                 log _b Y = x ,
where log _b denotes the logarithm to the base b.
This logarithmic statement is exactly
a restatement of the exponential
    relation b^x= y
       D E F I N I T I O N

Logarithm is the exponent that raises a base number to yield a given number;

So, for all real numbers b, y and x, if b > 0, y > 0, and  b not equal 1,

log_by = x , if and only if

              b^x = Y

Example1

Write log₅₍m) = 4 in exponential form.

                     Answer: 5⁴ = m

Example2

Write 7ⁿ = 49 in logarithmic form.

       Answer: log₇₍49)= n



Logarithmic expression with a base other than 10 can be converted into a base 10 expression in which log₁₀ is generally written as log for short.

    log _b Y   = log _Y/ log _b

There are three laws of logarithms.

Law #1: log_b(x × y) = log_bx + log_by

(where b > 0 and b not equal 1, x > 0, and y > 0)

The log of a product is the sum of the logs of the factors

Law #2: log_b(x/y) = log_b x - log_b y

(where b > 0 and b not equal 1, x > 0,

   and y > 0)

      The log of a quotient is the log of the numerator minus the log of the denominator

Law #3: log_b y^x = x × log_b y

(where b > 0 and b not equal 1, and y > 0)



Notice : 1.The base of logarithms b > 0 and never equals to 1 !

                   2. Logarithms   with base 10 are written simple   log.

                   3. Natural logarithms with base   e = 2.718 are written    as ln .

                   4. Relationship of logarithms with different bases can be expressed by the formula :

log_cY = log _bY / log _bC   ( From base c to base b )

             Examples :

    log_c Y = log Y/ log c ( from base c to base 10)

ln Y= log Y /log e

(from natural logarithms to log)

       NOTE :   There are simple properties of logarithms   that follow from the definition :

1.   log _bY= X ;  (   b ^X= Y ; b = Y^1/X )

    log _Y b = 1/X ;

( log _b Y) x ( log _Y b ) = X * 1/X=1



2. log _b (b ) = 1 ;        ( b ¹= b )

3. log _b1 = 0     ;           ( b⁰= 1 )

4. b^{( log}_b^P )    = P; ( Hence the expression   log _bP represents the exponent of the base b to get P )

5.   log _b ( b ^X ) = X ; ( What is the exponent of b to get   b ^X ?   Of cause, X ! )

6.   b ^log_b⁽b^log_b^{P )} = P ;   ( Hint : Evaluate the expression from right to left , beginning with   the content within the parantheses

      ⁽b^log_b^{P )}  )



What about following expressions ?

                e ^lnx= ? ; ln(exp( log_e e^x )) = ? ;

                ln( e ^x ) =? ; log_e ( e ^x ) = ?;        ( Hint : The answer is x for all the expressions !

                             WHY ? -- see the definition )

       Just  to remind you   :   Here's again the definition of Log:

      If   a^b = x, then  Log_a(x) = b.
When you read that, you say :"if a to the
b power equals x, then the Log to the base
a of x equals b."

                                      *      *

                                          *

Powers are easy to calculate using logarithms which are useful in obtaining values of exponential functions.

Generally, logarithms and exponentials do have several applications in different domains : growth of population, compound interest, radioactive decay , Richter's earthquake scale , light intensity ,healing of wounds ,air  pollution, noise formula , depreciation et c.

When we use a LOG SCALE?
   Try searching for things relating math
to music (the scale on the piano is a
logarithmic sort of scale, if you use both
white and black keys). Also look for
Richter scale (for earthquakes) and deciBel
scale (for sounds), materials on the
psychology of perception (how humans see
brightness of light, hear intensity of
sound, and so on) also would have lots
of references to log scales.
Do your own experiment: Have people look
at a 50 watt bulb, 100, 150, 200 without
telling them which is which. Ask them to
rate the brightness on a scale of 1 to
10.You'll probably see that the space
between 100 and 200 watts is closer to
the space between 50 and 100,showing that
we see brightness logarithmically. Log
scales are generally useful for things of
human perception like this (it's how we're
wired!), for things with a huge range
(like earthquakes, where strong ones are
a million times bigger than weak ones),
or for things where the ratio is more
important than the difference (musical
notes sound good together if their ratio
is a nice number).

Now we can go to some 'life problems'!

I consider 'compound interest '   as the closest to 'life math ' problems .

SIGNIFICANT NOTES TO

  'COMPOUND INTEREST'

The interest, as we know, can be paid in many

different ways- for example,once a year or many

times a year. If the interest is paid more

frequently than once per year and the interest

is not withdrawn,there
                  is a benefit to the

investor,since the interest earns interest.

 The effect is called compounding . You may

have noticed banks offering accounts that differ both

in interest rates and in compounding
                  methods.

 Some offer interest compounded annually, some

 quarterly, and others daily. Some even offer

continuous compounding
                  .

  Let see what is the difference between a bank

account advertising 8% compounded annually and

one offering 8% compounded quarterly? In both

 cases 8% is an annual
                  rate of interest.

 So, in first case when the interest is

compounded
                  annually investing $1000 We'll get :

 After 
                  one  year   B = 1000( 1.08 )  = 1,080

After two years    B = 1000( 1.08 )² = 1,166

After  t years B =
                  1000(1.08)^t =2,159 (t=10)

       General  formula: B = B₀ (1 + r)^t

 In
                  second case , when interest is compounded

 quarterly, i.e. every three months
                  will be

added  8/4 =2% , we'll get accordingly
                  :

After  one  year  B = 1000( 1.02 )⁴ =1,082

After two years   B = 1000( 1.02 )^{8 =}1,172

After t years B = 1000( 1.02 )^4t=2,354(t=10)

  
                  As you can see - it makes a difference !

Conclusion : In solving a problem involving

compound Interest,it
                  is important to be

clear
                  whether the interest rates are nominal

 rates or effective yields , as well as

whether compounding is continuous or not .

     General 
                  formula: B = B₀(1 + r/n)^nt

     Where,-- B balance
                  after  t  years

               B₀ Initial  deposit

  r- nominal  rate (if annual
                  rate is 8%

 r =0.08 )

   n -- amount of compounding times a

year

Calculating the total balance after one year


                  under each method shows that  in case of

quarterly compounding  we  earned  more because

 the interest earns
                  as the year goes by

( we  got  8.24% , so named effective annual

 yield instead of 8%
                  - annual rate, APR ).

  
                  It turns  out that  e is intimately
                  connected

 to continuous  compounding.

                   (1 + r/n)ⁿ = e

 
 Then , the general  formula of continuous 
        compounding Will  be :

          B = B₀e^rt

  Dear
                  reader !
 Now we'll try  together
                  to answer a 
very popular and  relevant  question:
  How  long would it take to double 
an investment?
 Letting  X - years the interest is 
compounded,
  r - the  interest rate, we wish to
solve :              
       B( 1 + r )^X= 2 B ; (1)
 Dividing by B
                  ,                  
                  
          ( 1 + r )^X= B ; (2)
 This  tells immediately, that the 
answer does not depend
 on the amount invested  B, but only 
on the rate ! 
                  
      We use the
                  logarithms and 
appropriate properties of
 them to solve the equation
                  (2) 
  X * (ln (1+r))= ln 2 ;
  X= ln2/ln(1+r)= 0.693/ln(1+r)                         
                  
     Suppose  r = 0.05 ( annual rate 
= 5%) , then                  
                     
 = 0.693/ln(1.05)=0.693/0.049
                  =14.1 
  So, it would take about 14  years to  double
         
                  any investment by annual rate of 5% !

 I believe,dear reader, you

would
                  prefer a  simpler

solution for such a life

problem,avoiding calculations

with
                  logarithms... If so, a

magic constant will serve a

way:Remember the

constant70!

 What you need to do is only

to divide  the constant by the


                  given rate (%) . Thus,

x = 70/rate. In the above

 mentioned case

 x=70/5 =14 years.

( As you see
                  we got the same

result)

What if the rate is 6%? How

long
                  would it take to double

some capital?

The answer you get
                  instantly

dividing70by 6; (70/6=12years.)

How we've figured out this

magic
                  constant ?

In the above used formula

 x =ln 2/ln (1.05)=0.693/0.05,

multiplying
                  the both sides by

100 we get 69.3/5 = 70/5 =14

Notice
                  that ln (1.05 )= 0.05,

as ln(1.06) =0.06,

as ln(1.03) =0.03
                  et c.)

Now, we'll try do define a

magic constant for instantly

solution of another problem :

 How long would it take to

triple
                  an invested capital?

Following the same way,and

knowing that ln(3)=1.099

we find that the magic

constant in this case = 110

So,by rate =5%,some investment

will
                  be tripled in 110/5 = 22

years.