NATURAL LOGARITHMS -- DEFINITION, SOLVED EXAMPLES, EXERCISES, LINKS TO FURTHER STUDY

Please study

Logarithm Formulas before Natural Logarithms

if you have not already done so.

Having the knowledge of the Formulae
is a prerequisite here.

Definition of Natural logarithm :

Unlike common Logarithm for
which the base is an integer (10),
the base of natural logarithm is
an irrational number.

This irrational number denoted by 'e'
is not a surd.

e is defined as the sum of the infinite series
e = 1 + 1⁄1! + 1⁄2! + 1⁄3! + 1⁄4! + .....
where, ! stands for factorial
(n! = product of natural numbers from 1 upto n)

The value of e = 2.71828.... which is a
non-terminating and non-repeating decimal.

In calculus, unless otherwise mentioned,
the base of logarithm is always e.

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Solved Example 1 of Natural Logarithms :

If log_a bc = x, log_b ca = y, log_c ab = z,
then prove that 1⁄(x + 1) + 1⁄(y + 1) + 1⁄(z + 1) = 1.

Solution to Example 1 of Natural Logarithms :

By data, log_a bc = x
By converting from Logarithmic Form to Exponential Form (See Formula 1), we get
a^x = bc
Multiplying both sides with a, we get
a x a^x = a x bc ⇒ a^{x + 1} = abc
By converting from Exponential Form to Logarithmic Form (See Formula 1), we get
x + 1 = log_a abc
Taking reciprocals, we get
Reciprocal of L.H.S. = 1⁄(x + 1);
Reciprocal of R.H.S. = 1⁄(log_a abc)
We know, the reciprocal of the log of a quantity to a base is equal to the log with the quantity and the base interchanged.(See Formula 10)
∴ Reciprocal of R.H.S. = log_abc a
∴ 1⁄(x + 1) = log_abc a.......(i)

Next two paragraphs are repetition of the procedure in the above paragraph with variables changed. You can use the word 'Similarly' and write Equations (ii) and (iii) directly. But I am givingbelow the repititions.Two more repetitions of the Application ofFormula 1 (both ways) and Formula 10 will help you rememberthem.

By data, log_b ca = y
By converting from logarithmic form to exponential form (See Formula 1), we get
b^y = ca
Multiplying both sides with b, we get
b x b^y = b x ca ⇒ b^{y + 1} = bca = abc
By converting from Exponential Form to Logarithmic Form (See Formula 1), we get
y + 1 = log_b abc
Taking reciprocals, we get
Reciprocal of L.H.S. = 1⁄(y + 1)
Reciprocal of R.H.S. = 1⁄(log_b abc)
We know, the reciprocal of the log of a quantity to a base is equal to the log with the quantity and the base interchanged.(See Formula 10)
∴ Reciprocal of R.H.S. = log_abc b
∴ 1⁄(y + 1) = log_abc b .......(ii)

By data, log_c ab = z
By converting from logarithmic form to exponential form (See Formula 1), we get
c^z = ab
Multiplying both sides with c, we get
c x c^z = c x ab ⇒ c^{z + 1} =ca b = abc
By converting from Exponential Form to Logarithmic Form (See Formula 1), we get
z + 1 = log_c abc
Taking reciprocals, we get
Reciprocal of L.H.S. = 1⁄(z + 1)
Reciprocal of R.H.S. = 1⁄(log_c abc)
We know, the reciprocal of the log of a quantity to a base is equal to the log with the quantity and the base interchanged.(See Formula 10)
∴ Reciprocal of R.H.S. = log_abc c
∴ 1⁄(z + 1) = log_abc c .......(iii)

(i) + (ii) + (iii) gives
1⁄(x + 1) + 1⁄(y + 1) + 1⁄(z + 1) = log_abc a + log_abc b + log_abc c
= log_abc abc
We know log of a quantity to the same base is 1. (See Formula 4)
∴ 1⁄(x + 1) + 1⁄(y + 1) + 1⁄(z + 1) = 1. (proved.)

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Solved Example 2 of Natural Logarithms :

If x = log _2a a, y = log _3a 2a, z = log _4a 3a,
then prove that 1 + xyz = 2yz

Solution to Example 2 of Natural Logarithms :

We know, log of a quantity to a base can be written asthe ratio of log of the quantity and log of the base.(See Formula 8)
x = log _2a a = (log a)⁄(log 2a)
y = log _3a 2a = (log 2a)⁄(log 3a)
z = log _4a 3a = (log 3a)⁄(log 4a)
xyz = {(log a)⁄(log 2a)}{(log 2a)⁄(log 3a)}{(log 3a)⁄(log 4a)}
= (log a)⁄(log 4a)
We know, log of a quantity to a base can be written asthe ratio of log of the quantity and log of the base.(See Formula 8)
Considering the reverse, the ratio of logs of two quantities can be written as a single log with the denominator quantity as base.
∴ xyz = log _4a a
Now to prove 1 + xyz = 2yz
L.H.S. = 1 + xyz = 1 + (log a)⁄(log 4a) = {(log 4a) + (log a)}⁄(log 4a)
We know, log of a product can be written as the sum of the logs of the factors of the product (See Formula 5)
Considering the reverse, the sum of the logs of different quantitiescan be written as log of product of the quantities.
∴ L.H.S. = {log (4a x a)}⁄(log 4a) = {log (4a²)}⁄(log 4a)
= {log (4a²)}⁄(log 4a) = {log (2a)²}⁄(log 4a)
We know, in log of a power (See Formula 7), the exponent multiplies the log.
∴ L.H.S. = {2 log (2a)}⁄(log 4a)
R.H.S. = 2yz = 2 {(log 2a)⁄(log 3a)}{(log 3a)⁄(log 4a)}
= 2 (log 2a)⁄(log 4a)
∴ L.H.S. = R.H.S. (proved.)

More Solved Examples : Natural Logarithms

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to more Solved Examples on Natural Logarithms.

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Exercise : Natural Logarithms

Solve the following problems on Natural Logarithms :

1. Find the value of (i) log₂₄₃ 9 (ii) log₃ √(243)
   
2. If (3.4)^x = (0.034)^y = 10000,
   show that 1⁄x - 1⁄y = 1⁄2
   

For Answer to problem1 of the above exercise on
Natural Logarithms see at the bottom of the page.

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Answers to Exercise : Natural Logarithms

Answers to problems on Natural Logarithms :

(1) (i) 2⁄5 (ii) 5⁄2