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.

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} =cab = 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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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 (4a^{2})}⁄(log 4a) = {log (4a^{2})}⁄(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

The following Link takes you to more Solved Examples on Natural Logarithms.

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Solve the following problems on Natural Logarithms :

Find the value of (i) log_{243} 9 (ii) log_{3} √(243)

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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