Logarithm Lecture

Logarithm June 18, 2009

Denition. Exponential and Logarithm are dened to be inverse of each

other. Hence we have y = loga x ⇔ x = ay Note here : Exponential is dened for a huge number of values of x for which log is not dened. Try to understand what I mean by this statement.

Denition.

From the exponential form, a > 0 & a 6= 1 else the exponential function becomes We can validate the denition conditions if we see the inverse of log a many to one function & hence log wont exist function that is Exponential function. x > 0 since ay is always positive Example 1. Solve logx x = 1 y = loga x is dened if x > 0, a > 0, & a 6= 1

Solution

logx x = 1 ⇒ x1 = x

At rst instance we will feel that x = x is true for all real values of x hence solutions set is set of real numbers R. But there is catch here. See the original problem. There x > 0 and x 6= 1 Hence solution set is x ∈ (0, 1) ∪ (1, ∞)

Example 2. Solve (x − 1) logx (2 − x) = 0 Solution (x − 1) logx (2 − x)

=

0

⇒ x − 1 = 0 or logx (2 − x) = 0 x = 1 or 2 − x = x0 = 1 x = 1 or x = 1

This example says that x = 1 seems like the solution. But check if log exist in the question for x = 1. It doesn't! Since by denition base 6= 1. Hence the problem has no solution. 1

2 Now we come to the important properties of Logarithm function which makes them very useful for calculations.

0.1 Properties: 1. (Multiplication Rule) loga (mn) = loga m + loga n Note here log or product becomes sum of the logs. Moreover you can go from RHS to LHS only if the added logs have the same base. 2. (Power rule) loga mr = rloga m Here the power over the number gets multiplied by the power. Power rule can be produced from Multiplication rule Think! 3. (Division rule) loga ( m n ) = loga m − loga n This rule can be produced from the above two rules. 1 p

4. (Power over base rule) logap m = loga m 5. Combining both 2 & 4 we get, logap mr = √

Example 3. Prove log10 ( 3 − Solution √ √

√

r loga m p

√ √ 2) = − log10 ( 3 + 2)

√ √ √ √ 1 = 3 − 2 = ( 3)2 − ( 2)2 = ( 3 − 2)( 3 + 2)

Now applying log on √ both sides we get, √ √ √ log10 1 = log10 ( 3 − 2) + log10 ( 3 + √ √ √ √ 0 = log10 ( 3 − 2) + log10 ( 3 + 2) √ √ √ √ log10 ( 3 − 2) = − log10 ( 3 + 2)

2)

Another method of solving this problem is

√ √ √ √ √ 1 √ = log10 ( 3+ 2)−1 = − log10 ( 3+ LHS = log10 ( 3− 2) = log10 √ 3+ 2 √ 2) = RHS

Example 4. Is the following statement true always base can be any valid base.

Solution

2 log x = log x2 here

If you see the RHS, for all real values of x except 0 its is well dened. But the LHS is dened only for x > 0. Hence the above statement is true only where both LHS and RHS are welldened, though they are part of an idendity. So the statement is true for x > 0 only.

0.2

3

Fundamental Identity

0.2 Fundamental Identity N loga b = bloga N

0.2.1 Special Case N logN b = b

Example 5. Find value of Solution:

25log125 23 2

2

2

25log125 23 = 52 log53 23 = 5 3 log5 23 = 5log5 23 3 = 23 3

0.3 Base Changing formula in dierent forms 1. Multiplication form : loga b · logb c = loga c (a) Special case : loga b · logb a = 1 (b) Can also be seen as : loga b = 2. Division form :

loga b = logc b loga c

1 logb a