Logarithm Properties

Logarithm properties are used to expand or compress a logarithm. Logarithms are exponents written differently.

Logarithm properties arise from exponent properties.

Exponents properties:
Product rule: am. an=(a)m+n
Quotient rule: a^m/aⁿ = a^m-n
Power of a Power rule : (a^m)ⁿ = (a)^(mn)

Let us see how the logarithm properties are derived from the properties of exponents.

1. Product rule:

log_a(m . n) = log_a(m )+ log_a(n)

The log of the product of two numbers m and n, with base ‘a’ is equal to the sum of log m and log n with the same base ‘a’. where a, m, n are positive integers and a≠1.

Proof:

log_a(m . n) = log_a(m )+ log_a(n)

Let x = log_a m and y = log_a n

Transform logarithmic equations to exponential equations.

We are showing the product’s property, therefore, we will multiply m by n.

mn = a^x a^y

Using the exponents product rule : a^m. aⁿ=(a)^m+n

mn = a^x+y

Taking the logarithm of base a on both sides,

log_a (mn) = log_a (a^x+y)

The logarithm of an exponential number, where its base is the same as the base of the log, is equal to the exponent.
i.e., log_a (aⁿ ) = n

Therefore, log_a(mn) = x + y ………….(1)

We know that x = log_a m and y = log_a n.

Substituting on equation (1), we get

log_a(mn) = log_a m + log_a n

2. Quotient rule:

log_a(m / n) = log_a(m ) – log_a(n)

The log of the quotient of two numbers m and n, with base ‘a’ is equal to the difference of log m and log n with the same base ‘a’. where a, m, n are positive integers and a≠1.

Proof:

log_a(m / n) = log_a(m ) – log_a(n)

Let x = log_a m and y = log_a n

Transform logarithmic equations to exponential equations.

We are showing the quotient property, therefore, we will divide m by n.

m/n = a^x / a^y

Using the exponents quotient rule : a^m/ aⁿ=(a)^m-n

m/n = a^x-y

Taking logarithm of base a on both sides,

log_a (m/n) = log_a (a^x-y)

The logarithm of an exponential number where its base is the same as the base of the log is equal to the exponent.
i.e., log_a (aⁿ ) = n

Therefore, log_a(m/n) = x – y ………….(1)

We know that x = log_a m and y = log_a n.

Substituting equation (1), we get

log_a(m/n) = log_a m – log_a n

3. Power rule

log_amⁿ = n log_am

The logarithm of m to the power n is equal to the product of n and the logarithm of m, where a, m, and n are positive integers and a ≠ 1.

Let us assume mⁿ = x

Now, find log_a x = n log_am

Assume, log_am = y ……..(1)

express (1) in exponents, we get

a^y = m

Taking n power on both sides, we get

(a^y )ⁿ= mⁿ = x

∴ a^yn = x

a^z = x (Put z = yn)

Write the above exponent in logarithms, we get

log_a x = z

Substitute for x and z:

log_a mⁿ = yn

Substitute for y, we get:

log_a (mⁿ) = n log_am

4. Change of base rule

Log_n m = log_x m/log_x n, m, n, x are positive integers and n,x ≠1.

This means, log_n m can be solved by finding the quotient of log_am / log_a n, a≠1.

Derivation

Let us assume log_n m = x, log_a m = y and log_a n = z

Exponential forms are written as m = n^x , m = a^y and n = a^z

From the above exponents, we can equate:

n^x = a^y

Also, substitute the value of n = a^z in the above equation, and we get

(a^z )^x = a^y

⟹ a^zx =a^y

⇒ zx = y or x =y/z

Substituting the values for x, y and z, we get,

Log_n m = log_x m/log_x n

Hence, the change-of-base property of the logarithm is finally derived.

We know how to evaluate logarithms with base 10 and with base ‘e’. 

If we apply the change of base property rule , we can easily calculate logarithms with any bases.

Several logarithm properties are derived from exponent rules and logarithm definition.

log 1 = 0

We know that, a⁰ = 1

If we write the above exponent as a logarithm, we get:

log_a1 = 0

log_aa = 1

We know that a¹ = a

If we write the above exponent as a logarithm, we get:

log_aa = 1

Inverse Property: a^{logₐ x} = x

i.e., Number raised to the logarithm.

Let us take , a^{logₐ x} = y …………..(1)

log_a (a^{log_a x} ) = log_a y (Taking logarithm on both sides)

log_a x (log_a a) = log_a y (Using power rule)

log_a x (1) = log_a y (Using log_a a = 1)

log_a x = log_a y

⇒ x =y

∴ x = a^{logₐ x}  (Using (1))

Hence proved.

Equality rule or One to One property of logarithm:

log_a x = log_a y iff x =y, a ≠ 1

We know that, a^x = a^y iff x=y, for all real numbers x and y.

Therefore, log_a x = log_a y iff x =y, a ≠ 1, for all real numbers x ≻0 and y≻0.

The log of the number with negative power:

Taking a number to the power of negative one gives its reciprocal

Therefore, m^-1 = 1/m

Just substitute n=−1, n =−1 into the log of power rule, i.e., log_a (mⁿ)= n log_am

We get, log_a (m^-1)= – log_am

Therefore, log_a (m^-1) = – log_am

Reciprocal rule:

If a and b are the positive numbers other than 1, then;

log_a b = 1/log_ba

We use the change of base rule to prove this rule.

Change of base is: log_a b = log_b b / log_b a

implies log_a b = 1 / log_b a ( Since log_b b = 1)

Hence proved

Notes:

These properties apply to a log with any base. They are applicable for log, ln, loga, etc..

Logarithmic properties are used to expand or compress logarithms.

log 1 = 0 irrespective of the base.