Lia's Math – Home page

Logarithm – properties (logarithms of a product between two or more numbers and logarithm of the sum of two or more logarithms with same bases) – pg.4/10

Posted in Math Theory by Lia on 04/08/2013

pg. 1, 2, 3, 4, 5678910

 

3.   Logarithm of a product between two or more numbers and Logarithm of the sum of two or more logarithms with same bases.

 

The logarithm of a product between two or more numbers is equal with the sum of the numbers’ logarithmic expression.

The sum of two or more logarithmic expression is the reverse to the logarithm of a product.

 

Let’s assume that:

[3]   {{log_m}p = n} and {{log_m}q = r } or if we use the exponential form we have

[4]   {m^n=p} and {m^r = q}.

The product of exponential expressions in [4] is: {pq = {m^n}{}*{}{m^r}}  or we can write:

[5]  {pq = {m^{n + r}}} (see Exponents (Powers) §2)

By transforming the exponential expression in [5] into a logarithmic expression, we have:

       {{n + r} = {log_m}(pq)}

 From [3] we have: {{n + r} = {log_m}p +{log_m}q}

This means that:

[6]  (logarithm of product)

{log_m}(pq) = {log_m}p + {log_m}q

OR

[7]  (sum of logarithms)

{log_m}p + {log_m}q = {log_m}(pq)

 

When there are more numbers to multiply or more logarithms with same base to add, we have:

[8]    (logarithm of product)

{{log_m}(pqrs) = {log_m}p + {log_m}q + {log_m}r + {log_m}s} 

[9]    (sum of logarithms)

 {{log_m}p + {log_m}q + {log_m}r + {log_m}s = {log_m}(pqrs)} 

 

Example:

1.

  • (logarithm of product)

 {log_2}(4*8*32) = {log_2}4 + {log_2}8 + {log_2}32= {log_2}(2^2) + {log_2}(2^3) + {log_2}(2^5) = 2 + 3 + 5 = 10

  • (sum of logarithms)

{log_2}4 + {log_2}8 + {log_2}32 = {log_2}(4*8*32) = {log_2}1024 = {log_2}(2^10) = 10

 

See  Prime factorization of a number to learn how we did the prime factorization.

 

2.

  • (logarithm of product)

{log_2}(4ab) = {log_2}4 + {log_2}a + {log_2}b= {log_2}(2^2) + {log_2}a + {log_2}b = 2 + {log_2}a + {log_2}b

  • (sum of logarithms)

{log_2}4 + {log_2}a + {log_2}b = {log_2}(4ab) = {log_2}({2^2}ab)

 

 3.

  •  (logarithm of product)

{log_2}(8{a^3}{b^6}) = {log_2}(2^3) + {log_2}(a^3) + {log_2}(b^6)= 3{log_2}2 + 3{log_2}a + 6{log_2}b = 3(1 + {log_2}a + 2{log_2}b)

  •  (sum of logarithms)

{log_2}8 + {log_2}(a^3) + {log_2}(b^6) = 3({log_2}2 + {log_2}a + {log_2}(b^2)) = 3{log_2}(2a{b^2})

 

pg. 1, 2, 3, 4, 5678910

 

 

Tagged with: ,

Leave a Reply