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Can a Logarithm be to any base?

  • 30-10-2014 1:53am
    #1
    Registered Users, Registered Users 2 Posts: 252 ✭✭


    Hi,

    As the title suggests can a logarithm be to any base or must it be between 0 and 10 inclusive?

    I acknowledge this is a very basic question but for some reason I have always shied away when it has come to Logarithms and I want to try and tackle that head on now.


Comments

  • Registered Users, Registered Users 2 Posts: 2,345 ✭✭✭Kavrocks


    Googling brings up a very good answer.


  • Registered Users, Registered Users 2 Posts: 252 ✭✭Chickentown


    Ok thanks, that is a good answer, I had googled already just thought I would come on here and air out some of my issues.

    So in a nutshell, when I take the logarithm of some number 'x' to some base 'b', I am finding the power 'b' must be raised to in order to 'get x'.

    In the study of electronics I have come across this formula for calculating gain in dB; 20log10(*gain) ....//I have used the '*' here for illustrative purposes.

    So what this now says to me is: To find gain in dB I need to multiply 20 by the power which 10 must be raised to in order to get *gain.

    So if *gain = 100 this implies gain in dB = 40.
    if *gain = 1000 this implies gain in dB = 60.
    if *gain = 1000000 this implies gain in dB = 120.

    So what I achieve by this is that I can represent large variances in gain on a graph in a meaningful way since the graph will be 'scaled' correctly?


  • Registered Users, Registered Users 2 Posts: 5,633 ✭✭✭TheBody


    Ok thanks, that is a good answer, I had googled already just thought I would come on here and air out some of my issues.

    So in a nutshell, when I take the logarithm of some number 'x' to some base 'b', I am finding the power 'b' must be raised to in order to 'get x'.

    In the study of electronics I have come across this formula for calculating gain in dB; 20log10(*gain) ....//I have used the '*' here for illustrative purposes.

    So what this now says to me is: To find gain in dB I need to multiply 20 by the power which 10 must be raised to in order to get *gain.

    So if *gain = 100 this implies gain in dB = 40.
    if *gain = 1000 this implies gain in dB = 60.
    if *gain = 1000000 this implies gain in dB = 120.

    So what I achieve by this is that I can represent large variances in gain on a graph in a meaningful way since the graph will be 'scaled' correctly?

    That's spot on. Well done.


  • Registered Users, Registered Users 2 Posts: 252 ✭✭Chickentown


    I appreciate the help guys, thanks.


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