Results 1 to 7 of 7

Math Help - Logarithimiccs

  1. #1
    Newbie
    Joined
    Jun 2006
    Posts
    4

    Logarithimiccs

    2 Problems I need help on

    Log(5)2/Log(5)X=X/LogX

    and

    Log4Log(2)X=1

    both solved without a calculator
    Last edited by scfan000; June 5th 2006 at 06:48 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    I'm confused about your notation. Is this what you mean for the first one? \frac{\log_{5}2}{\log_{5}x}=\frac{x}{\log x}?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jun 2006
    Posts
    4
    yes it is
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by scfan000
    2 Problems I need help on

    Log(5)2/Log(5)X=X/LogX

    and

    Log4Log(2)X=1

    both solved without a calculator
    I think you will have to adopt a clearer notation here.

    I'm guessing here, do you intend that

    <br />
Log(5)2=\log_5(2)<br />
?

    Then do you intend that LogX denote the common (log to base 10)
    or natrural log of X?

    I am also mystified by what you want "Log4Log(2)X" to represent.

    Might I suggest that you use "log_n(x)" to denote the log to the base
    n of x, be explicit in all instances about the base, and rewrite your questions
    accordingly?

    RonL
    Last edited by CaptainBlack; June 5th 2006 at 07:01 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Jameson
    I'm confused about your notation. Is this what you mean for the first one? \frac{\log_{5}2}{\log_{5}x}=\frac{x}{\log x}?

    To solve:

    \frac{\log_{5}2}{\log_{5}x}=\frac{x}{\log x},

    first turn it upside down (this is a personal preference, it can be done this way up):

    \frac{\log_{5}(x)}{\log_{5}(2)}=\frac{\log(x)}{x},

    then apply the change of base rule for logs to the left hand side to reduce
    the all the logs to the same base as that on the right hand side.

    RonL
    Last edited by CaptainBlack; June 5th 2006 at 07:15 AM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Jun 2006
    Posts
    4
    i c it now. and i can figure out the problem from there
    sry about the notation. i just don't quite know how to yet
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Jun 2006
    Posts
    5

    haj

    You need to solve:
    Log(5)2/Log(5)X=X/LogX

    In your notation:
    Log(5)2 can be written also as Log2/Log5
    and similary
    Log(5)X = LogX/Log5

    When we put this into equation we get:
    Log2/LogX=X/LogX

    From this: X=Log2
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum