Results 1 to 5 of 5

Math Help - Logarithms.

  1. #1
    Member Awsom Guy's Avatar
    Joined
    Jun 2009
    From
    Mars
    Posts
    186

    Logarithms.

    Find the values of x for which:

    (log x)(log x^2) + log x^3 - 5 = 0
    All logs are to the base ten.

    I cannpt do this.
    This is what I did.
    log x + log x^2 + log x^3 - 5 = 0
    log x (1+x+x^2) - 5=0
    log x (1+x+x^2) = 5

    I got stuck there, but I am sure this method is wrong.
    Please help.
    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Oct 2009
    From
    Brisbane
    Posts
    311
    Thanks
    2
    Quote Originally Posted by Awsom Guy View Post
    Find the values of x for which:

    (log x)(log x^2) + log x^3 - 5 = 0
    All logs are to the base ten.

    I cannpt do this.
    This is what I did.
    log x + log x^2 + log x^3 - 5 = 0
    log x (1+x+x^2) - 5=0
    log x (1+x+x^2) = 5

    I got stuck there, but I am sure this method is wrong.
    Please help.
    Thanks
    You need to learn your basic log rules:
    1. log a + log b = log (ab) NOT log(a+b) which you tried to use in your second line. (Also in your first line your first + sign should be multiplication)
    2. log a -log b = log(a/b)
    and
    3. log(a^n) = n log a

    So: your first term:
    (log x)(log x^2) = (log x)* 2log x = 2 (log x)^2

    second term: (log x^3) = 3 log x

    So you have:
    2 (log x)^2 + 3 log x -5 = 0

    (Hint: Think quadratic!!) and go from there.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by Awsom Guy View Post
    Find the values of x for which:

    (log x)(log x^2) + log x^3 - 5 = 0
    All logs are to the base ten.

    I cannpt do this.
    This is what I did.

    log x + log x^2 + log x^3 - 5 = 0

    log x (1+x+x^2) - 5=0 ----- ???How did you arrive here??

    log x (1+x+x^2) = 5

    I got stuck there, but I am sure this method is wrong.
    Please help.
    Thanks

    One of basic properties of logarithms to any base is \log_ax^n=n\log_ax , so \log x+\log x^2+\log x^3=\log x+2\log x+3\log x=6\log x ...now continue and solve your problem.

    Tonio
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Oct 2009
    From
    Brisbane
    Posts
    311
    Thanks
    2
    Quote Originally Posted by tonio View Post
    One of basic properties of logarithms to any base is \log_ax^n=n\log_ax , so \log x+\log x^2+\log x^3=\log x+2\log x+3\log x=6\log x ...now continue and solve your problem.

    Tonio
    But the sign between the first two brackets in the original equation is multiplication not addition.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member Awsom Guy's Avatar
    Joined
    Jun 2009
    From
    Mars
    Posts
    186
    Tonio,
    I thought I could factorise that, and I now know that this is wrong.'Thanks to Debsta.
    Thanks
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. logarithms help
    Posted in the Algebra Forum
    Replies: 4
    Last Post: April 6th 2010, 06:29 PM
  2. Logarithms
    Posted in the Algebra Forum
    Replies: 2
    Last Post: April 6th 2010, 04:46 PM
  3. logarithms
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: March 24th 2010, 04:26 AM
  4. Logarithms
    Posted in the Algebra Forum
    Replies: 3
    Last Post: March 18th 2010, 02:52 PM
  5. logarithms
    Posted in the Algebra Forum
    Replies: 4
    Last Post: April 16th 2008, 09:55 AM

Search Tags


/mathhelpforum @mathhelpforum