Results 1 to 10 of 10

Math Help - log9 (4) + log3(x) = 3 solve for x

  1. #1
    Member
    Joined
    Jul 2009
    From
    London
    Posts
    109

    log9 (4) + log3(x) = 3 solve for x

    log9 (4) + log3(x) = 3 solve for x

    by this I mean log base 9 (4) etc

    The different bases makes this harder for me.

    This is my (incorrect) attempt

    log3(x) + log3(4) = 3
    -------
    log3(9)


    log3(x) + log3(4/9) = 3

    log3(4/9) = log(4/9) / log(3) = -0.35

    log3(x) = 3 -- 0.35 = 3.35

    3^3.35 = x = 39.66 which is incorrect.

    Where did I go wrong and how do I solve this?

    Angus
    Last edited by angypangy; May 2nd 2011 at 02:41 AM. Reason: mistake
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,390
    Thanks
    1476
    Awards
    1
    Quote Originally Posted by angypangy View Post
    log9 (4) + log3(x) = 3 solve for x
    Use these facts: \log_9(4)=\frac{\ln(4)}{\ln(9)}~\&~\log_3(x)=\frac  {\ln(x)}{\ln(3)}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,667
    Thanks
    298
    Awards
    1
    Quote Originally Posted by angypangy View Post
    log9 (4) + log3(x) = 3 solve for x

    by this I mean log base 9 (4) etc

    The different bases makes this harder for me.

    This is my (incorrect) attempt

    log3(x) + log3(4) = 3
    -------
    log3(9)


    log3(x) + log3(4/9) = 3

    log3(4/9) = log(4/9) / log(3) = -0.35

    log3(x) = 3 -- 0.35 = 3.35

    3^3.35 = x = 39.66 which is incorrect.

    Where did I go wrong and how do I solve this?

    Angus
    Your second equation confused me for a moment. It didn't print out the way you wanted it to. Here it is, just for clarity:
    log_3(x) + \frac{log_3(4)}{log_3(9)} = 3

    You applied your log rules incorrectly in the next line.
    \frac{log_3(4)}{log_3(9)} \neq log_3 \left ( \frac{4}{9} \right )

    But note that log_3(9) = log_3(3^2). What can you do with this?

    The rest of the method is correct, but unless you have any conflicting instructions, I'd solve it exactly...ie leave it in terms of log3(4). (Which you can actually simplify a bit.) There is no need to use a calculator on this one.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Jul 2009
    From
    London
    Posts
    109
    Ah I looked at the log9 and thought I must be missing some trick with 3^2 = 9

    OK so you just work out the logs. OK, no problem. Thanks.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Jul 2009
    From
    London
    Posts
    109
    Is this correct?

    log3(4)
    -------
    log3(9)

    = log3(4) - log3(9) = log3(4) - 2*log3(3) = log3(4) - 2
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,667
    Thanks
    298
    Awards
    1
    Quote Originally Posted by angypangy View Post
    Is this correct?

    log3(4)
    -------
    log3(9)

    = log3(4) - log3(9) = log3(4) - 2*log3(3) = log3(4) - 2
    This is correct, but I'd do it a bit differently. First of all, yes log_3(9) = 2. So....
    \frac{log_3(4)}{log_3(9)} = \frac{1}{2} log_3(4)

    The reason I would do so is that you can also simplify log_3(4) = log_3(2^2)

    See what that gives you.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,547
    Thanks
    539
    Hello, angypangy!


    I solved it like this . . .





    . .




    . .

    . .



    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Jul 2009
    From
    London
    Posts
    109
    My final working out was like this:

    log9(4) = log3(4)/log3(9) = log3(2^2)/log3(3^2) = 2*log3(2)/2*log3(3) - cancels down to log3(2)

    => log3(2) + log3(x) = 3

    => 3^3 = 2x => 2x = 27 => x = 27/2

    Many thanks for your help.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member
    Joined
    Nov 2007
    From
    Trumbull Ct
    Posts
    888
    Thanks
    25
    Hi angypangy. I did this slightly differently The logB9(4)is converted to B3
    log B9 (9) =1 Log B3 (9) =2 therefor log B9 (4) = 1/2 log B3 (4)

    The final equations logB3 (2) + logB3(x) = log B3 (27)
    log B3 (2x) = log B3 (27)
    2x =27 x=27/2


    bjh
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by angypangy View Post
    My final working out was like this:

    log9(4) = log3(4)/log3(9) = log3(2^2)/log3(3^2) = 2*log3(2)/2*log3(3) - cancels down to log3(2)

    => log3(2) + log3(x) = 3

    => 3^3 = 2x => 2x = 27 => x = 27/2

    Many thanks for your help.
    I would solve it similarly...

    log_94=y\Rightarrow\ 4=9^y\Rightarrow2^2=3^{2y}\Rightarrow\ 2=3^y

    \Rightarrow\ log_32=y=log_94

    which is an alternative to changing base, due to the 9 being "3 squared".

    Then

    log_32+log_3x=3=log_327

    log_32x=log_327

    2x=27
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. need to solve summation equation to solve sum(x2)
    Posted in the Statistics Forum
    Replies: 2
    Last Post: July 16th 2010, 10:29 PM
  2. X^2+xlog6+(log2)(log3)=0?
    Posted in the Algebra Forum
    Replies: 1
    Last Post: July 14th 2010, 05:41 PM
  3. solve in C2
    Posted in the Algebra Forum
    Replies: 0
    Last Post: March 2nd 2010, 10:28 AM
  4. Replies: 5
    Last Post: December 8th 2009, 01:38 PM
  5. Replies: 1
    Last Post: June 9th 2009, 10:37 PM

Search Tags


/mathhelpforum @mathhelpforum