Results 1 to 4 of 4

Math Help - Log

  1. #1
    Member
    Joined
    Sep 2006
    Posts
    134

    Log

    Help please!
    Attached Thumbnails Attached Thumbnails Log-log1.jpg  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member General's Avatar
    Joined
    Jan 2010
    From
    Kuwait
    Posts
    562
    since \log_{3}(3)=1, \log_{9}(9)=1 and 1=\log_{4}(4), the equation can be written like the following:
    2\log_{4}(3x)-\log_{4}(x)=\log_{4}(4)-\frac{1}{2}\log_{4}(x^2)
    since a\log(b)=\log(b)^a, then:
    \log_{4}(3x)^2-\log_{4}(x)=\log_{4}(4)-\log_{4}(x^2)^{\frac{1}{2}}
    since (x^2)^{\frac{1}{2}}=x and \log(a)-\log(b)=\log(\frac{a}{b})
    then the equation can be written like:
    \log_{4}(\frac{(3x)^2}{x})=\log_{4}(\frac{4}{x})
    \log_{4}(9x)=\log_{4}(\frac{4}{x})

    Now, Solve the last equation.
    Last edited by General; January 19th 2010 at 07:24 AM. Reason: typo..I did not see the 2 in the first ;s
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

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

    2\log_4(3x)\log_3(3) - \log_4(x) \;=\;1 - \tfrac{1}{2}\log_4(x^2)\log_9(9)

    I have it down to this and I'm lost: . 2\log_4(3x) \:=\:1 . .
    . . . Keep going!

    \text{You have: }\;\;\log_4(3x)^2 \;=\;1

    . . . . . . . . \log_4(9x^2) \;=\;1

    . . . . . . . . . . . 9x^2 \;=\;4^1

    . . . . . . . . . . . x^2 \;=\;\frac{4}{9}

    . . . . . . . . . . . . x \;=\;\frac{2}{3}


    (The negative square root is an extraneous root.)

    Last edited by Soroban; January 19th 2010 at 07:27 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Quote Originally Posted by reiward View Post
    Help please!

    Assuming that what you wrote is correct (I didn't check), we get:

    2\log_4(3x)=1\Longrightarrow\log_4(3x)^2=1\Longrig  htarrow (3x)^2=4^1 ...take it from here

    Tonio
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum