Results 1 to 4 of 4

Math Help - I'm having some trouble with a logarithmic equation

  1. #1
    Newbie
    Joined
    Feb 2009
    From
    Sweden
    Posts
    3

    I'm having some trouble with a logarithmic equation

    I have tried to solve it in several different ways, but can't get it right. So I need to see the working. Hope someone can help me, thanks.

    Solve for x in the logarithmic equation:
    log7 - log(4x + 5) + log(2x - 3) = 0
    Give exact answers and be sure to check for extraneous solutions.
    The key says x = 13/5
    Last edited by Dirtsa; February 7th 2009 at 08:23 AM. Reason: solved
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,875
    Thanks
    656
    Quote Originally Posted by Dirtsa View Post
    I have tried to solve it in several different ways, but can't get it right. So I need to see the working. Hope someone can help me, thanks.

    Solve for x in the logarithmic equation:
    log7 - log(4x + 5) + log(2x - 3) = 0
    Give exact answers and be sure to check for extraneous solutions.
    The key says x = 13/5
    \log{7} - \log(4x + 5) + \log(2x - 3) = 0<br />

    \log\left[\frac{7(2x-3)}{4x+5}\right] = 0

    \frac{7(2x-3)}{4x+5} = 1

    7(2x-3) = 4x+5

    finish
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by Dirtsa View Post
    I have tried to solve it in several different ways, but can't get it right. So I need to see the working. Hope someone can help me, thanks.

    Solve for x in the logarithmic equation:
    log7 - log(4x + 5) + log(2x - 3) = 0
    Give exact answers and be sure to check for extraneous solutions.
    The key says x = 13/5
    Use the log-rules

    1. \log(a)-\log(b)=\log\left(\frac ab\right) and

    2. \log(a)+\log(b)=\log\left( a\cdot b\right)


    3. \log7 - \log(4x + 5) + \log(2x - 3) = 0~\implies~ \log\left(\dfrac7{4x+5} \cdot (2x-3)\right)=0

    Now use the logs (the RHS of the equation is a log too) with the base 10:

    \dfrac7{4x+5} \cdot (2x-3) = 1~\implies~14x-21 = 4x+5

    Solve for x.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Feb 2009
    From
    Sweden
    Posts
    3

    thank you

    thanks for the help
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Having Trouble Solving Logarithmic Equation
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: October 17th 2011, 03:40 PM
  2. logarithmic equation
    Posted in the Algebra Forum
    Replies: 1
    Last Post: May 13th 2009, 12:55 PM
  3. logarithmic differentiation trouble
    Posted in the Calculus Forum
    Replies: 5
    Last Post: January 23rd 2009, 05:16 PM
  4. Logarithmic equation
    Posted in the Calculus Forum
    Replies: 7
    Last Post: September 29th 2008, 06:32 PM
  5. Trouble with logarithmic Integral
    Posted in the Calculus Forum
    Replies: 3
    Last Post: February 24th 2008, 10:22 AM

Search Tags


/mathhelpforum @mathhelpforum