Results 1 to 7 of 7

Math Help - Logs

  1. #1
    Newbie
    Joined
    Apr 2010
    Posts
    3

    Logs

    Brand new to this so aplogies if anything is wrong:

    Any help with this would be appreciated:

    Solve x for the equation:

    1/2 log5 (x^2-1) = 1/4 + 1/2log5(x-1)

    The log5 is log to the base 5

    Hope this makes sense
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Dec 2009
    From
    1111
    Posts
    872
    Thanks
    3
    Quote Originally Posted by wtopping View Post
    Brand new to this so aplogies if anything is wrong:

    Any help with this would be appreciated:

    Solve x for the equation:

    1/2 log5 (x^2-1) = 1/4 + 1/2log5(x-1)

    The log5 is log to the base 5

    Hope this makes sense
    Dear wtopping,

    So your equation is, \frac{1}{2}log_{5}(x^2-1)=\frac{1}{4}+\frac{1}{2}log_{5}(x-1)

    Am I correct??
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Apr 2010
    Posts
    3
    Yes, exactly that, thanks for the quick reply,

    Wayne
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Dec 2009
    From
    1111
    Posts
    872
    Thanks
    3
    Quote Originally Posted by wtopping View Post
    Yes, exactly that, thanks for the quick reply,

    Wayne
    Dear wtopping,

    Let me give you some ideas....

    First you can factorize (x^2-1). Then try using the logarithm identity, logA-logB=log\frac{A}{B}

    I hope you could continue from here!! In any case please don't hesitate to reply!!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Apr 2010
    Posts
    3
    Having come back to maths after a 15 year break i seem unable to see the woods for the trees. I think i can see that i need to bring the right hand side log to the left of = but stuck after that. I think the 1/2 is confusing me!!!

    Its been a long time
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Dec 2009
    From
    1111
    Posts
    872
    Thanks
    3
    Quote Originally Posted by wtopping View Post
    Having come back to maths after a 15 year break i seem unable to see the woods for the trees. I think i can see that i need to bring the right hand side log to the left of = but stuck after that. I think the 1/2 is confusing me!!!

    Its been a long time
    Dear wtopping,

    Ok. After taking the right hand side log to the left hand side try to apply the log identity I have given you. For this you have take 1/2 out of the bracket first.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,553
    Thanks
    542
    Hello, wtopping!

    Solve for x\!:\;\;\tfrac{1}{2}\log_5(x^2-1) \:=\:\tfrac{1}{4}+ \tfrac{1}{2}\log_5(x-1)

    Note that: {\color{blue}x \neq 1}

    Don't like fractions? . . . Eliminate them!


    Multiply by 4: . 2\log_5(x^2-1) \;=\;1 + 2\log_5(x-1)

    . . 2\log_5(x^2-1) - 2\log_5(x-1) \;=\;1 \quad\Longrightarrow\quad 2\bigg[\log_5(x^2-1) - \log_5(x-1)\bigg] \;=\;1

    . . 2\log_5\left[\frac{x^2-1}{x-1}\right] \;=\;1 \quad\Longrightarrow\quad 2\log_5\left[\frac{(x-1)(x+1)}{x-1}\right] \;=\;1


    . . 2\log_5(x+1) \;=\;1 \quad\Longrightarrow\quad \log_5(x+1) \:=\:\tfrac{1}{2}


    . . x + 1 \;=\;5^{\frac{1}{2}} \quad\Longrightarrow\quad x \;=\;\sqrt{5} - 1

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 9
    Last Post: February 22nd 2011, 05:39 PM
  2. Logs
    Posted in the Algebra Forum
    Replies: 4
    Last Post: April 24th 2010, 07:52 AM
  3. Logs
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: October 10th 2009, 06:08 PM
  4. Dealing with Logs and Natural Logs
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: April 14th 2008, 06:18 AM
  5. several questions-logs/natural logs
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: November 12th 2007, 08:58 PM

Search Tags


/mathhelpforum @mathhelpforum