Results 1 to 9 of 9

Math Help - Solving Log Equations

  1. #1
    Newbie
    Joined
    May 2008
    Posts
    17

    Solving Log Equations

    I'm having problems with these equations in particular:

    (logx^1/2)-(logx^1/3)=log2

    and

    log(base4)(x+2)+log(base4)(x-3)=log(base4)9

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member Jonboy's Avatar
    Joined
    May 2007
    From
    South Point
    Posts
    193
    log\cdot x^{\frac{1}{2}} - log \cdot x^{\frac{1}{3}} = log\,2

    Focus on the left of the equal sign. When you subtract logs, and they have like bases, you may divide what you are taking the log of.

    So: log\,\frac{x^{\frac{1}{2}}}{x^{\frac{1}{3}}} = log\,2

    Now the logs cancel, and you're left with: \frac{\sqrt{x}}{\sqrt[3]{x}} = 2

    Now cube both sides: \frac{\sqrt{x}\cdot\sqrt{x}\cdot \sqrt{x}}{x} = \frac{x\cdot \sqrt{x}}{x} = 8

    That simplifies to: \sqrt{x} = 8

    Square both sides, and so x is 64.

    Plug it back in to make sure it's correct.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    May 2008
    Posts
    17
    Thanks a lot. That helps. Could you show me how to do the second one too? I think it involves factoring and I couldn't figure it out.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member Jonboy's Avatar
    Joined
    May 2007
    From
    South Point
    Posts
    193
    log_{4}\,(x + 2) + log_{4}\,(x - 3) = log_{4}\,9

    Once again, focus on the left side of the equal sign. When you add logs with like bases, you may multiply what you are taking the log of.

    So: log_{4}\,(x + 2)(x-3) = log_{4}\,9

    Logs cancel: (x + 2)\,(x - 4) = 9 \Rightarrow x^2 -2x - 17 = 0

    Use the quadratic formula.

    I got x =\pm 3\sqrt{2} + 1

    Check to see if both solutions work.
    Last edited by Jonboy; May 25th 2008 at 06:04 PM. Reason: simple mistake, then simplifying
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member Jonboy's Avatar
    Joined
    May 2007
    From
    South Point
    Posts
    193
    and when i say "plug it back in..." i mean in the original problem. make sure you have no extraneous, or false, solutions.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    May 2008
    Posts
    17
    So:

    Logs cancel:

    How did you go from log(base4)(x+2)(x-4) to (x+2)(x-4)? shouldnt it still be (x+2)(x-4)?

    If you the part that confuses me is if you have (x+2)(x-4) because then you have an x factor to deal with also.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    May 2008
    Posts
    17
    or wait sorry, originally it was (x+2)(x-3)
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member Jonboy's Avatar
    Joined
    May 2007
    From
    South Point
    Posts
    193
    i messed up. got too wishful.

    i edited my post. i think it's correct now.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    May 2008
    Posts
    17
    Thanks So Much
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 6
    Last Post: November 30th 2011, 02:41 AM
  2. Solving two equations
    Posted in the Algebra Forum
    Replies: 7
    Last Post: November 21st 2011, 03:55 PM
  3. Solving equations
    Posted in the Algebra Forum
    Replies: 29
    Last Post: November 18th 2011, 07:05 AM
  4. Solving these equations...
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: April 27th 2009, 06:06 AM
  5. solving equations
    Posted in the Algebra Forum
    Replies: 3
    Last Post: May 16th 2006, 04:58 PM

Search Tags


/mathhelpforum @mathhelpforum