Results 1 to 3 of 3

Thread: Log Equations

  1. #1
    MHF Contributor
    Joined
    Jul 2008
    From
    NYC
    Posts
    1,489

    Log Equations

    Solve each log equation below. Express irrational solutions in exact form and as a decimal rounded to 3 decimal places.

    (1) -2 log_4 (x) = log_4 (9)

    (2) ln (x + 1) - ln x = 2

    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Sep 2007
    Posts
    528
    Awards
    1
    Quote Originally Posted by magentarita View Post
    Solve each log equation below. Express irrational solutions in exact form and as a decimal rounded to 3 decimal places.

    (1) -2 log_4 (x) = log_4 (9)

    (2) ln (x + 1) - ln x = 2
    1.)
    $\displaystyle -2 \log _4 (x) = \log_4 (9)$
    A general rule is: $\displaystyle a \log (x) = \log (x^a)$
    Hence: $\displaystyle \log_4(x^{-2}) = \log_4(9)$
    $\displaystyle \log_4\left(\frac{1}{x^2}\right) = \log_4(9)$
    As both sides are to the same logarithm base, we can equate the changing variable hence:
    $\displaystyle \frac{1}{x^2} = 9$
    Now solve for $\displaystyle x$


    2.)
    $\displaystyle \ln (x+1) - \ln x = 2$
    A general rule is: $\displaystyle \ln (a) - \ln (b) = \ln \left(\frac{a}{b}\right)$
    Hence: $\displaystyle \ln \left(\frac{x+1}{x}\right) = 2$
    A general rule is: $\displaystyle e^{\ln(ax)} = ax$
    Hence: $\displaystyle e^{\ln \left(\frac{x+1}{x}\right)} = e^{2}$
    $\displaystyle \frac{x+1}{x} = e^{2}$
    Now solve for $\displaystyle x$ using algebra skills.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Jul 2008
    From
    NYC
    Posts
    1,489

    Air...

    Great reply as always. Where have you been, Air?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: Apr 7th 2010, 02:22 PM
  2. Equi Potential Equations or Euler Type Equations
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: Apr 2nd 2010, 02:58 AM
  3. Replies: 3
    Last Post: Feb 27th 2009, 07:05 PM
  4. Replies: 1
    Last Post: Sep 1st 2007, 06:35 AM
  5. Replies: 1
    Last Post: Jul 29th 2007, 02:37 PM

Search Tags


/mathhelpforum @mathhelpforum