Results 1 to 4 of 4

Math Help - Solving Without a Calculator?

  1. #1
    Newbie
    Joined
    Jul 2010
    From
    New Jersey
    Posts
    20

    Solving Without a Calculator?

    Evaluate:

    log4 (1/8)

    how can i solve without a calculator. i know this answer is -1.5

    = log(1/8)
    log 4
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Mar 2009
    Posts
    378
    I would not use a change of base rule here. What you want is \frac{1}{8} in terms of 4's, i.e., \frac{1}{2}\cdot \frac{1}{4} = \left(\frac{1}{4}\right)^{0.5}\cdot \frac{1}{4}. Now you can use your log rules to calculate the actual value.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,915
    Thanks
    779
    Hello, danielh9103!

    Evaluate: . \log_4\!\left(\frac{1}{8}\right)

    Let: . \log_4\!\left(\frac{1}{8}\right) \:=\:P

    Then: . 4^P \:=\:\dfrac{1}{8} \quad\Rightarrow\quad (2^2)^P \:=\:\dfrac{1}{2^3} \quad\Rightarrow\quad 2^{2P} \:=\:2^{-3}

    Hence: . 2P \:=\:-3 \quad\Rightarrow\quad P \:=\:-\frac{3}{2}


    Therefore: . \log_4(\frac{1}{8}) \;=\;-\frac{3}{2}

    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1
    log(1/8)=log(1/2^3)=log(2^(-3))=-3log2=-3/2
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: May 6th 2012, 03:16 PM
  2. solving logs with calculator
    Posted in the Calculators Forum
    Replies: 3
    Last Post: November 21st 2009, 05:04 PM
  3. solving trigo without calculator
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: June 27th 2009, 02:39 AM
  4. Replies: 4
    Last Post: March 20th 2009, 03:25 AM
  5. logarithms/calculator/solving
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: November 18th 2007, 02:30 PM

Search Tags


/mathhelpforum @mathhelpforum