Results 1 to 6 of 6

Thread: Logarithms

  1. #1
    Newbie
    Joined
    Jun 2006
    Posts
    12

    Logarithms

    Find the value oflog 2^1/2+1(3-2^1/2)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    5
    Quote Originally Posted by Joyce
    Find the value of log 2^1/2+1(3-2^1/2)
    You need to use brackets to make this unambiguous.

    This could mean:

    $\displaystyle
    \log(2^{1/2}+1)\times (3-2^{1/2})
    $,

    $\displaystyle
    \log[(2^{1/2}+1)\times (3-2^{1/2})]
    $

    or

    $\displaystyle
    \log((2^1/2+1)(3-2^{1/2}))
    $

    or any number of other things

    RonL
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    849
    Hello,Joyce!

    I'd like to help, but I can't read the problem . . .


    Find the value oflog 2^1/2+1(3-2^1/2)

    Did you even look at what showed up on the screen?


    By the way, there should be no exponent in your log statement.

    You probably meant something like: $\displaystyle \log_2\left(\frac{1}{2}\right)$

    If you see $\displaystyle \log_28$, do not write it as: log 2^8, which means $\displaystyle \log 2^8$

    The "2" is a subscript, written 'below the line'.
    . . You can write: log_2(8) . . . okay?

    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Dec 2005
    From
    Kolkata, India
    Posts
    19
    I think he meant
    $\displaystyle
    \log_{\sqrt{2}+1}(3-\sqrt{2})
    $
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    5
    Quote Originally Posted by shubhadeep
    I think he meant
    $\displaystyle
    \log_{\sqrt{2}+1}(3-\sqrt{2})
    $
    Good idea, I would give you +rep for that but I can't give you more
    +rep today

    Then by the change of logarithm base rule, this can be written:

    $\displaystyle
    \log_{\sqrt{2}+1}(3-\sqrt{2})=\frac{\log_e(3-\sqrt{2})}{\log_e(\sqrt{2}+1)}$$\displaystyle
    \approx 0.5231


    $

    RonL
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    849
    I too like Shubadeep's interpretation.

    But I hope that there are typos in the original problem.

    If it was: .$\displaystyle \log_{1+\sqrt{2}}(3 + 2\sqrt{2}) $ . . . the answer is $\displaystyle 2$.


    Otherwise, I see no point to the problem
    . . other than using the Base-Change Formula
    . . which could have been practiced with $\displaystyle \log_{1.586}(2.414)$

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. logarithms help
    Posted in the Algebra Forum
    Replies: 4
    Last Post: Apr 6th 2010, 06:29 PM
  2. Logarithms
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Apr 6th 2010, 04:46 PM
  3. logarithms
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: Mar 24th 2010, 04:26 AM
  4. Logarithms
    Posted in the Algebra Forum
    Replies: 3
    Last Post: Mar 18th 2010, 02:52 PM
  5. logarithms
    Posted in the Algebra Forum
    Replies: 4
    Last Post: Apr 16th 2008, 09:55 AM

Search Tags


/mathhelpforum @mathhelpforum