Results 1 to 6 of 6
Like Tree2Thanks
  • 1 Post By Plato
  • 1 Post By HallsofIvy

Thread: logaithm help.

  1. #1
    Newbie
    Joined
    Mar 2012
    From
    europe
    Posts
    3

    logarithm help.



    How do you solve this? Why does the answer become squareroot(3)-1??
    Last edited by Iamnothere; Mar 27th 2012 at 01:58 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,800
    Thanks
    2829
    Awards
    1

    Re: logaithm help.

    Quote Originally Posted by Iamnothere View Post
    How do you solve this? Why does the answer become squareroot(3)-1??
    Because $\displaystyle \sqrt3>1$ we must rewwrite $\displaystyle \log_{25}(1-\sqrt3)^2=2\log_{25}(\sqrt3-1)$

    Note $\displaystyle 25^{\log_{25}(\sqrt3-1)}=\sqrt3 -1$
    Thanks from Iamnothere
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,802
    Thanks
    3035

    Re: logaithm help.

    $\displaystyle 5= 25^{1/2}$ so
    $\displaystyle 5^{log_{25}(a)}= (25^{1/2})^{log_{25}(a)}= 25^{(1/2)log_{25}(a)}= 25^{log_{25}(\sqrt{a})}= \sqrt{a}$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Mar 2012
    From
    europe
    Posts
    3

    Re: logaithm help.

    Do you mind me asking where does that rule come from? The one with sqrt(3)>1, I'm a little bit confused.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,802
    Thanks
    3035

    Re: logaithm help.

    Quote Originally Posted by Iamnothere View Post
    Do you mind me asking where does that rule come from? The one with sqrt(3)>1, I'm a little bit confused.
    ??? You will accept 3> 1 won't you? And square root is an increasing function so $\displaystyle \sqrt{3}> \sqrt{1}= 1$. In fact, it is not at all difficult to use a calculator to see that $\displaystyle \sqrt{3}= 1.7320508075688772935274463415059... > 1$.

    Or do you mean "$\displaystyle ln(1- \sqrt{3})^2)= 2ln(\sqrt{3}- 1)$"?

    Obviously both $\displaystyle (1- \sqrt{3})^2$ and $\displaystyle (\sqrt{3}- 1)^2$ give the same thing and $\displaystyle log(a^2)= 2log(a)$ provided that a is positive becauser log(a) is not defined for negative a.
    Last edited by HallsofIvy; Mar 27th 2012 at 02:04 PM.
    Thanks from Iamnothere
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Mar 2012
    From
    europe
    Posts
    3

    Re: logaithm help.

    Emm, I think you got me a little bit wrong here, I was just wondering why did we have to write sqrt(3)-1 instead of 1-sqrt(3).

    Edit: Ah yeah, I totally forgot about looking at that part. Thank you.
    Last edited by Iamnothere; Mar 27th 2012 at 02:07 PM.
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum