Page 1 of 2 12 LastLast
Results 1 to 15 of 17

Math Help - Logs and Radicals

  1. #1
    Newbie
    Joined
    Dec 2010
    Posts
    10

    Logs and Radicals

    -1 = log(2+radical(5))
    I don't understand how its equal to -1?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,919
    Thanks
    1762
    Awards
    1
    Quote Originally Posted by HELPHELPHELP View Post
    -1 = log(2+radical(5))
    I don't understand how its equal to -1?
    There must be more to this problem than you posted.
    Please always post the entire question.
    You are quite right is cannot be -1 as written.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    What base is the log?

    Base 10 doesn't equal -1.

    Base e doesn't equal -1.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Dec 2010
    Posts
    10
    This the entire question, but someone please explain to me not solve, how does that part of the equation equal -1?
    Attached Thumbnails Attached Thumbnails Logs and Radicals-extra-credit.jpg  
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    This \displaystyle f(2)=4+log(2+\sqrt{5})=4.62696 equals on my calculator.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,919
    Thanks
    1762
    Awards
    1
    Now that makes more sense.
    If \log_b(2+\sqrt5)=-1 then 2+\sqrt5=b^{-1}.

    Can you solve that?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Dec 2010
    Posts
    10
    how would i find b?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Dec 2010
    Posts
    10
    Quote Originally Posted by Plato View Post
    Now that makes more sense.
    If \log_b(2+\sqrt5)=-1 then 2+\sqrt5=b^{-1}.

    Can you solve that?
    How would i find b? So in this case b is not 10 right, so that would mean it would come out as -1 in the calculator right?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    \displaystyle b^{-1}=\frac{1}{b}

    \displaystyle \frac{1}{b}=2+\sqrt{5} How would you solve for b?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    Dec 2010
    Posts
    10
    k
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Newbie
    Joined
    Dec 2010
    Posts
    10
    so now do i use log 1/2+radical 5 to solve for f(-2)?
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Newbie
    Joined
    Dec 2010
    Posts
    10
    Such as log#(-2+radical5)?
    #= 1/2+radical 5
    Follow Math Help Forum on Facebook and Google+

  13. #13
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    \displaystyle f(-2)=4+log_{\frac{1}{2+\sqrt{5}}}(-2+\sqrt{5})=x
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Newbie
    Joined
    Dec 2010
    Posts
    10
    ok i got that, but how do you find the exponent of a radical?
    Follow Math Help Forum on Facebook and Google+

  15. #15
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    \displaystyle log_a(b)=x\Rightarrow a^x=b
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Replies: 9
    Last Post: February 22nd 2011, 06:39 PM
  2. Radicals
    Posted in the Algebra Forum
    Replies: 8
    Last Post: August 3rd 2009, 09:12 AM
  3. Dealing with Logs and Natural Logs
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: April 14th 2008, 07:18 AM
  4. several questions-logs/natural logs
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: November 12th 2007, 09:58 PM
  5. radicals
    Posted in the Algebra Forum
    Replies: 2
    Last Post: January 11th 2007, 10:03 PM

Search Tags


/mathhelpforum @mathhelpforum