Results 1 to 10 of 10
Like Tree3Thanks
  • 2 Post By Prove It
  • 1 Post By bjhopper

Math Help - Logarithms

  1. #1
    Junior Member
    Joined
    Mar 2013
    From
    Unreals
    Posts
    44

    Logarithms

    Hi!

    Solve:
    1. 2^{x - 1} = 3

    \log_2 2^{x - 1} = \log_2 3

    (x - 1)\log_2 2 = \log_2 3

    x - 1 = \log_2 3

    x = \log_2 3 + 1
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    10,969
    Thanks
    1009

    Re: Logarithms

    What are you actually asking? If it's correct? Because everything you've done is fine
    Thanks from Unreal and topsquark
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Nov 2007
    From
    Trumbull Ct
    Posts
    888
    Thanks
    25

    Re: Logarithms

    Hi unreal,
    Why use base 2 and not base 10 or base e.Othewise you will have to change base to solve for x
    Thanks from abualabed
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    14,973
    Thanks
    1121

    Re: Logarithms

    I disagree with that. In order to get a specific numerical value for x, yes, you would have to change to base 10 or base e.

    But " x= log_2(3)+ 1" is a perfectly good solution to the equation.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Nov 2007
    From
    Trumbull Ct
    Posts
    888
    Thanks
    25

    Re: Logarithms

    Quote Originally Posted by HallsofIvy View Post
    I disagree with that. In order to get a specific numerical value for x, yes, you would have to change to base 10 or base e.

    But " x= log_2(3)+ 1" is a perfectly good solution to the equation.


    But the poster has not solved for x
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,386
    Thanks
    1476
    Awards
    1

    Re: Logarithms

    Quote Originally Posted by Unreal View Post
    Solve:
    1. 2^{x - 1} = 3
    x = \log_2 3 + 1
    Quote Originally Posted by bjhopper View Post
    But the poster has not solved for x
    Why in the world would you say that?

    What do you mean by a solution?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member
    Joined
    Nov 2007
    From
    Trumbull Ct
    Posts
    888
    Thanks
    25

    Re: Logarithms

    Quote Originally Posted by Plato View Post
    Why in the world would you say that?

    What do you mean by a solution?

    See post 3.I think the OP does not know how to find the numerical value of x
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,386
    Thanks
    1476
    Awards
    1

    Re: Logarithms

    Quote Originally Posted by bjhopper View Post
    See post 3.I think the OP does not know how to find the numerical value of x
    So, is place of "But the poster has not solved for x" you really meant "But the poster has no numerical value for x"?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor

    Joined
    Apr 2005
    Posts
    14,973
    Thanks
    1121

    Re: Logarithms

    Quote Originally Posted by bjhopper View Post
    But the poster has not solved for x
    Of course he has. He has x equal to a specific number, ln_3(2)+ 1. Perhaps you don't like the way the number is written, but that has nothing to do with "solving". Are you saying that he should solve for \frac{log(2)}{log(3)}+ 1= \frac{0.3010}{0.47712}+ 1= 0.63093+ 1= 1.06093? Because that is NOT the correct answer- it is only approximately correct.

    It is the same as saying that the solutions to x^2= 2 are \sqrt{2} and -\sqrt{2}, NOT "1.41421" and "-1.41421".
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Super Member
    Joined
    Nov 2007
    From
    Trumbull Ct
    Posts
    888
    Thanks
    25

    Re: Logarithms

    Quote Originally Posted by HallsofIvy View Post
    Of course he has. He has x equal to a specific number, ln_3(2)+ 1. Perhaps you don't like the way the number is written, but that has nothing to do with "solving". Are you saying that he should solve for \frac{log(2)}{log(3)}+ 1= \frac{0.3010}{0.47712}+ 1= 0.63093+ 1= 1.06093? Because that is NOT the correct answer- it is only approximately correct.

    It is the same as saying that the solutions to x^2= 2 are \sqrt{2} and -\sqrt{2}, NOT "1.41421" and "-1.41421".


    You've jumbled a few digits.Iam saying only that the OP should use base 10 0r base e to find the numerical value of x
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: November 17th 2012, 10:51 AM
  2. Logarithms
    Posted in the Algebra Forum
    Replies: 4
    Last Post: October 29th 2010, 12:35 PM
  3. Logarithms
    Posted in the Algebra Forum
    Replies: 7
    Last Post: April 17th 2010, 09:31 PM
  4. logarithms
    Posted in the Algebra Forum
    Replies: 1
    Last Post: April 7th 2010, 07:40 AM
  5. Logarithms
    Posted in the Algebra Forum
    Replies: 2
    Last Post: November 24th 2008, 04:17 PM

Search Tags


/mathhelpforum @mathhelpforum