Results 1 to 3 of 3

Math Help - Solving exponential to different bases and a constant term

  1. #1
    Newbie
    Joined
    Mar 2009
    Posts
    3

    Solving exponential to different bases and a constant term

    Trying to solve for x
    2^(0.5x)= [3^(-0.5x)] + 5/3

    I know, the usual, take logs of both sides and bring down the exponent, but I'm having a problem with the RHS of the equation, since I can't take logs seperately, i.e. log(a+b) is NOT equal to log(a)+log(b)

    I actually know (graphically) that the answer is x=2, and I know that's correct from substituting. Any ideas on how to approach it (besides trial and error)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by skylabel View Post
    Trying to solve for x
    2^(0.5x)= [3^(-0.5x)] + 5/3

    I know, the usual, take logs of both sides and bring down the exponent, but I'm having a problem with the RHS of the equation, since I can't take logs seperately, i.e. log(a+b) is NOT equal to log(a)+log(b)

    I actually know (graphically) that the answer is x=2, and I know that's correct from substituting. Any ideas on how to approach it (besides trial and error)
    If you know that there's a solution and that you have to find it by hand, it makes sense to guess-and-check even values of x. And x = 2 would naturally be the first value to check .....

    Off-hand, I don't see any simple algebraic approach to the question.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Mar 2009
    Posts
    31
    Hello!

    If you will work with the formula a little you will be able to write it like this:



     (\sqrt3)^x * [(\sqrt2)^x -5/3 ]=1 (1)

    as you can observe (\sqrt3)^x > 0 =>  (\sqrt2)^x -5/3 >0 =>

    (\sqrt2)^x > 5/3 => x>1,5 (2)

    The second step is to observe that with x>1 => (\sqrt3)^x >1 => (\sqrt2)^x -5/3 < 1 in order for equation (1) to be verified.

    From the condition  (\sqrt2)^x -5/3 < 1 => (\sqrt2)^x<8/3 => x<2.5 (3)


    From the condition (2) and (3) we have 1 < x< 3 . The only solution for x in this interval in order for (\sqrt2)^x and (\sqrt3)^x to be rational numbers is x=2;

    To demonstrate that this is the only solution it,s easy to study the monotony of the function in interval ( 1,5 ; 2,5)

    Have a nice day!
    Last edited by Hush_Hush; March 21st 2009 at 02:30 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. First order ODE with trignometric function and constant term
    Posted in the Differential Equations Forum
    Replies: 5
    Last Post: September 21st 2010, 11:12 AM
  2. Constant term problem
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: April 1st 2010, 01:16 PM
  3. 2nd ODE -- Finding Particular solution with A constant term !!
    Posted in the Differential Equations Forum
    Replies: 4
    Last Post: November 14th 2009, 05:21 AM
  4. Solving for constant in exponential
    Posted in the Algebra Forum
    Replies: 7
    Last Post: November 14th 2008, 07:03 PM
  5. Constant term
    Posted in the Algebra Forum
    Replies: 5
    Last Post: April 3rd 2008, 12:44 PM

Search Tags


/mathhelpforum @mathhelpforum