Results 1 to 7 of 7
Like Tree3Thanks
  • 1 Post By Plato
  • 1 Post By Soroban
  • 1 Post By SlipEternal

Math Help - Solve x if b=(a^x+a^-x)/2

  1. #1
    Junior Member
    Joined
    May 2014
    From
    India
    Posts
    74
    Thanks
    3

    Question Solve x if b=(a^x+a^-x)/2

    <br />
\\b=\frac{a^x+a^{-x}}{2}\text{Find x}\\<br /> <br />
    I took it to x=\log_a (2b-a^{-x})

    The choices are
    a) log_a(b+\sqrt{b^2+1})
    b) log_a(b+\sqrt{b^2-1})
    c) log_b(a+\sqrt{a^2+1})
    d)none of these.

    Is it none of these?
    Can anyone help me? Or at least a hint?
    Last edited by NameIsHidden; May 30th 2014 at 04:21 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1574
    Awards
    1

    Re: Solve x if b=(a^x+a^-x)/2

    Quote Originally Posted by NameIsHidden View Post
    $
    \\b=\frac{a^x+a^{-x}}{2}\text{Find x}\\$
    I took it to x=\log_a (2b-a^{-x})

    The choices are
    a) log_a(b+\sqrt{b^2+1})
    b) log_a(b+\sqrt{b^2-1})
    c) log_b(a+\sqrt{a^2+1})
    d)none of these.
    Write as $a^{2x}-2ba^x+1=0$ solve for $a^x$.

    BTW use $\$a^x\$$ not [ tex][/tex].
    Thanks from NameIsHidden
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    May 2014
    From
    India
    Posts
    74
    Thanks
    3

    Re: Solve x if b=(a^x+a^-x)/2

    Quote Originally Posted by Plato View Post
    Write as $a^{2x}-2ba^x+1=0$ solve for $a^x$.

    BTW use $\$a^x\$$ not [ tex][/tex].
    $a^x=\frac{1}{a^x-2b}$

    If I substitute this in given equation, I end up with weird results.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,678
    Thanks
    611

    Re: Solve x if b=(a^x+a^-x)/2

    Hello, NameIsHidden!

    b\:=\:\frac{a^x+a^{-x}}{2}.\;\text{ Find }x

    (a)\;\log_a(b+\sqrt{b^2+1})
    (b)\;\log_a(b+\sqrt{b^2-1})
    (c)\;\log_b(a+\sqrt{a^2+1})
    (d)\text{ none of these}

    I took it to x\:=\:\log_a (2b-a^{-x})
    You haven't solve for x.

    We have: . b \;=\;\frac{a^x + a^{-x}}{2}

    Multiply by 2: . 2b \;=\;a^x + a^{-x}

    Multiply by a^x\!:\;2ba^x \;=\;a^{2x} + 1 \quad\Rightarrowquad a^{2x} - 2ba^x + 1 \;=\;0

    Quadratic Formula: . a^x \;=\;\frac{2b \pm\sqrt{4b^2-4}}{2} \quad\Rightarrow\quad a^x \;=\;b \pm\sqrt{b^2-1}

    Take logs, base a: . \log_a(a^x) \;=\;\log_a(b \pm \sqrt{b^2-1)

    Therefore: . x \;=\;\log_a(b+ \sqrt{b-1})\qquad \text{answer (b)}
    Thanks from NameIsHidden
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    May 2014
    From
    India
    Posts
    74
    Thanks
    3

    Re: Solve x if b=(a^x+a^-x)/2

    Quote Originally Posted by Soroban View Post
    Hello, NameIsHidden!


    We have: . b \;=\;\frac{a^x + a^{-x}}{2}

    Multiply by 2: . 2b \;=\;a^x + a^{-x}

    Multiply by a^x\!:\;2ba^x \;=\;a^{2x} + 1 \quad\Rightarrowquad a^{2x} - 2ba^x + 1 \;=\;0

    Quadratic Formula: . a^x \;=\;\frac{2b \pm\sqrt{4b^2-4}}{2} \quad\Rightarrow\quad a^x \;=\;b \pm\sqrt{b^2-1}

    Take logs, base a: . \log_a(a^x) \;=\;\log_a(b \pm \sqrt{b^2-1)

    Therefore: . x \;=\;\log_a(b+ \sqrt{b-1})\qquad \text{answer (b)}
    I wonder, do I need quadratic equation for this.

    BTW thanks for your help.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,790
    Thanks
    693

    Re: Solve x if b=(a^x+a^-x)/2

    Quote Originally Posted by NameIsHidden View Post
    I wonder, do I need quadratic equation for this.

    BTW thanks for your help.
    You never need the quadratic equation if you know other methods for solving quadratics. You can also try factoring or completing the square (completing the square is similar to using the quadratic equation, since the derivation of the quadratic equation involves completing the square of a general quadratic). However, in general, an equation x = f(x) is called an implicit definition for x. When you write x = \log_a(2b-a^{-x}), you do not give an explicit definition for x. In order to have an explicit definition (to find the actual value for a variable), you need that variable on one side of the equation and the other side to be a function that does not include the variable you solved for. So, with Plato's and Soroban's solution, they have x on one side of the equation and a function that does NOT involve x on the other (making their solution an explicit definition for x, allowing them to find an actual value for it).
    Thanks from Shakarri
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,368
    Thanks
    1313

    Re: Solve x if b=(a^x+a^-x)/2

    You don't need to use the quadratic formula or even write it as a quadratic equation (if that was your question).

    We can write a^x= e^{ln(a^x)}= e^{xln(a)} and a^{-x}= e^{ln(a^{-x})}= e^{-xln(a)}

    So \frac{a^x+ a^{-x}}{2}= \frac{e^{xln(a)}+ e^{-xln(a)}}{2}= cosh(xln(a))= b
    so that x= \frac{cosh^{-1}(b)}{ln(a)}
    Of course, that isn't one of the forms given but you can write cosh^{-1}(b) in terms of a logarithm, basically by solving a quadratic equation as you have been told!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. need to solve summation equation to solve sum(x2)
    Posted in the Statistics Forum
    Replies: 2
    Last Post: July 16th 2010, 10:29 PM
  2. Solve y'' +2y' + 5y = cos(2t).
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: July 15th 2010, 09:08 AM
  3. Replies: 1
    Last Post: June 9th 2009, 10:37 PM
  4. solve for m and n ?
    Posted in the Algebra Forum
    Replies: 1
    Last Post: October 20th 2008, 12:37 PM
  5. x^4-x^2+2=0 Solve. Need help :(
    Posted in the Algebra Forum
    Replies: 2
    Last Post: October 6th 2008, 06:04 PM

Search Tags


/mathhelpforum @mathhelpforum