$(x*(log_{2}(x))^2)/2 = x^{3/2}$ how to solve it?












1












$begingroup$


Is there a manual solution for this equation? Or I should use Wolfram:
result from Wolfram.






logarithms







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Well, as you can see, the Wolfram Alpha results at a solution involving some special functions (namely the $W$ function). So it seems like there is likely no algebraic solution.
    $endgroup$
    – Matti P.
    Jan 2 at 10:55








  • 1




    $begingroup$
    You could try by approaching this with a substitution like $x=2^y$, $e^y$ or something similar.
    $endgroup$
    – Matti P.
    Jan 2 at 10:58












  • $begingroup$
    Are you looking for only an approximate solution?
    $endgroup$
    – Matti P.
    Jan 2 at 10:59










  • $begingroup$
    @MattiP. thanks, your idea with $x = 2^y$ was great!
    $endgroup$
    – Hmmman
    Jan 2 at 11:09










  • $begingroup$
    see herehttps://planetmath.org/approximationofthelogfunction
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 2 at 11:37
















1












$begingroup$


Is there a manual solution for this equation? Or I should use Wolfram:
result from Wolfram.






logarithms







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Well, as you can see, the Wolfram Alpha results at a solution involving some special functions (namely the $W$ function). So it seems like there is likely no algebraic solution.
    $endgroup$
    – Matti P.
    Jan 2 at 10:55








  • 1




    $begingroup$
    You could try by approaching this with a substitution like $x=2^y$, $e^y$ or something similar.
    $endgroup$
    – Matti P.
    Jan 2 at 10:58












  • $begingroup$
    Are you looking for only an approximate solution?
    $endgroup$
    – Matti P.
    Jan 2 at 10:59










  • $begingroup$
    @MattiP. thanks, your idea with $x = 2^y$ was great!
    $endgroup$
    – Hmmman
    Jan 2 at 11:09










  • $begingroup$
    see herehttps://planetmath.org/approximationofthelogfunction
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 2 at 11:37














1












1








1





$begingroup$


Is there a manual solution for this equation? Or I should use Wolfram:
result from Wolfram.






logarithms







share|cite|improve this question











$endgroup$




Is there a manual solution for this equation? Or I should use Wolfram:
result from Wolfram.






logarithms




logarithms






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 2 at 10:53









Bernard

122k740116




122k740116










asked Jan 2 at 10:51









HmmmanHmmman

175




175








  • 1




    $begingroup$
    Well, as you can see, the Wolfram Alpha results at a solution involving some special functions (namely the $W$ function). So it seems like there is likely no algebraic solution.
    $endgroup$
    – Matti P.
    Jan 2 at 10:55








  • 1




    $begingroup$
    You could try by approaching this with a substitution like $x=2^y$, $e^y$ or something similar.
    $endgroup$
    – Matti P.
    Jan 2 at 10:58












  • $begingroup$
    Are you looking for only an approximate solution?
    $endgroup$
    – Matti P.
    Jan 2 at 10:59










  • $begingroup$
    @MattiP. thanks, your idea with $x = 2^y$ was great!
    $endgroup$
    – Hmmman
    Jan 2 at 11:09










  • $begingroup$
    see herehttps://planetmath.org/approximationofthelogfunction
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 2 at 11:37














  • 1




    $begingroup$
    Well, as you can see, the Wolfram Alpha results at a solution involving some special functions (namely the $W$ function). So it seems like there is likely no algebraic solution.
    $endgroup$
    – Matti P.
    Jan 2 at 10:55








  • 1




    $begingroup$
    You could try by approaching this with a substitution like $x=2^y$, $e^y$ or something similar.
    $endgroup$
    – Matti P.
    Jan 2 at 10:58












  • $begingroup$
    Are you looking for only an approximate solution?
    $endgroup$
    – Matti P.
    Jan 2 at 10:59










  • $begingroup$
    @MattiP. thanks, your idea with $x = 2^y$ was great!
    $endgroup$
    – Hmmman
    Jan 2 at 11:09










  • $begingroup$
    see herehttps://planetmath.org/approximationofthelogfunction
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 2 at 11:37








1




1




$begingroup$
Well, as you can see, the Wolfram Alpha results at a solution involving some special functions (namely the $W$ function). So it seems like there is likely no algebraic solution.
$endgroup$
– Matti P.
Jan 2 at 10:55






$begingroup$
Well, as you can see, the Wolfram Alpha results at a solution involving some special functions (namely the $W$ function). So it seems like there is likely no algebraic solution.
$endgroup$
– Matti P.
Jan 2 at 10:55






1




1




$begingroup$
You could try by approaching this with a substitution like $x=2^y$, $e^y$ or something similar.
$endgroup$
– Matti P.
Jan 2 at 10:58






$begingroup$
You could try by approaching this with a substitution like $x=2^y$, $e^y$ or something similar.
$endgroup$
– Matti P.
Jan 2 at 10:58














$begingroup$
Are you looking for only an approximate solution?
$endgroup$
– Matti P.
Jan 2 at 10:59




$begingroup$
Are you looking for only an approximate solution?
$endgroup$
– Matti P.
Jan 2 at 10:59












$begingroup$
@MattiP. thanks, your idea with $x = 2^y$ was great!
$endgroup$
– Hmmman
Jan 2 at 11:09




$begingroup$
@MattiP. thanks, your idea with $x = 2^y$ was great!
$endgroup$
– Hmmman
Jan 2 at 11:09












$begingroup$
see herehttps://planetmath.org/approximationofthelogfunction
$endgroup$
– Dr. Sonnhard Graubner
Jan 2 at 11:37




$begingroup$
see herehttps://planetmath.org/approximationofthelogfunction
$endgroup$
– Dr. Sonnhard Graubner
Jan 2 at 11:37










3 Answers
3






active

oldest

votes


















2












$begingroup$

There is no algebraic solution, but you can use numeric methods like Newton-Raphson method or the Lambert $W$ function:




The Lambert $W$ function is defined by $z=W(ze^z)$ with the domain as $mathbb{C}$ and there are two possible functions $W_0$ with $textrm{Re}(W_0(x))geq -1$ and $W_{-1}$ with $textrm{Re}(W_{-1}) < -1$.




A graph of Lambert W function



Then:
begin{eqnarray}
frac{1}{2}x left(log_2(x)right)^2 &=& x^{3/2} \
frac{1}{2}x^{-1/2} left(log_2(x)right)^2 &=& 1 \
frac{1}{sqrt{2}}x^{-1/4}log_2(x) &=& pm 1 \
x^{-1/4}frac{ln(x)}{ln(2)} &=& pm sqrt{2} \
x^{-1/4}ln(x) &=& pm sqrt{2}ln(2) \
e^{(-1/4)ln(x)}ln(x) &=& pm sqrt{2}ln(2) \
e^{(-1/4)ln(x)}left(-frac{1}{4}ln(x)right) &=& pm frac{1}{4}sqrt{2}ln(2) \
end{eqnarray}
Then with Lambert function:
begin{eqnarray}
-frac{1}{4}ln(x) &=& Wleft(pmfrac{1}{4}sqrt{2}ln(2)right) \
x &=& exp leftlbrace-4Wleft(pmfrac{1}{4}sqrt{2}ln(2)right)rightrbrace \
end{eqnarray}
Then the solutions are:
begin{eqnarray}
x_1 &=& exp leftlbrace-4W_0left(frac{1}{4}sqrt{2}ln(2)right)rightrbrace = 0.44836908960ldots \
x_2 &=& exp leftlbrace-4W_0left(-frac{1}{4}sqrt{2}ln(2)right)rightrbrace=4 \
x_3 &=& exp leftlbrace-4W_{-1}left(frac{1}{4}sqrt{2}ln(2)right)rightrbrace=-153792.65205358ldots-i101297.96245405ldots \
x_4 &=& exp leftlbrace-4W_{-1}left(-frac{1}{4}sqrt{2}ln(2) right)rightrbrace=6380.45994697086ldots \
end{eqnarray}



The function is numeric too, but it is very elegant.



There is an article in wikipedia for this Lambert W function.






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    Thanks for your exhaustive explanation!!!
    $endgroup$
    – Hmmman
    Jan 2 at 12:48






  • 2




    $begingroup$
    This is a nicely detailed and illustrated solution ! May I confess that I missed the complex root ? $to +1$ for sure.
    $endgroup$
    – Claude Leibovici
    Jan 3 at 5:29



















3












$begingroup$

This equation has three real solutions:
$$xapprox 0.4483690898$$
$$x=4$$
$$xapprox 6380.459941$$
This can be obtained by the Newton Raphson method.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    How did you find them? I asked how to solve, I know answers by myself!
    $endgroup$
    – Hmmman
    Jan 2 at 11:22






  • 1




    $begingroup$
    see here sosmath.com/calculus/diff/der07/der07.html
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 2 at 11:25






  • 1




    $begingroup$
    Thanks, but I was just trying to solve it without computer. Maybe using some approximation formulas for $(log_{2}(x))^2$, with rough approximation. Seems like it s impossible!
    $endgroup$
    – Hmmman
    Jan 2 at 11:31



















2












$begingroup$

Welcome to the world of Lambert function !



Using Matti P.'s suggestion $x=2^y$, the equation becomes
$$2^{y-1} y^2=2^{frac{3y}{2}}$$ that is to say
$$y^2=2^{frac{y}{2}+1}=2 times2^{frac{y}{2}}$$ Now, let $2^{frac{y}{2}}=t$, that is to say $y=frac{2 t}{log (2)}$ to make the equation
$$e^{2 t} left(frac{2 t^2}{log ^2(2)}-e^tright)=0implies e^t=frac{2 t^2}{log ^2(2)}$$ and then the three roots
$$t_1=-2 W_0left(frac{log (2)}{2 sqrt{2}}right)qquad t_2=-2 W_0left(-frac{log (2)}{2 sqrt{2}}right)qquad t_3=-2 W_{-1}left(-frac{log (2)}{2 sqrt{2}}right)$$



It have been faster to use $x=2^{e^t}$ for the same result.






share|cite|improve this answer











$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3059336%2fx-log-2x2-2-x3-2-how-to-solve-it%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    There is no algebraic solution, but you can use numeric methods like Newton-Raphson method or the Lambert $W$ function:




    The Lambert $W$ function is defined by $z=W(ze^z)$ with the domain as $mathbb{C}$ and there are two possible functions $W_0$ with $textrm{Re}(W_0(x))geq -1$ and $W_{-1}$ with $textrm{Re}(W_{-1}) < -1$.




    A graph of Lambert W function



    Then:
    begin{eqnarray}
    frac{1}{2}x left(log_2(x)right)^2 &=& x^{3/2} \
    frac{1}{2}x^{-1/2} left(log_2(x)right)^2 &=& 1 \
    frac{1}{sqrt{2}}x^{-1/4}log_2(x) &=& pm 1 \
    x^{-1/4}frac{ln(x)}{ln(2)} &=& pm sqrt{2} \
    x^{-1/4}ln(x) &=& pm sqrt{2}ln(2) \
    e^{(-1/4)ln(x)}ln(x) &=& pm sqrt{2}ln(2) \
    e^{(-1/4)ln(x)}left(-frac{1}{4}ln(x)right) &=& pm frac{1}{4}sqrt{2}ln(2) \
    end{eqnarray}
    Then with Lambert function:
    begin{eqnarray}
    -frac{1}{4}ln(x) &=& Wleft(pmfrac{1}{4}sqrt{2}ln(2)right) \
    x &=& exp leftlbrace-4Wleft(pmfrac{1}{4}sqrt{2}ln(2)right)rightrbrace \
    end{eqnarray}
    Then the solutions are:
    begin{eqnarray}
    x_1 &=& exp leftlbrace-4W_0left(frac{1}{4}sqrt{2}ln(2)right)rightrbrace = 0.44836908960ldots \
    x_2 &=& exp leftlbrace-4W_0left(-frac{1}{4}sqrt{2}ln(2)right)rightrbrace=4 \
    x_3 &=& exp leftlbrace-4W_{-1}left(frac{1}{4}sqrt{2}ln(2)right)rightrbrace=-153792.65205358ldots-i101297.96245405ldots \
    x_4 &=& exp leftlbrace-4W_{-1}left(-frac{1}{4}sqrt{2}ln(2) right)rightrbrace=6380.45994697086ldots \
    end{eqnarray}



    The function is numeric too, but it is very elegant.



    There is an article in wikipedia for this Lambert W function.






    share|cite|improve this answer











    $endgroup$









    • 2




      $begingroup$
      Thanks for your exhaustive explanation!!!
      $endgroup$
      – Hmmman
      Jan 2 at 12:48






    • 2




      $begingroup$
      This is a nicely detailed and illustrated solution ! May I confess that I missed the complex root ? $to +1$ for sure.
      $endgroup$
      – Claude Leibovici
      Jan 3 at 5:29
















    2












    $begingroup$

    There is no algebraic solution, but you can use numeric methods like Newton-Raphson method or the Lambert $W$ function:




    The Lambert $W$ function is defined by $z=W(ze^z)$ with the domain as $mathbb{C}$ and there are two possible functions $W_0$ with $textrm{Re}(W_0(x))geq -1$ and $W_{-1}$ with $textrm{Re}(W_{-1}) < -1$.




    A graph of Lambert W function



    Then:
    begin{eqnarray}
    frac{1}{2}x left(log_2(x)right)^2 &=& x^{3/2} \
    frac{1}{2}x^{-1/2} left(log_2(x)right)^2 &=& 1 \
    frac{1}{sqrt{2}}x^{-1/4}log_2(x) &=& pm 1 \
    x^{-1/4}frac{ln(x)}{ln(2)} &=& pm sqrt{2} \
    x^{-1/4}ln(x) &=& pm sqrt{2}ln(2) \
    e^{(-1/4)ln(x)}ln(x) &=& pm sqrt{2}ln(2) \
    e^{(-1/4)ln(x)}left(-frac{1}{4}ln(x)right) &=& pm frac{1}{4}sqrt{2}ln(2) \
    end{eqnarray}
    Then with Lambert function:
    begin{eqnarray}
    -frac{1}{4}ln(x) &=& Wleft(pmfrac{1}{4}sqrt{2}ln(2)right) \
    x &=& exp leftlbrace-4Wleft(pmfrac{1}{4}sqrt{2}ln(2)right)rightrbrace \
    end{eqnarray}
    Then the solutions are:
    begin{eqnarray}
    x_1 &=& exp leftlbrace-4W_0left(frac{1}{4}sqrt{2}ln(2)right)rightrbrace = 0.44836908960ldots \
    x_2 &=& exp leftlbrace-4W_0left(-frac{1}{4}sqrt{2}ln(2)right)rightrbrace=4 \
    x_3 &=& exp leftlbrace-4W_{-1}left(frac{1}{4}sqrt{2}ln(2)right)rightrbrace=-153792.65205358ldots-i101297.96245405ldots \
    x_4 &=& exp leftlbrace-4W_{-1}left(-frac{1}{4}sqrt{2}ln(2) right)rightrbrace=6380.45994697086ldots \
    end{eqnarray}



    The function is numeric too, but it is very elegant.



    There is an article in wikipedia for this Lambert W function.






    share|cite|improve this answer











    $endgroup$









    • 2




      $begingroup$
      Thanks for your exhaustive explanation!!!
      $endgroup$
      – Hmmman
      Jan 2 at 12:48






    • 2




      $begingroup$
      This is a nicely detailed and illustrated solution ! May I confess that I missed the complex root ? $to +1$ for sure.
      $endgroup$
      – Claude Leibovici
      Jan 3 at 5:29














    2












    2








    2





    $begingroup$

    There is no algebraic solution, but you can use numeric methods like Newton-Raphson method or the Lambert $W$ function:




    The Lambert $W$ function is defined by $z=W(ze^z)$ with the domain as $mathbb{C}$ and there are two possible functions $W_0$ with $textrm{Re}(W_0(x))geq -1$ and $W_{-1}$ with $textrm{Re}(W_{-1}) < -1$.




    A graph of Lambert W function



    Then:
    begin{eqnarray}
    frac{1}{2}x left(log_2(x)right)^2 &=& x^{3/2} \
    frac{1}{2}x^{-1/2} left(log_2(x)right)^2 &=& 1 \
    frac{1}{sqrt{2}}x^{-1/4}log_2(x) &=& pm 1 \
    x^{-1/4}frac{ln(x)}{ln(2)} &=& pm sqrt{2} \
    x^{-1/4}ln(x) &=& pm sqrt{2}ln(2) \
    e^{(-1/4)ln(x)}ln(x) &=& pm sqrt{2}ln(2) \
    e^{(-1/4)ln(x)}left(-frac{1}{4}ln(x)right) &=& pm frac{1}{4}sqrt{2}ln(2) \
    end{eqnarray}
    Then with Lambert function:
    begin{eqnarray}
    -frac{1}{4}ln(x) &=& Wleft(pmfrac{1}{4}sqrt{2}ln(2)right) \
    x &=& exp leftlbrace-4Wleft(pmfrac{1}{4}sqrt{2}ln(2)right)rightrbrace \
    end{eqnarray}
    Then the solutions are:
    begin{eqnarray}
    x_1 &=& exp leftlbrace-4W_0left(frac{1}{4}sqrt{2}ln(2)right)rightrbrace = 0.44836908960ldots \
    x_2 &=& exp leftlbrace-4W_0left(-frac{1}{4}sqrt{2}ln(2)right)rightrbrace=4 \
    x_3 &=& exp leftlbrace-4W_{-1}left(frac{1}{4}sqrt{2}ln(2)right)rightrbrace=-153792.65205358ldots-i101297.96245405ldots \
    x_4 &=& exp leftlbrace-4W_{-1}left(-frac{1}{4}sqrt{2}ln(2) right)rightrbrace=6380.45994697086ldots \
    end{eqnarray}



    The function is numeric too, but it is very elegant.



    There is an article in wikipedia for this Lambert W function.






    share|cite|improve this answer











    $endgroup$



    There is no algebraic solution, but you can use numeric methods like Newton-Raphson method or the Lambert $W$ function:




    The Lambert $W$ function is defined by $z=W(ze^z)$ with the domain as $mathbb{C}$ and there are two possible functions $W_0$ with $textrm{Re}(W_0(x))geq -1$ and $W_{-1}$ with $textrm{Re}(W_{-1}) < -1$.




    A graph of Lambert W function



    Then:
    begin{eqnarray}
    frac{1}{2}x left(log_2(x)right)^2 &=& x^{3/2} \
    frac{1}{2}x^{-1/2} left(log_2(x)right)^2 &=& 1 \
    frac{1}{sqrt{2}}x^{-1/4}log_2(x) &=& pm 1 \
    x^{-1/4}frac{ln(x)}{ln(2)} &=& pm sqrt{2} \
    x^{-1/4}ln(x) &=& pm sqrt{2}ln(2) \
    e^{(-1/4)ln(x)}ln(x) &=& pm sqrt{2}ln(2) \
    e^{(-1/4)ln(x)}left(-frac{1}{4}ln(x)right) &=& pm frac{1}{4}sqrt{2}ln(2) \
    end{eqnarray}
    Then with Lambert function:
    begin{eqnarray}
    -frac{1}{4}ln(x) &=& Wleft(pmfrac{1}{4}sqrt{2}ln(2)right) \
    x &=& exp leftlbrace-4Wleft(pmfrac{1}{4}sqrt{2}ln(2)right)rightrbrace \
    end{eqnarray}
    Then the solutions are:
    begin{eqnarray}
    x_1 &=& exp leftlbrace-4W_0left(frac{1}{4}sqrt{2}ln(2)right)rightrbrace = 0.44836908960ldots \
    x_2 &=& exp leftlbrace-4W_0left(-frac{1}{4}sqrt{2}ln(2)right)rightrbrace=4 \
    x_3 &=& exp leftlbrace-4W_{-1}left(frac{1}{4}sqrt{2}ln(2)right)rightrbrace=-153792.65205358ldots-i101297.96245405ldots \
    x_4 &=& exp leftlbrace-4W_{-1}left(-frac{1}{4}sqrt{2}ln(2) right)rightrbrace=6380.45994697086ldots \
    end{eqnarray}



    The function is numeric too, but it is very elegant.



    There is an article in wikipedia for this Lambert W function.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 23 at 1:33

























    answered Jan 2 at 12:42









    El boritoEl borito

    666216




    666216








    • 2




      $begingroup$
      Thanks for your exhaustive explanation!!!
      $endgroup$
      – Hmmman
      Jan 2 at 12:48






    • 2




      $begingroup$
      This is a nicely detailed and illustrated solution ! May I confess that I missed the complex root ? $to +1$ for sure.
      $endgroup$
      – Claude Leibovici
      Jan 3 at 5:29














    • 2




      $begingroup$
      Thanks for your exhaustive explanation!!!
      $endgroup$
      – Hmmman
      Jan 2 at 12:48






    • 2




      $begingroup$
      This is a nicely detailed and illustrated solution ! May I confess that I missed the complex root ? $to +1$ for sure.
      $endgroup$
      – Claude Leibovici
      Jan 3 at 5:29








    2




    2




    $begingroup$
    Thanks for your exhaustive explanation!!!
    $endgroup$
    – Hmmman
    Jan 2 at 12:48




    $begingroup$
    Thanks for your exhaustive explanation!!!
    $endgroup$
    – Hmmman
    Jan 2 at 12:48




    2




    2




    $begingroup$
    This is a nicely detailed and illustrated solution ! May I confess that I missed the complex root ? $to +1$ for sure.
    $endgroup$
    – Claude Leibovici
    Jan 3 at 5:29




    $begingroup$
    This is a nicely detailed and illustrated solution ! May I confess that I missed the complex root ? $to +1$ for sure.
    $endgroup$
    – Claude Leibovici
    Jan 3 at 5:29











    3












    $begingroup$

    This equation has three real solutions:
    $$xapprox 0.4483690898$$
    $$x=4$$
    $$xapprox 6380.459941$$
    This can be obtained by the Newton Raphson method.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      How did you find them? I asked how to solve, I know answers by myself!
      $endgroup$
      – Hmmman
      Jan 2 at 11:22






    • 1




      $begingroup$
      see here sosmath.com/calculus/diff/der07/der07.html
      $endgroup$
      – Dr. Sonnhard Graubner
      Jan 2 at 11:25






    • 1




      $begingroup$
      Thanks, but I was just trying to solve it without computer. Maybe using some approximation formulas for $(log_{2}(x))^2$, with rough approximation. Seems like it s impossible!
      $endgroup$
      – Hmmman
      Jan 2 at 11:31
















    3












    $begingroup$

    This equation has three real solutions:
    $$xapprox 0.4483690898$$
    $$x=4$$
    $$xapprox 6380.459941$$
    This can be obtained by the Newton Raphson method.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      How did you find them? I asked how to solve, I know answers by myself!
      $endgroup$
      – Hmmman
      Jan 2 at 11:22






    • 1




      $begingroup$
      see here sosmath.com/calculus/diff/der07/der07.html
      $endgroup$
      – Dr. Sonnhard Graubner
      Jan 2 at 11:25






    • 1




      $begingroup$
      Thanks, but I was just trying to solve it without computer. Maybe using some approximation formulas for $(log_{2}(x))^2$, with rough approximation. Seems like it s impossible!
      $endgroup$
      – Hmmman
      Jan 2 at 11:31














    3












    3








    3





    $begingroup$

    This equation has three real solutions:
    $$xapprox 0.4483690898$$
    $$x=4$$
    $$xapprox 6380.459941$$
    This can be obtained by the Newton Raphson method.






    share|cite|improve this answer











    $endgroup$



    This equation has three real solutions:
    $$xapprox 0.4483690898$$
    $$x=4$$
    $$xapprox 6380.459941$$
    This can be obtained by the Newton Raphson method.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 2 at 11:23

























    answered Jan 2 at 11:18









    Dr. Sonnhard GraubnerDr. Sonnhard Graubner

    76.7k42866




    76.7k42866








    • 1




      $begingroup$
      How did you find them? I asked how to solve, I know answers by myself!
      $endgroup$
      – Hmmman
      Jan 2 at 11:22






    • 1




      $begingroup$
      see here sosmath.com/calculus/diff/der07/der07.html
      $endgroup$
      – Dr. Sonnhard Graubner
      Jan 2 at 11:25






    • 1




      $begingroup$
      Thanks, but I was just trying to solve it without computer. Maybe using some approximation formulas for $(log_{2}(x))^2$, with rough approximation. Seems like it s impossible!
      $endgroup$
      – Hmmman
      Jan 2 at 11:31














    • 1




      $begingroup$
      How did you find them? I asked how to solve, I know answers by myself!
      $endgroup$
      – Hmmman
      Jan 2 at 11:22






    • 1




      $begingroup$
      see here sosmath.com/calculus/diff/der07/der07.html
      $endgroup$
      – Dr. Sonnhard Graubner
      Jan 2 at 11:25






    • 1




      $begingroup$
      Thanks, but I was just trying to solve it without computer. Maybe using some approximation formulas for $(log_{2}(x))^2$, with rough approximation. Seems like it s impossible!
      $endgroup$
      – Hmmman
      Jan 2 at 11:31








    1




    1




    $begingroup$
    How did you find them? I asked how to solve, I know answers by myself!
    $endgroup$
    – Hmmman
    Jan 2 at 11:22




    $begingroup$
    How did you find them? I asked how to solve, I know answers by myself!
    $endgroup$
    – Hmmman
    Jan 2 at 11:22




    1




    1




    $begingroup$
    see here sosmath.com/calculus/diff/der07/der07.html
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 2 at 11:25




    $begingroup$
    see here sosmath.com/calculus/diff/der07/der07.html
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 2 at 11:25




    1




    1




    $begingroup$
    Thanks, but I was just trying to solve it without computer. Maybe using some approximation formulas for $(log_{2}(x))^2$, with rough approximation. Seems like it s impossible!
    $endgroup$
    – Hmmman
    Jan 2 at 11:31




    $begingroup$
    Thanks, but I was just trying to solve it without computer. Maybe using some approximation formulas for $(log_{2}(x))^2$, with rough approximation. Seems like it s impossible!
    $endgroup$
    – Hmmman
    Jan 2 at 11:31











    2












    $begingroup$

    Welcome to the world of Lambert function !



    Using Matti P.'s suggestion $x=2^y$, the equation becomes
    $$2^{y-1} y^2=2^{frac{3y}{2}}$$ that is to say
    $$y^2=2^{frac{y}{2}+1}=2 times2^{frac{y}{2}}$$ Now, let $2^{frac{y}{2}}=t$, that is to say $y=frac{2 t}{log (2)}$ to make the equation
    $$e^{2 t} left(frac{2 t^2}{log ^2(2)}-e^tright)=0implies e^t=frac{2 t^2}{log ^2(2)}$$ and then the three roots
    $$t_1=-2 W_0left(frac{log (2)}{2 sqrt{2}}right)qquad t_2=-2 W_0left(-frac{log (2)}{2 sqrt{2}}right)qquad t_3=-2 W_{-1}left(-frac{log (2)}{2 sqrt{2}}right)$$



    It have been faster to use $x=2^{e^t}$ for the same result.






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      Welcome to the world of Lambert function !



      Using Matti P.'s suggestion $x=2^y$, the equation becomes
      $$2^{y-1} y^2=2^{frac{3y}{2}}$$ that is to say
      $$y^2=2^{frac{y}{2}+1}=2 times2^{frac{y}{2}}$$ Now, let $2^{frac{y}{2}}=t$, that is to say $y=frac{2 t}{log (2)}$ to make the equation
      $$e^{2 t} left(frac{2 t^2}{log ^2(2)}-e^tright)=0implies e^t=frac{2 t^2}{log ^2(2)}$$ and then the three roots
      $$t_1=-2 W_0left(frac{log (2)}{2 sqrt{2}}right)qquad t_2=-2 W_0left(-frac{log (2)}{2 sqrt{2}}right)qquad t_3=-2 W_{-1}left(-frac{log (2)}{2 sqrt{2}}right)$$



      It have been faster to use $x=2^{e^t}$ for the same result.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        Welcome to the world of Lambert function !



        Using Matti P.'s suggestion $x=2^y$, the equation becomes
        $$2^{y-1} y^2=2^{frac{3y}{2}}$$ that is to say
        $$y^2=2^{frac{y}{2}+1}=2 times2^{frac{y}{2}}$$ Now, let $2^{frac{y}{2}}=t$, that is to say $y=frac{2 t}{log (2)}$ to make the equation
        $$e^{2 t} left(frac{2 t^2}{log ^2(2)}-e^tright)=0implies e^t=frac{2 t^2}{log ^2(2)}$$ and then the three roots
        $$t_1=-2 W_0left(frac{log (2)}{2 sqrt{2}}right)qquad t_2=-2 W_0left(-frac{log (2)}{2 sqrt{2}}right)qquad t_3=-2 W_{-1}left(-frac{log (2)}{2 sqrt{2}}right)$$



        It have been faster to use $x=2^{e^t}$ for the same result.






        share|cite|improve this answer











        $endgroup$



        Welcome to the world of Lambert function !



        Using Matti P.'s suggestion $x=2^y$, the equation becomes
        $$2^{y-1} y^2=2^{frac{3y}{2}}$$ that is to say
        $$y^2=2^{frac{y}{2}+1}=2 times2^{frac{y}{2}}$$ Now, let $2^{frac{y}{2}}=t$, that is to say $y=frac{2 t}{log (2)}$ to make the equation
        $$e^{2 t} left(frac{2 t^2}{log ^2(2)}-e^tright)=0implies e^t=frac{2 t^2}{log ^2(2)}$$ and then the three roots
        $$t_1=-2 W_0left(frac{log (2)}{2 sqrt{2}}right)qquad t_2=-2 W_0left(-frac{log (2)}{2 sqrt{2}}right)qquad t_3=-2 W_{-1}left(-frac{log (2)}{2 sqrt{2}}right)$$



        It have been faster to use $x=2^{e^t}$ for the same result.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 3 at 5:30

























        answered Jan 2 at 11:49









        Claude LeiboviciClaude Leibovici

        123k1157134




        123k1157134






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3059336%2fx-log-2x2-2-x3-2-how-to-solve-it%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Quarter-circle Tiles

            Image
            up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2
            Read more

            build a pushdown automaton that recognizes the reverse language of a given pushdown automaton?

            Image
            0 I think it's similiar to NFAs. I replace the finite states of the given automaton for start-states for my new automaton. I do it with $epsilon$ -transition from the start state to the actual finite states of the given automaton. The start state of the given automaton will be my accepting(finite) state of my new one. All transitions will be reversed of course to recognize the reversed word. But how do I have to change the stack of the pushdown automaton? That's tricky. formal-languages automata context-free-grammar share | cite | improve this question edited Nov 29 '18 at 14:20 joee
            Read more

            Mont Emei

            Image
            Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog
            Read more