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












1












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






logarithms







share|cite|improve this question











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  • 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











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    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






























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