What are the steps to Solve $N cdot log_2N = 10^6$ algebraically?












1












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I can write code that can solve it by just going through every integer but what are the steps to solve it with basic algebra. I seem to end up in a circular fashion of just writing it in terms of other things.






algebra-precalculus logarithms







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  • $begingroup$
    If you mean "Solve $Nlog_2(N) = 10^6$," then please format this as $Nlog_2(N) = 10^6$
    $endgroup$
    – Namaste
    Dec 19 '18 at 1:48












  • $begingroup$
    @amWhy thanks the first link for formatting did not have it listed that way
    $endgroup$
    – Dustin.Shropshire
    Dec 19 '18 at 1:50










  • $begingroup$
    I suspect there is no algebraic solution.
    $endgroup$
    – Ethan Bolker
    Dec 19 '18 at 1:54










  • $begingroup$
    N is gonna be one huge number! Given $$Nlog_2 N = 10^6$$ $$N^N = 2^{10^6}$$ $$N^N = 2^{1,000,000}$$ where 1 million = 1,000,000
    $endgroup$
    – Christopher Marley
    Dec 19 '18 at 2:52








  • 1




    $begingroup$
    @ChristopherMarley: Not so huge. Note that $N lt 10^6$.
    $endgroup$
    – hardmath
    Dec 20 '18 at 6:37
















1












$begingroup$


I can write code that can solve it by just going through every integer but what are the steps to solve it with basic algebra. I seem to end up in a circular fashion of just writing it in terms of other things.






algebra-precalculus logarithms







share|cite|improve this question











$endgroup$












  • $begingroup$
    If you mean "Solve $Nlog_2(N) = 10^6$," then please format this as $Nlog_2(N) = 10^6$
    $endgroup$
    – Namaste
    Dec 19 '18 at 1:48












  • $begingroup$
    @amWhy thanks the first link for formatting did not have it listed that way
    $endgroup$
    – Dustin.Shropshire
    Dec 19 '18 at 1:50










  • $begingroup$
    I suspect there is no algebraic solution.
    $endgroup$
    – Ethan Bolker
    Dec 19 '18 at 1:54










  • $begingroup$
    N is gonna be one huge number! Given $$Nlog_2 N = 10^6$$ $$N^N = 2^{10^6}$$ $$N^N = 2^{1,000,000}$$ where 1 million = 1,000,000
    $endgroup$
    – Christopher Marley
    Dec 19 '18 at 2:52








  • 1




    $begingroup$
    @ChristopherMarley: Not so huge. Note that $N lt 10^6$.
    $endgroup$
    – hardmath
    Dec 20 '18 at 6:37














1












1








1





$begingroup$


I can write code that can solve it by just going through every integer but what are the steps to solve it with basic algebra. I seem to end up in a circular fashion of just writing it in terms of other things.






algebra-precalculus logarithms







share|cite|improve this question











$endgroup$




I can write code that can solve it by just going through every integer but what are the steps to solve it with basic algebra. I seem to end up in a circular fashion of just writing it in terms of other things.






algebra-precalculus logarithms




algebra-precalculus logarithms






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 19 '18 at 1:52









Namaste

1




1










asked Dec 19 '18 at 1:44









Dustin.ShropshireDustin.Shropshire

62




62












  • $begingroup$
    If you mean "Solve $Nlog_2(N) = 10^6$," then please format this as $Nlog_2(N) = 10^6$
    $endgroup$
    – Namaste
    Dec 19 '18 at 1:48












  • $begingroup$
    @amWhy thanks the first link for formatting did not have it listed that way
    $endgroup$
    – Dustin.Shropshire
    Dec 19 '18 at 1:50










  • $begingroup$
    I suspect there is no algebraic solution.
    $endgroup$
    – Ethan Bolker
    Dec 19 '18 at 1:54










  • $begingroup$
    N is gonna be one huge number! Given $$Nlog_2 N = 10^6$$ $$N^N = 2^{10^6}$$ $$N^N = 2^{1,000,000}$$ where 1 million = 1,000,000
    $endgroup$
    – Christopher Marley
    Dec 19 '18 at 2:52








  • 1




    $begingroup$
    @ChristopherMarley: Not so huge. Note that $N lt 10^6$.
    $endgroup$
    – hardmath
    Dec 20 '18 at 6:37


















  • $begingroup$
    If you mean "Solve $Nlog_2(N) = 10^6$," then please format this as $Nlog_2(N) = 10^6$
    $endgroup$
    – Namaste
    Dec 19 '18 at 1:48












  • $begingroup$
    @amWhy thanks the first link for formatting did not have it listed that way
    $endgroup$
    – Dustin.Shropshire
    Dec 19 '18 at 1:50










  • $begingroup$
    I suspect there is no algebraic solution.
    $endgroup$
    – Ethan Bolker
    Dec 19 '18 at 1:54










  • $begingroup$
    N is gonna be one huge number! Given $$Nlog_2 N = 10^6$$ $$N^N = 2^{10^6}$$ $$N^N = 2^{1,000,000}$$ where 1 million = 1,000,000
    $endgroup$
    – Christopher Marley
    Dec 19 '18 at 2:52








  • 1




    $begingroup$
    @ChristopherMarley: Not so huge. Note that $N lt 10^6$.
    $endgroup$
    – hardmath
    Dec 20 '18 at 6:37
















$begingroup$
If you mean "Solve $Nlog_2(N) = 10^6$," then please format this as $Nlog_2(N) = 10^6$
$endgroup$
– Namaste
Dec 19 '18 at 1:48






$begingroup$
If you mean "Solve $Nlog_2(N) = 10^6$," then please format this as $Nlog_2(N) = 10^6$
$endgroup$
– Namaste
Dec 19 '18 at 1:48














$begingroup$
@amWhy thanks the first link for formatting did not have it listed that way
$endgroup$
– Dustin.Shropshire
Dec 19 '18 at 1:50




$begingroup$
@amWhy thanks the first link for formatting did not have it listed that way
$endgroup$
– Dustin.Shropshire
Dec 19 '18 at 1:50












$begingroup$
I suspect there is no algebraic solution.
$endgroup$
– Ethan Bolker
Dec 19 '18 at 1:54




$begingroup$
I suspect there is no algebraic solution.
$endgroup$
– Ethan Bolker
Dec 19 '18 at 1:54












$begingroup$
N is gonna be one huge number! Given $$Nlog_2 N = 10^6$$ $$N^N = 2^{10^6}$$ $$N^N = 2^{1,000,000}$$ where 1 million = 1,000,000
$endgroup$
– Christopher Marley
Dec 19 '18 at 2:52






$begingroup$
N is gonna be one huge number! Given $$Nlog_2 N = 10^6$$ $$N^N = 2^{10^6}$$ $$N^N = 2^{1,000,000}$$ where 1 million = 1,000,000
$endgroup$
– Christopher Marley
Dec 19 '18 at 2:52






1




1




$begingroup$
@ChristopherMarley: Not so huge. Note that $N lt 10^6$.
$endgroup$
– hardmath
Dec 20 '18 at 6:37




$begingroup$
@ChristopherMarley: Not so huge. Note that $N lt 10^6$.
$endgroup$
– hardmath
Dec 20 '18 at 6:37










1 Answer
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$begingroup$

$f : xmapsto x log_2x$ clearly is a strictly increasing countinuous function over $[1,infty)$ since both $xmapsto x$ and $xmapsto log_2x$ are and are positive on $[1,infty)$. Given that $f(1) = 0$ and that $underset{+infty}{lim}f = +infty$, the intermediate value theorem enables us to know there is a unique solution to:
$$
(E_a) : f(x) = a, ain [1,infty)
$$
Let $g : xmapsto x ln x$ to make calculaions easier. Similarly to what was previously said, $(E'_a) : g(x) = a, ain [1,infty)$ has a unique solution $xin[1,infty)$. Note that the solution to $(E'_{aln2})$ is the solution to $(E_a)$.



Then, as SmileyCraft thought, this is a plain use case of Lambert W function which is defined as the reciprocal of $w mapsto we^w$. We have $xln x = a$ so $e^{ln x}ln x = a$ and $ln x = W(a)$ and $x = e^{W(a)}$ which yields $x = frac{a}{W(a)}$ (because, by definition $W(a)e^{W(a)} = a$).



We thus solved $(E'_a)$ for all $ain [1,infty)$:
$$
x ln x = a Leftrightarrow x = frac{a}{W(a)}
$$
Then, given what was said previously:
$$
xlog_2x = a Leftrightarrow x =frac{aln 2}{W(aln 2)}
$$
Algebra will not get you much further than that.






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    $begingroup$

    $f : xmapsto x log_2x$ clearly is a strictly increasing countinuous function over $[1,infty)$ since both $xmapsto x$ and $xmapsto log_2x$ are and are positive on $[1,infty)$. Given that $f(1) = 0$ and that $underset{+infty}{lim}f = +infty$, the intermediate value theorem enables us to know there is a unique solution to:
    $$
    (E_a) : f(x) = a, ain [1,infty)
    $$
    Let $g : xmapsto x ln x$ to make calculaions easier. Similarly to what was previously said, $(E'_a) : g(x) = a, ain [1,infty)$ has a unique solution $xin[1,infty)$. Note that the solution to $(E'_{aln2})$ is the solution to $(E_a)$.



    Then, as SmileyCraft thought, this is a plain use case of Lambert W function which is defined as the reciprocal of $w mapsto we^w$. We have $xln x = a$ so $e^{ln x}ln x = a$ and $ln x = W(a)$ and $x = e^{W(a)}$ which yields $x = frac{a}{W(a)}$ (because, by definition $W(a)e^{W(a)} = a$).



    We thus solved $(E'_a)$ for all $ain [1,infty)$:
    $$
    x ln x = a Leftrightarrow x = frac{a}{W(a)}
    $$
    Then, given what was said previously:
    $$
    xlog_2x = a Leftrightarrow x =frac{aln 2}{W(aln 2)}
    $$
    Algebra will not get you much further than that.






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      $f : xmapsto x log_2x$ clearly is a strictly increasing countinuous function over $[1,infty)$ since both $xmapsto x$ and $xmapsto log_2x$ are and are positive on $[1,infty)$. Given that $f(1) = 0$ and that $underset{+infty}{lim}f = +infty$, the intermediate value theorem enables us to know there is a unique solution to:
      $$
      (E_a) : f(x) = a, ain [1,infty)
      $$
      Let $g : xmapsto x ln x$ to make calculaions easier. Similarly to what was previously said, $(E'_a) : g(x) = a, ain [1,infty)$ has a unique solution $xin[1,infty)$. Note that the solution to $(E'_{aln2})$ is the solution to $(E_a)$.



      Then, as SmileyCraft thought, this is a plain use case of Lambert W function which is defined as the reciprocal of $w mapsto we^w$. We have $xln x = a$ so $e^{ln x}ln x = a$ and $ln x = W(a)$ and $x = e^{W(a)}$ which yields $x = frac{a}{W(a)}$ (because, by definition $W(a)e^{W(a)} = a$).



      We thus solved $(E'_a)$ for all $ain [1,infty)$:
      $$
      x ln x = a Leftrightarrow x = frac{a}{W(a)}
      $$
      Then, given what was said previously:
      $$
      xlog_2x = a Leftrightarrow x =frac{aln 2}{W(aln 2)}
      $$
      Algebra will not get you much further than that.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        $f : xmapsto x log_2x$ clearly is a strictly increasing countinuous function over $[1,infty)$ since both $xmapsto x$ and $xmapsto log_2x$ are and are positive on $[1,infty)$. Given that $f(1) = 0$ and that $underset{+infty}{lim}f = +infty$, the intermediate value theorem enables us to know there is a unique solution to:
        $$
        (E_a) : f(x) = a, ain [1,infty)
        $$
        Let $g : xmapsto x ln x$ to make calculaions easier. Similarly to what was previously said, $(E'_a) : g(x) = a, ain [1,infty)$ has a unique solution $xin[1,infty)$. Note that the solution to $(E'_{aln2})$ is the solution to $(E_a)$.



        Then, as SmileyCraft thought, this is a plain use case of Lambert W function which is defined as the reciprocal of $w mapsto we^w$. We have $xln x = a$ so $e^{ln x}ln x = a$ and $ln x = W(a)$ and $x = e^{W(a)}$ which yields $x = frac{a}{W(a)}$ (because, by definition $W(a)e^{W(a)} = a$).



        We thus solved $(E'_a)$ for all $ain [1,infty)$:
        $$
        x ln x = a Leftrightarrow x = frac{a}{W(a)}
        $$
        Then, given what was said previously:
        $$
        xlog_2x = a Leftrightarrow x =frac{aln 2}{W(aln 2)}
        $$
        Algebra will not get you much further than that.






        share|cite|improve this answer











        $endgroup$



        $f : xmapsto x log_2x$ clearly is a strictly increasing countinuous function over $[1,infty)$ since both $xmapsto x$ and $xmapsto log_2x$ are and are positive on $[1,infty)$. Given that $f(1) = 0$ and that $underset{+infty}{lim}f = +infty$, the intermediate value theorem enables us to know there is a unique solution to:
        $$
        (E_a) : f(x) = a, ain [1,infty)
        $$
        Let $g : xmapsto x ln x$ to make calculaions easier. Similarly to what was previously said, $(E'_a) : g(x) = a, ain [1,infty)$ has a unique solution $xin[1,infty)$. Note that the solution to $(E'_{aln2})$ is the solution to $(E_a)$.



        Then, as SmileyCraft thought, this is a plain use case of Lambert W function which is defined as the reciprocal of $w mapsto we^w$. We have $xln x = a$ so $e^{ln x}ln x = a$ and $ln x = W(a)$ and $x = e^{W(a)}$ which yields $x = frac{a}{W(a)}$ (because, by definition $W(a)e^{W(a)} = a$).



        We thus solved $(E'_a)$ for all $ain [1,infty)$:
        $$
        x ln x = a Leftrightarrow x = frac{a}{W(a)}
        $$
        Then, given what was said previously:
        $$
        xlog_2x = a Leftrightarrow x =frac{aln 2}{W(aln 2)}
        $$
        Algebra will not get you much further than that.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 19 '18 at 8:44

























        answered Dec 19 '18 at 8:29









        Bill O'HaranBill O'Haran

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