Prove that the product of the roots of $x^{log_{2016}x}*sqrt{2016}=x^{2016}$ is a Natural Number [duplicate]












0












$begingroup$



This question already has an answer here:




  • The product of the two roots of $sqrt{2014}x^{log_{2014} x}=x^{2014}$ is an integer. Find its units digit

    4 answers




Prove that the product of the roots of $x^{log_{2016}x}*sqrt{2016}=x^{2016}$ is a Natural Number.



This is my solution:



by putting $log_{2016}$ on both sides we get:
$log^2{_{2016}}x-2016log_{2016}x=-log_{2016}sqrt{2016}$



then by putting $log_{2016}x$ in front of bracket on left side, then removing $log_{2016}$ from both sides and squaring the equation, and $t=log_{2016}x$
$t^2-4032t+2016^2-2016=0$



solving for $t_{1/2}$ we get:
$x_1 = 2016^{12left(168+sqrt{14}right)}$
$x_2 = 2016^{12left(168-sqrt{14}right)}$
$x_1 * x_2 = 2016^{4032}$



Is my solution correct? (I think I might have a mistake because I haven't solved this type of problem before, and overall I just started practicing). And if correct is there a better/easier way to solve it?






algebra-precalculus







share|cite|improve this question









$endgroup$



marked as duplicate by астон вілла олоф мэллбэрг, Community♦ Nov 30 '18 at 23:13


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


















  • $begingroup$
    this question is a near duplicate.
    $endgroup$
    – lulu
    Nov 30 '18 at 22:25










  • $begingroup$
    Oh it actually is the same. My bad I guess. I searched for my problem and nothing came up. That link is the exact same problem but with 2014 instead of 2016.
    $endgroup$
    – Pero
    Nov 30 '18 at 22:29










  • $begingroup$
    Do I delete my post now? Or learn to solve it from the link and answer my question? Or what?
    $endgroup$
    – Pero
    Nov 30 '18 at 22:30










  • $begingroup$
    Sounds fine to me!
    $endgroup$
    – Mostafa Ayaz
    Nov 30 '18 at 22:31










  • $begingroup$
    Leave your question as is. We will close it as a duplicate. I have added one above.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Nov 30 '18 at 23:09


















0












$begingroup$



This question already has an answer here:




  • The product of the two roots of $sqrt{2014}x^{log_{2014} x}=x^{2014}$ is an integer. Find its units digit

    4 answers




Prove that the product of the roots of $x^{log_{2016}x}*sqrt{2016}=x^{2016}$ is a Natural Number.



This is my solution:



by putting $log_{2016}$ on both sides we get:
$log^2{_{2016}}x-2016log_{2016}x=-log_{2016}sqrt{2016}$



then by putting $log_{2016}x$ in front of bracket on left side, then removing $log_{2016}$ from both sides and squaring the equation, and $t=log_{2016}x$
$t^2-4032t+2016^2-2016=0$



solving for $t_{1/2}$ we get:
$x_1 = 2016^{12left(168+sqrt{14}right)}$
$x_2 = 2016^{12left(168-sqrt{14}right)}$
$x_1 * x_2 = 2016^{4032}$



Is my solution correct? (I think I might have a mistake because I haven't solved this type of problem before, and overall I just started practicing). And if correct is there a better/easier way to solve it?






algebra-precalculus







share|cite|improve this question









$endgroup$



marked as duplicate by астон вілла олоф мэллбэрг, Community♦ Nov 30 '18 at 23:13


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


















  • $begingroup$
    this question is a near duplicate.
    $endgroup$
    – lulu
    Nov 30 '18 at 22:25










  • $begingroup$
    Oh it actually is the same. My bad I guess. I searched for my problem and nothing came up. That link is the exact same problem but with 2014 instead of 2016.
    $endgroup$
    – Pero
    Nov 30 '18 at 22:29










  • $begingroup$
    Do I delete my post now? Or learn to solve it from the link and answer my question? Or what?
    $endgroup$
    – Pero
    Nov 30 '18 at 22:30










  • $begingroup$
    Sounds fine to me!
    $endgroup$
    – Mostafa Ayaz
    Nov 30 '18 at 22:31










  • $begingroup$
    Leave your question as is. We will close it as a duplicate. I have added one above.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Nov 30 '18 at 23:09
















0












0








0





$begingroup$



This question already has an answer here:




  • The product of the two roots of $sqrt{2014}x^{log_{2014} x}=x^{2014}$ is an integer. Find its units digit

    4 answers




Prove that the product of the roots of $x^{log_{2016}x}*sqrt{2016}=x^{2016}$ is a Natural Number.



This is my solution:



by putting $log_{2016}$ on both sides we get:
$log^2{_{2016}}x-2016log_{2016}x=-log_{2016}sqrt{2016}$



then by putting $log_{2016}x$ in front of bracket on left side, then removing $log_{2016}$ from both sides and squaring the equation, and $t=log_{2016}x$
$t^2-4032t+2016^2-2016=0$



solving for $t_{1/2}$ we get:
$x_1 = 2016^{12left(168+sqrt{14}right)}$
$x_2 = 2016^{12left(168-sqrt{14}right)}$
$x_1 * x_2 = 2016^{4032}$



Is my solution correct? (I think I might have a mistake because I haven't solved this type of problem before, and overall I just started practicing). And if correct is there a better/easier way to solve it?






algebra-precalculus







share|cite|improve this question









$endgroup$





This question already has an answer here:




  • The product of the two roots of $sqrt{2014}x^{log_{2014} x}=x^{2014}$ is an integer. Find its units digit

    4 answers




Prove that the product of the roots of $x^{log_{2016}x}*sqrt{2016}=x^{2016}$ is a Natural Number.



This is my solution:



by putting $log_{2016}$ on both sides we get:
$log^2{_{2016}}x-2016log_{2016}x=-log_{2016}sqrt{2016}$



then by putting $log_{2016}x$ in front of bracket on left side, then removing $log_{2016}$ from both sides and squaring the equation, and $t=log_{2016}x$
$t^2-4032t+2016^2-2016=0$



solving for $t_{1/2}$ we get:
$x_1 = 2016^{12left(168+sqrt{14}right)}$
$x_2 = 2016^{12left(168-sqrt{14}right)}$
$x_1 * x_2 = 2016^{4032}$



Is my solution correct? (I think I might have a mistake because I haven't solved this type of problem before, and overall I just started practicing). And if correct is there a better/easier way to solve it?





This question already has an answer here:




  • The product of the two roots of $sqrt{2014}x^{log_{2014} x}=x^{2014}$ is an integer. Find its units digit

    4 answers







algebra-precalculus




algebra-precalculus






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 30 '18 at 22:24









PeroPero

1497




1497




marked as duplicate by астон вілла олоф мэллбэрг, Community♦ Nov 30 '18 at 23:13


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









marked as duplicate by астон вілла олоф мэллбэрг, Community♦ Nov 30 '18 at 23:13


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • $begingroup$
    this question is a near duplicate.
    $endgroup$
    – lulu
    Nov 30 '18 at 22:25










  • $begingroup$
    Oh it actually is the same. My bad I guess. I searched for my problem and nothing came up. That link is the exact same problem but with 2014 instead of 2016.
    $endgroup$
    – Pero
    Nov 30 '18 at 22:29










  • $begingroup$
    Do I delete my post now? Or learn to solve it from the link and answer my question? Or what?
    $endgroup$
    – Pero
    Nov 30 '18 at 22:30










  • $begingroup$
    Sounds fine to me!
    $endgroup$
    – Mostafa Ayaz
    Nov 30 '18 at 22:31










  • $begingroup$
    Leave your question as is. We will close it as a duplicate. I have added one above.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Nov 30 '18 at 23:09




















  • $begingroup$
    this question is a near duplicate.
    $endgroup$
    – lulu
    Nov 30 '18 at 22:25










  • $begingroup$
    Oh it actually is the same. My bad I guess. I searched for my problem and nothing came up. That link is the exact same problem but with 2014 instead of 2016.
    $endgroup$
    – Pero
    Nov 30 '18 at 22:29










  • $begingroup$
    Do I delete my post now? Or learn to solve it from the link and answer my question? Or what?
    $endgroup$
    – Pero
    Nov 30 '18 at 22:30










  • $begingroup$
    Sounds fine to me!
    $endgroup$
    – Mostafa Ayaz
    Nov 30 '18 at 22:31










  • $begingroup$
    Leave your question as is. We will close it as a duplicate. I have added one above.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Nov 30 '18 at 23:09


















$begingroup$
this question is a near duplicate.
$endgroup$
– lulu
Nov 30 '18 at 22:25




$begingroup$
this question is a near duplicate.
$endgroup$
– lulu
Nov 30 '18 at 22:25












$begingroup$
Oh it actually is the same. My bad I guess. I searched for my problem and nothing came up. That link is the exact same problem but with 2014 instead of 2016.
$endgroup$
– Pero
Nov 30 '18 at 22:29




$begingroup$
Oh it actually is the same. My bad I guess. I searched for my problem and nothing came up. That link is the exact same problem but with 2014 instead of 2016.
$endgroup$
– Pero
Nov 30 '18 at 22:29












$begingroup$
Do I delete my post now? Or learn to solve it from the link and answer my question? Or what?
$endgroup$
– Pero
Nov 30 '18 at 22:30




$begingroup$
Do I delete my post now? Or learn to solve it from the link and answer my question? Or what?
$endgroup$
– Pero
Nov 30 '18 at 22:30












$begingroup$
Sounds fine to me!
$endgroup$
– Mostafa Ayaz
Nov 30 '18 at 22:31




$begingroup$
Sounds fine to me!
$endgroup$
– Mostafa Ayaz
Nov 30 '18 at 22:31












$begingroup$
Leave your question as is. We will close it as a duplicate. I have added one above.
$endgroup$
– астон вілла олоф мэллбэрг
Nov 30 '18 at 23:09






$begingroup$
Leave your question as is. We will close it as a duplicate. I have added one above.
$endgroup$
– астон вілла олоф мэллбэрг
Nov 30 '18 at 23:09












2 Answers
2






active

oldest

votes


















0












$begingroup$

We have that



$$y=log_{2016}x iff 2016^y=x implies x^y=2016^{y^2}$$



$$x^{log_{2016}x}cdot sqrt{2016}=x^{2016} iff 2016^{y^2}cdot sqrt{2016}=2016^{2016y}$$



$$2016^{(y^2-2016y)} =frac{1}{sqrt{2016}} iff y^2-2016y=frac12$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Apparently, I could've solved it a lot easier. Thanks to @lulu for providing the link.



    $x^{log_{2016}x}*sqrt{2016}=x^{2016}$



    By taking $log_{2016}$ of both sides we get:



    $log^2{_{2016}}x-2016log_{2016}x+1/2=0$
    $t^2-2016t+1/2=0$



    Let r, s be the roots of the original problem, we have:



    $log_{2016} rs = log_{2016} r + log_{2016} s = 2016 => rs = 2016^{2016}$






    share|cite|improve this answer









    $endgroup$




















      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0












      $begingroup$

      We have that



      $$y=log_{2016}x iff 2016^y=x implies x^y=2016^{y^2}$$



      $$x^{log_{2016}x}cdot sqrt{2016}=x^{2016} iff 2016^{y^2}cdot sqrt{2016}=2016^{2016y}$$



      $$2016^{(y^2-2016y)} =frac{1}{sqrt{2016}} iff y^2-2016y=frac12$$






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        We have that



        $$y=log_{2016}x iff 2016^y=x implies x^y=2016^{y^2}$$



        $$x^{log_{2016}x}cdot sqrt{2016}=x^{2016} iff 2016^{y^2}cdot sqrt{2016}=2016^{2016y}$$



        $$2016^{(y^2-2016y)} =frac{1}{sqrt{2016}} iff y^2-2016y=frac12$$






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          We have that



          $$y=log_{2016}x iff 2016^y=x implies x^y=2016^{y^2}$$



          $$x^{log_{2016}x}cdot sqrt{2016}=x^{2016} iff 2016^{y^2}cdot sqrt{2016}=2016^{2016y}$$



          $$2016^{(y^2-2016y)} =frac{1}{sqrt{2016}} iff y^2-2016y=frac12$$






          share|cite|improve this answer









          $endgroup$



          We have that



          $$y=log_{2016}x iff 2016^y=x implies x^y=2016^{y^2}$$



          $$x^{log_{2016}x}cdot sqrt{2016}=x^{2016} iff 2016^{y^2}cdot sqrt{2016}=2016^{2016y}$$



          $$2016^{(y^2-2016y)} =frac{1}{sqrt{2016}} iff y^2-2016y=frac12$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 30 '18 at 22:31









          gimusigimusi

          92.8k84494




          92.8k84494























              0












              $begingroup$

              Apparently, I could've solved it a lot easier. Thanks to @lulu for providing the link.



              $x^{log_{2016}x}*sqrt{2016}=x^{2016}$



              By taking $log_{2016}$ of both sides we get:



              $log^2{_{2016}}x-2016log_{2016}x+1/2=0$
              $t^2-2016t+1/2=0$



              Let r, s be the roots of the original problem, we have:



              $log_{2016} rs = log_{2016} r + log_{2016} s = 2016 => rs = 2016^{2016}$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Apparently, I could've solved it a lot easier. Thanks to @lulu for providing the link.



                $x^{log_{2016}x}*sqrt{2016}=x^{2016}$



                By taking $log_{2016}$ of both sides we get:



                $log^2{_{2016}}x-2016log_{2016}x+1/2=0$
                $t^2-2016t+1/2=0$



                Let r, s be the roots of the original problem, we have:



                $log_{2016} rs = log_{2016} r + log_{2016} s = 2016 => rs = 2016^{2016}$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Apparently, I could've solved it a lot easier. Thanks to @lulu for providing the link.



                  $x^{log_{2016}x}*sqrt{2016}=x^{2016}$



                  By taking $log_{2016}$ of both sides we get:



                  $log^2{_{2016}}x-2016log_{2016}x+1/2=0$
                  $t^2-2016t+1/2=0$



                  Let r, s be the roots of the original problem, we have:



                  $log_{2016} rs = log_{2016} r + log_{2016} s = 2016 => rs = 2016^{2016}$






                  share|cite|improve this answer









                  $endgroup$



                  Apparently, I could've solved it a lot easier. Thanks to @lulu for providing the link.



                  $x^{log_{2016}x}*sqrt{2016}=x^{2016}$



                  By taking $log_{2016}$ of both sides we get:



                  $log^2{_{2016}}x-2016log_{2016}x+1/2=0$
                  $t^2-2016t+1/2=0$



                  Let r, s be the roots of the original problem, we have:



                  $log_{2016} rs = log_{2016} r + log_{2016} s = 2016 => rs = 2016^{2016}$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 30 '18 at 23:11









                  PeroPero

                  1497




                  1497















                      -

                      Popular posts from this blog

                      How to change which sound is reproduced for terminal bell?

                      Image
                      3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question
                      Read more

                      Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

                      Image
                      up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx
                      Read more

                      Can I use Tabulator js library in my java Spring + Thymeleaf project?

                      Image
                      0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
                      Read more