Trouble factoring with $ln$











up vote
0
down vote

favorite












I have the equation



$$0 = 0.5x^frac {1}{2}(3ln(x) + 2) $$



How do I interpret the root inside the brackets?



The solutions are $x = 0$ and $x = e^{-frac{2}{3}}$, but I have absolutely no idea how that last one was found. Could anybody explain it to me?






logarithms factoring







share|cite|improve this question




















  • 1




    Recall the definition of $ln x$. $$ln x = y iff e^y = x$$ Also, the first solution is incorrect since $x = 0$ is not within the domain $x in (0, +infty)$ in which $ln x$ is defined.
    – KM101
    Nov 19 at 15:36

















up vote
0
down vote

favorite












I have the equation



$$0 = 0.5x^frac {1}{2}(3ln(x) + 2) $$



How do I interpret the root inside the brackets?



The solutions are $x = 0$ and $x = e^{-frac{2}{3}}$, but I have absolutely no idea how that last one was found. Could anybody explain it to me?






logarithms factoring







share|cite|improve this question




















  • 1




    Recall the definition of $ln x$. $$ln x = y iff e^y = x$$ Also, the first solution is incorrect since $x = 0$ is not within the domain $x in (0, +infty)$ in which $ln x$ is defined.
    – KM101
    Nov 19 at 15:36















up vote
0
down vote

favorite









up vote
0
down vote

favorite











I have the equation



$$0 = 0.5x^frac {1}{2}(3ln(x) + 2) $$



How do I interpret the root inside the brackets?



The solutions are $x = 0$ and $x = e^{-frac{2}{3}}$, but I have absolutely no idea how that last one was found. Could anybody explain it to me?






logarithms factoring







share|cite|improve this question















I have the equation



$$0 = 0.5x^frac {1}{2}(3ln(x) + 2) $$



How do I interpret the root inside the brackets?



The solutions are $x = 0$ and $x = e^{-frac{2}{3}}$, but I have absolutely no idea how that last one was found. Could anybody explain it to me?






logarithms factoring




logarithms factoring






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 19 at 15:31









Glorfindel

3,38471730




3,38471730










asked Nov 19 at 15:20









M Do

54




54








  • 1




    Recall the definition of $ln x$. $$ln x = y iff e^y = x$$ Also, the first solution is incorrect since $x = 0$ is not within the domain $x in (0, +infty)$ in which $ln x$ is defined.
    – KM101
    Nov 19 at 15:36
















  • 1




    Recall the definition of $ln x$. $$ln x = y iff e^y = x$$ Also, the first solution is incorrect since $x = 0$ is not within the domain $x in (0, +infty)$ in which $ln x$ is defined.
    – KM101
    Nov 19 at 15:36










1




1




Recall the definition of $ln x$. $$ln x = y iff e^y = x$$ Also, the first solution is incorrect since $x = 0$ is not within the domain $x in (0, +infty)$ in which $ln x$ is defined.
– KM101
Nov 19 at 15:36






Recall the definition of $ln x$. $$ln x = y iff e^y = x$$ Also, the first solution is incorrect since $x = 0$ is not within the domain $x in (0, +infty)$ in which $ln x$ is defined.
– KM101
Nov 19 at 15:36












3 Answers
3






active

oldest

votes

















up vote
1
down vote













$$0 = 0.5x^{frac{1}{2}}(3ln(x)+2)$$



Set either factor equal to $0$. So, you get



$$0.5x^{frac{1}{2}} = 0 implies x = 0$$



or



$$3ln(x)+2 = 0 implies 3ln x = -2 implies ln x = -frac{2}{3} implies x = e^{-frac{2}{3}}$$



However, since $ln x$ is defined for all $x > 0$, the first solution is discarded, leaving you with only one solution.






share|cite|improve this answer




























    up vote
    0
    down vote













    We have, $$ 0 = 0.5x^frac {1}{2}(3ln(x) + 2)$$
    Then,
    $$ 0.5x^frac {1}{2} quad mbox{or}quad 3ln(x) + 2=0$$
    hence,
    $$ x=0quad mbox{or}quad ln(x)=frac{-2}{3}Longrightarrow x=e^frac{-2}{3}$$
    And since $ln(x)$ is only defined on $Bbb{R}^*_+$ then the only solution is:
    $$ x=e^frac{-2}{3}$$






    share|cite|improve this answer



















    • 1




      $x$ cannot be zero because $ln x$ is undefined there
      – Vasya
      Nov 19 at 15:31


















    up vote
    0
    down vote













    $3 log x +2=0;$



    Since $x >0$ (why?) , we set



    $x=e^y$ , $y>0$, real.



    Then:



    $3 log (e^y) +2=0;$



    $3y +2=0;$



    $y=-2/3.$



    Since $x=e^y$, we get



    $x=e^{-2/3}.$






    share|cite|improve this answer





















      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',
      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%2f3005059%2ftrouble-factoring-with-ln%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








      up vote
      1
      down vote













      $$0 = 0.5x^{frac{1}{2}}(3ln(x)+2)$$



      Set either factor equal to $0$. So, you get



      $$0.5x^{frac{1}{2}} = 0 implies x = 0$$



      or



      $$3ln(x)+2 = 0 implies 3ln x = -2 implies ln x = -frac{2}{3} implies x = e^{-frac{2}{3}}$$



      However, since $ln x$ is defined for all $x > 0$, the first solution is discarded, leaving you with only one solution.






      share|cite|improve this answer

























        up vote
        1
        down vote













        $$0 = 0.5x^{frac{1}{2}}(3ln(x)+2)$$



        Set either factor equal to $0$. So, you get



        $$0.5x^{frac{1}{2}} = 0 implies x = 0$$



        or



        $$3ln(x)+2 = 0 implies 3ln x = -2 implies ln x = -frac{2}{3} implies x = e^{-frac{2}{3}}$$



        However, since $ln x$ is defined for all $x > 0$, the first solution is discarded, leaving you with only one solution.






        share|cite|improve this answer























          up vote
          1
          down vote










          up vote
          1
          down vote









          $$0 = 0.5x^{frac{1}{2}}(3ln(x)+2)$$



          Set either factor equal to $0$. So, you get



          $$0.5x^{frac{1}{2}} = 0 implies x = 0$$



          or



          $$3ln(x)+2 = 0 implies 3ln x = -2 implies ln x = -frac{2}{3} implies x = e^{-frac{2}{3}}$$



          However, since $ln x$ is defined for all $x > 0$, the first solution is discarded, leaving you with only one solution.






          share|cite|improve this answer












          $$0 = 0.5x^{frac{1}{2}}(3ln(x)+2)$$



          Set either factor equal to $0$. So, you get



          $$0.5x^{frac{1}{2}} = 0 implies x = 0$$



          or



          $$3ln(x)+2 = 0 implies 3ln x = -2 implies ln x = -frac{2}{3} implies x = e^{-frac{2}{3}}$$



          However, since $ln x$ is defined for all $x > 0$, the first solution is discarded, leaving you with only one solution.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 19 at 15:34









          KM101

          2,928416




          2,928416






















              up vote
              0
              down vote













              We have, $$ 0 = 0.5x^frac {1}{2}(3ln(x) + 2)$$
              Then,
              $$ 0.5x^frac {1}{2} quad mbox{or}quad 3ln(x) + 2=0$$
              hence,
              $$ x=0quad mbox{or}quad ln(x)=frac{-2}{3}Longrightarrow x=e^frac{-2}{3}$$
              And since $ln(x)$ is only defined on $Bbb{R}^*_+$ then the only solution is:
              $$ x=e^frac{-2}{3}$$






              share|cite|improve this answer



















              • 1




                $x$ cannot be zero because $ln x$ is undefined there
                – Vasya
                Nov 19 at 15:31















              up vote
              0
              down vote













              We have, $$ 0 = 0.5x^frac {1}{2}(3ln(x) + 2)$$
              Then,
              $$ 0.5x^frac {1}{2} quad mbox{or}quad 3ln(x) + 2=0$$
              hence,
              $$ x=0quad mbox{or}quad ln(x)=frac{-2}{3}Longrightarrow x=e^frac{-2}{3}$$
              And since $ln(x)$ is only defined on $Bbb{R}^*_+$ then the only solution is:
              $$ x=e^frac{-2}{3}$$






              share|cite|improve this answer



















              • 1




                $x$ cannot be zero because $ln x$ is undefined there
                – Vasya
                Nov 19 at 15:31













              up vote
              0
              down vote










              up vote
              0
              down vote









              We have, $$ 0 = 0.5x^frac {1}{2}(3ln(x) + 2)$$
              Then,
              $$ 0.5x^frac {1}{2} quad mbox{or}quad 3ln(x) + 2=0$$
              hence,
              $$ x=0quad mbox{or}quad ln(x)=frac{-2}{3}Longrightarrow x=e^frac{-2}{3}$$
              And since $ln(x)$ is only defined on $Bbb{R}^*_+$ then the only solution is:
              $$ x=e^frac{-2}{3}$$






              share|cite|improve this answer














              We have, $$ 0 = 0.5x^frac {1}{2}(3ln(x) + 2)$$
              Then,
              $$ 0.5x^frac {1}{2} quad mbox{or}quad 3ln(x) + 2=0$$
              hence,
              $$ x=0quad mbox{or}quad ln(x)=frac{-2}{3}Longrightarrow x=e^frac{-2}{3}$$
              And since $ln(x)$ is only defined on $Bbb{R}^*_+$ then the only solution is:
              $$ x=e^frac{-2}{3}$$







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Nov 19 at 16:42

























              answered Nov 19 at 15:27









              hamza boulahia

              954319




              954319








              • 1




                $x$ cannot be zero because $ln x$ is undefined there
                – Vasya
                Nov 19 at 15:31














              • 1




                $x$ cannot be zero because $ln x$ is undefined there
                – Vasya
                Nov 19 at 15:31








              1




              1




              $x$ cannot be zero because $ln x$ is undefined there
              – Vasya
              Nov 19 at 15:31




              $x$ cannot be zero because $ln x$ is undefined there
              – Vasya
              Nov 19 at 15:31










              up vote
              0
              down vote













              $3 log x +2=0;$



              Since $x >0$ (why?) , we set



              $x=e^y$ , $y>0$, real.



              Then:



              $3 log (e^y) +2=0;$



              $3y +2=0;$



              $y=-2/3.$



              Since $x=e^y$, we get



              $x=e^{-2/3}.$






              share|cite|improve this answer

























                up vote
                0
                down vote













                $3 log x +2=0;$



                Since $x >0$ (why?) , we set



                $x=e^y$ , $y>0$, real.



                Then:



                $3 log (e^y) +2=0;$



                $3y +2=0;$



                $y=-2/3.$



                Since $x=e^y$, we get



                $x=e^{-2/3}.$






                share|cite|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  $3 log x +2=0;$



                  Since $x >0$ (why?) , we set



                  $x=e^y$ , $y>0$, real.



                  Then:



                  $3 log (e^y) +2=0;$



                  $3y +2=0;$



                  $y=-2/3.$



                  Since $x=e^y$, we get



                  $x=e^{-2/3}.$






                  share|cite|improve this answer












                  $3 log x +2=0;$



                  Since $x >0$ (why?) , we set



                  $x=e^y$ , $y>0$, real.



                  Then:



                  $3 log (e^y) +2=0;$



                  $3y +2=0;$



                  $y=-2/3.$



                  Since $x=e^y$, we get



                  $x=e^{-2/3}.$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 19 at 16:51









                  Peter Szilas

                  10.3k2720




                  10.3k2720






























                      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.





                      Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                      Please pay close attention to the following guidance:


                      • 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.


                      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%2f3005059%2ftrouble-factoring-with-ln%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

                      Ellipse (mathématiques)

                      Image
                      Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
                      Read more

                      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

                      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