Trouble factoring with $ln$











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0
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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







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  • 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






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edited Nov 19 at 15:31









Glorfindel

3,38471730




3,38471730










asked Nov 19 at 15:20









M Do

124




124








  • 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





















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      3 Answers
      3






      active

      oldest

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      3 Answers
      3






      active

      oldest

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      active

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      active

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      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

          3,609417




          3,609417






















              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

              964319




              964319








              • 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

                  10.5k2720




                  10.5k2720






























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