Problem with system of logarythmic equation.












1












$begingroup$


So I have this problem with a system of logarytmic equations. Specifically how to get rid of the (ln(x))^2 in order to solve this one.
I know you can get This as a solution for the first equation.
My problem is how to use this in the second one.
I can also provide the solution for the problem, however I don't know which steps to do how to get there.Solution fo the problem



Thanks in advance



ln(x) + ln(y^2)=4 ;
(ln(x))^2 - 3ln(xy)=-5






logarithms systems-of-equations nonlinear-system







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    So I have this problem with a system of logarytmic equations. Specifically how to get rid of the (ln(x))^2 in order to solve this one.
    I know you can get This as a solution for the first equation.
    My problem is how to use this in the second one.
    I can also provide the solution for the problem, however I don't know which steps to do how to get there.Solution fo the problem



    Thanks in advance



    ln(x) + ln(y^2)=4 ;
    (ln(x))^2 - 3ln(xy)=-5






    logarithms systems-of-equations nonlinear-system







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      So I have this problem with a system of logarytmic equations. Specifically how to get rid of the (ln(x))^2 in order to solve this one.
      I know you can get This as a solution for the first equation.
      My problem is how to use this in the second one.
      I can also provide the solution for the problem, however I don't know which steps to do how to get there.Solution fo the problem



      Thanks in advance



      ln(x) + ln(y^2)=4 ;
      (ln(x))^2 - 3ln(xy)=-5






      logarithms systems-of-equations nonlinear-system







      share|cite|improve this question









      $endgroup$




      So I have this problem with a system of logarytmic equations. Specifically how to get rid of the (ln(x))^2 in order to solve this one.
      I know you can get This as a solution for the first equation.
      My problem is how to use this in the second one.
      I can also provide the solution for the problem, however I don't know which steps to do how to get there.Solution fo the problem



      Thanks in advance



      ln(x) + ln(y^2)=4 ;
      (ln(x))^2 - 3ln(xy)=-5






      logarithms systems-of-equations nonlinear-system




      logarithms systems-of-equations nonlinear-system






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 30 '18 at 9:54









      user630424user630424

      82




      82






















          2 Answers
          2






          active

          oldest

          votes


















          0












          $begingroup$

          $x>0$ and from here $y>0$.



          Let $ln{y}=t$.



          Thus, $ln{x}=4-2t$ and we obtain:
          $$(4-2t)^2-3(2-2t)-3t=-5$$ or
          $$4t^2-13t+9=0.$$
          Can you end it now?



          I got the following answer.
          $$left{(e^2,e),left(frac{1}{sqrt{e}},e^{frac{9}{4}}right)right}$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Yes! Thank you very much Sir! I can get that same solution now.
            $endgroup$
            – user630424
            Dec 30 '18 at 10:10










          • $begingroup$
            @user630424 You are welcome!
            $endgroup$
            – Michael Rozenberg
            Dec 30 '18 at 10:11



















          0












          $begingroup$

          Assume $x$ and $y$ to be positive. Then
          $$log()+log(y^2)=4 implies log(x)+2log(y)=4 tag1$$
          $$log^2(x)-3log(xy)=5 implies log^2(x)-3 log(x)-3log(y)=5 tag 2$$



          From $(1)$, express $log(y)$ as a function of $log(x)$. Replace in $(2)$ and you have a quadratic equation in $log(x)$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I see, when we then substitute ln(x) = t we get the same equation as Michael's. Thank you vey much for your help!
            $endgroup$
            – user630424
            Dec 30 '18 at 10:12










          • $begingroup$
            @user630424. You are welcome !
            $endgroup$
            – Claude Leibovici
            Dec 30 '18 at 10:28











          Your Answer





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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          0












          $begingroup$

          $x>0$ and from here $y>0$.



          Let $ln{y}=t$.



          Thus, $ln{x}=4-2t$ and we obtain:
          $$(4-2t)^2-3(2-2t)-3t=-5$$ or
          $$4t^2-13t+9=0.$$
          Can you end it now?



          I got the following answer.
          $$left{(e^2,e),left(frac{1}{sqrt{e}},e^{frac{9}{4}}right)right}$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Yes! Thank you very much Sir! I can get that same solution now.
            $endgroup$
            – user630424
            Dec 30 '18 at 10:10










          • $begingroup$
            @user630424 You are welcome!
            $endgroup$
            – Michael Rozenberg
            Dec 30 '18 at 10:11
















          0












          $begingroup$

          $x>0$ and from here $y>0$.



          Let $ln{y}=t$.



          Thus, $ln{x}=4-2t$ and we obtain:
          $$(4-2t)^2-3(2-2t)-3t=-5$$ or
          $$4t^2-13t+9=0.$$
          Can you end it now?



          I got the following answer.
          $$left{(e^2,e),left(frac{1}{sqrt{e}},e^{frac{9}{4}}right)right}$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Yes! Thank you very much Sir! I can get that same solution now.
            $endgroup$
            – user630424
            Dec 30 '18 at 10:10










          • $begingroup$
            @user630424 You are welcome!
            $endgroup$
            – Michael Rozenberg
            Dec 30 '18 at 10:11














          0












          0








          0





          $begingroup$

          $x>0$ and from here $y>0$.



          Let $ln{y}=t$.



          Thus, $ln{x}=4-2t$ and we obtain:
          $$(4-2t)^2-3(2-2t)-3t=-5$$ or
          $$4t^2-13t+9=0.$$
          Can you end it now?



          I got the following answer.
          $$left{(e^2,e),left(frac{1}{sqrt{e}},e^{frac{9}{4}}right)right}$$






          share|cite|improve this answer









          $endgroup$



          $x>0$ and from here $y>0$.



          Let $ln{y}=t$.



          Thus, $ln{x}=4-2t$ and we obtain:
          $$(4-2t)^2-3(2-2t)-3t=-5$$ or
          $$4t^2-13t+9=0.$$
          Can you end it now?



          I got the following answer.
          $$left{(e^2,e),left(frac{1}{sqrt{e}},e^{frac{9}{4}}right)right}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 30 '18 at 9:59









          Michael RozenbergMichael Rozenberg

          105k1892198




          105k1892198












          • $begingroup$
            Yes! Thank you very much Sir! I can get that same solution now.
            $endgroup$
            – user630424
            Dec 30 '18 at 10:10










          • $begingroup$
            @user630424 You are welcome!
            $endgroup$
            – Michael Rozenberg
            Dec 30 '18 at 10:11


















          • $begingroup$
            Yes! Thank you very much Sir! I can get that same solution now.
            $endgroup$
            – user630424
            Dec 30 '18 at 10:10










          • $begingroup$
            @user630424 You are welcome!
            $endgroup$
            – Michael Rozenberg
            Dec 30 '18 at 10:11
















          $begingroup$
          Yes! Thank you very much Sir! I can get that same solution now.
          $endgroup$
          – user630424
          Dec 30 '18 at 10:10




          $begingroup$
          Yes! Thank you very much Sir! I can get that same solution now.
          $endgroup$
          – user630424
          Dec 30 '18 at 10:10












          $begingroup$
          @user630424 You are welcome!
          $endgroup$
          – Michael Rozenberg
          Dec 30 '18 at 10:11




          $begingroup$
          @user630424 You are welcome!
          $endgroup$
          – Michael Rozenberg
          Dec 30 '18 at 10:11











          0












          $begingroup$

          Assume $x$ and $y$ to be positive. Then
          $$log()+log(y^2)=4 implies log(x)+2log(y)=4 tag1$$
          $$log^2(x)-3log(xy)=5 implies log^2(x)-3 log(x)-3log(y)=5 tag 2$$



          From $(1)$, express $log(y)$ as a function of $log(x)$. Replace in $(2)$ and you have a quadratic equation in $log(x)$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I see, when we then substitute ln(x) = t we get the same equation as Michael's. Thank you vey much for your help!
            $endgroup$
            – user630424
            Dec 30 '18 at 10:12










          • $begingroup$
            @user630424. You are welcome !
            $endgroup$
            – Claude Leibovici
            Dec 30 '18 at 10:28
















          0












          $begingroup$

          Assume $x$ and $y$ to be positive. Then
          $$log()+log(y^2)=4 implies log(x)+2log(y)=4 tag1$$
          $$log^2(x)-3log(xy)=5 implies log^2(x)-3 log(x)-3log(y)=5 tag 2$$



          From $(1)$, express $log(y)$ as a function of $log(x)$. Replace in $(2)$ and you have a quadratic equation in $log(x)$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I see, when we then substitute ln(x) = t we get the same equation as Michael's. Thank you vey much for your help!
            $endgroup$
            – user630424
            Dec 30 '18 at 10:12










          • $begingroup$
            @user630424. You are welcome !
            $endgroup$
            – Claude Leibovici
            Dec 30 '18 at 10:28














          0












          0








          0





          $begingroup$

          Assume $x$ and $y$ to be positive. Then
          $$log()+log(y^2)=4 implies log(x)+2log(y)=4 tag1$$
          $$log^2(x)-3log(xy)=5 implies log^2(x)-3 log(x)-3log(y)=5 tag 2$$



          From $(1)$, express $log(y)$ as a function of $log(x)$. Replace in $(2)$ and you have a quadratic equation in $log(x)$.






          share|cite|improve this answer









          $endgroup$



          Assume $x$ and $y$ to be positive. Then
          $$log()+log(y^2)=4 implies log(x)+2log(y)=4 tag1$$
          $$log^2(x)-3log(xy)=5 implies log^2(x)-3 log(x)-3log(y)=5 tag 2$$



          From $(1)$, express $log(y)$ as a function of $log(x)$. Replace in $(2)$ and you have a quadratic equation in $log(x)$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 30 '18 at 10:03









          Claude LeiboviciClaude Leibovici

          122k1157134




          122k1157134












          • $begingroup$
            I see, when we then substitute ln(x) = t we get the same equation as Michael's. Thank you vey much for your help!
            $endgroup$
            – user630424
            Dec 30 '18 at 10:12










          • $begingroup$
            @user630424. You are welcome !
            $endgroup$
            – Claude Leibovici
            Dec 30 '18 at 10:28


















          • $begingroup$
            I see, when we then substitute ln(x) = t we get the same equation as Michael's. Thank you vey much for your help!
            $endgroup$
            – user630424
            Dec 30 '18 at 10:12










          • $begingroup$
            @user630424. You are welcome !
            $endgroup$
            – Claude Leibovici
            Dec 30 '18 at 10:28
















          $begingroup$
          I see, when we then substitute ln(x) = t we get the same equation as Michael's. Thank you vey much for your help!
          $endgroup$
          – user630424
          Dec 30 '18 at 10:12




          $begingroup$
          I see, when we then substitute ln(x) = t we get the same equation as Michael's. Thank you vey much for your help!
          $endgroup$
          – user630424
          Dec 30 '18 at 10:12












          $begingroup$
          @user630424. You are welcome !
          $endgroup$
          – Claude Leibovici
          Dec 30 '18 at 10:28




          $begingroup$
          @user630424. You are welcome !
          $endgroup$
          – Claude Leibovici
          Dec 30 '18 at 10:28


















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