Logarithm Equation - Lost What to do with a squared logarithm?












0












$begingroup$


Struggling with a logarithm problem here:




Solve the quation for x:



$x ^{2 - (lgx)^2-lgx^2}-1/x = 0$




My initial thought was to present the equation in such a way as to get rid of the 'x' base and use the exponents solemnly. I.e.



$x ^{2 - (lgx)^2-lgx^2}-1/x = 0$



Becomes: $x ^{2 - (lgx)^2-lgx^2}-x^{-1} = 0$
Which leads to: $2-(lgx)^{2} -lgx^{2} +1 = 0 $



However, from here on, I am absolutely lost in regards to what to do. I have no clue what to do with the squared logarithm, and in turn, no idea how to solve this.



The answers given by the textbook are:
x1=1 ; x2=10 ; x3=0.001



I must admit, I'm even double lost as to how to reach 3 answers, haha.



I am preparing for my math exam (and I am studying on my own, no tutor) so any help would be much appreciated!



Thank you in advance and happy holiday preparations! :)






calculus logarithms







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  • $begingroup$
    can you clarify your notation: is $lgx := log (x)$? and $(lgx)^2 := log^2 (x)$
    $endgroup$
    – Bo5man
    Dec 18 '18 at 21:18


















0












$begingroup$


Struggling with a logarithm problem here:




Solve the quation for x:



$x ^{2 - (lgx)^2-lgx^2}-1/x = 0$




My initial thought was to present the equation in such a way as to get rid of the 'x' base and use the exponents solemnly. I.e.



$x ^{2 - (lgx)^2-lgx^2}-1/x = 0$



Becomes: $x ^{2 - (lgx)^2-lgx^2}-x^{-1} = 0$
Which leads to: $2-(lgx)^{2} -lgx^{2} +1 = 0 $



However, from here on, I am absolutely lost in regards to what to do. I have no clue what to do with the squared logarithm, and in turn, no idea how to solve this.



The answers given by the textbook are:
x1=1 ; x2=10 ; x3=0.001



I must admit, I'm even double lost as to how to reach 3 answers, haha.



I am preparing for my math exam (and I am studying on my own, no tutor) so any help would be much appreciated!



Thank you in advance and happy holiday preparations! :)






calculus logarithms







share|cite|improve this question









$endgroup$












  • $begingroup$
    can you clarify your notation: is $lgx := log (x)$? and $(lgx)^2 := log^2 (x)$
    $endgroup$
    – Bo5man
    Dec 18 '18 at 21:18
















0












0








0





$begingroup$


Struggling with a logarithm problem here:




Solve the quation for x:



$x ^{2 - (lgx)^2-lgx^2}-1/x = 0$




My initial thought was to present the equation in such a way as to get rid of the 'x' base and use the exponents solemnly. I.e.



$x ^{2 - (lgx)^2-lgx^2}-1/x = 0$



Becomes: $x ^{2 - (lgx)^2-lgx^2}-x^{-1} = 0$
Which leads to: $2-(lgx)^{2} -lgx^{2} +1 = 0 $



However, from here on, I am absolutely lost in regards to what to do. I have no clue what to do with the squared logarithm, and in turn, no idea how to solve this.



The answers given by the textbook are:
x1=1 ; x2=10 ; x3=0.001



I must admit, I'm even double lost as to how to reach 3 answers, haha.



I am preparing for my math exam (and I am studying on my own, no tutor) so any help would be much appreciated!



Thank you in advance and happy holiday preparations! :)






calculus logarithms







share|cite|improve this question









$endgroup$




Struggling with a logarithm problem here:




Solve the quation for x:



$x ^{2 - (lgx)^2-lgx^2}-1/x = 0$




My initial thought was to present the equation in such a way as to get rid of the 'x' base and use the exponents solemnly. I.e.



$x ^{2 - (lgx)^2-lgx^2}-1/x = 0$



Becomes: $x ^{2 - (lgx)^2-lgx^2}-x^{-1} = 0$
Which leads to: $2-(lgx)^{2} -lgx^{2} +1 = 0 $



However, from here on, I am absolutely lost in regards to what to do. I have no clue what to do with the squared logarithm, and in turn, no idea how to solve this.



The answers given by the textbook are:
x1=1 ; x2=10 ; x3=0.001



I must admit, I'm even double lost as to how to reach 3 answers, haha.



I am preparing for my math exam (and I am studying on my own, no tutor) so any help would be much appreciated!



Thank you in advance and happy holiday preparations! :)






calculus logarithms




calculus logarithms






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asked Dec 18 '18 at 21:10









VRTVRT

957




957












  • $begingroup$
    can you clarify your notation: is $lgx := log (x)$? and $(lgx)^2 := log^2 (x)$
    $endgroup$
    – Bo5man
    Dec 18 '18 at 21:18




















  • $begingroup$
    can you clarify your notation: is $lgx := log (x)$? and $(lgx)^2 := log^2 (x)$
    $endgroup$
    – Bo5man
    Dec 18 '18 at 21:18


















$begingroup$
can you clarify your notation: is $lgx := log (x)$? and $(lgx)^2 := log^2 (x)$
$endgroup$
– Bo5man
Dec 18 '18 at 21:18






$begingroup$
can you clarify your notation: is $lgx := log (x)$? and $(lgx)^2 := log^2 (x)$
$endgroup$
– Bo5man
Dec 18 '18 at 21:18












3 Answers
3






active

oldest

votes


















1












$begingroup$

Note that $x=1$ is always a solution since $1$ raised to any power is $1=1/x$.



Take $log x=A$, you get $(log x)^2+log(x^2)=(log x)^2+2log(x)=A^2+2A=3$



$displaystyleimplies(A+3)(A-1)=0, A=-3,1\therefore x=10^{-3},10$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for taking the time!
    $endgroup$
    – VRT
    Dec 18 '18 at 21:30



















1












$begingroup$

Hint: clearly $x ne 0$ so multiplying by $x$ we get $x ^{3 - (lg x)^2-lg x^2} = 1$. So either $x=1$ or $3 - (lg x)^2-lg x^2=0$. The last equation you can solve by taking $y=lg x$ and solving $3-y^2-2y=0$.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Recall that $log(x^2) = 2log x$; so, you have to solve
    $$
    3 - (log x)^2 - 2log x = 0
    $$
    for $x>0$. Set $y = log x$. Then, the original equation is equivalent to solving
    $$
    3 - y^2 -2y =0
    $$
    (and then, given a solution $y^ast$, $x^ast = e^{y^ast}$ will be a solution of the original equation). But this is now a quadratic equation, which you know how to handle.



    Important: when you "got rid of $x$ in the base", you implicitly simplified by $log x$. So you may have discarded a solution, $x=1$ (for which $log x=0$). You have to check separately whether this was a solution of the original equation.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you so so much for taking the time! This makes perfect sense!
      $endgroup$
      – VRT
      Dec 18 '18 at 21:30










    • $begingroup$
      @VRT You're welcome.
      $endgroup$
      – Clement C.
      Dec 18 '18 at 21:34











    Your Answer





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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Note that $x=1$ is always a solution since $1$ raised to any power is $1=1/x$.



    Take $log x=A$, you get $(log x)^2+log(x^2)=(log x)^2+2log(x)=A^2+2A=3$



    $displaystyleimplies(A+3)(A-1)=0, A=-3,1\therefore x=10^{-3},10$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you for taking the time!
      $endgroup$
      – VRT
      Dec 18 '18 at 21:30
















    1












    $begingroup$

    Note that $x=1$ is always a solution since $1$ raised to any power is $1=1/x$.



    Take $log x=A$, you get $(log x)^2+log(x^2)=(log x)^2+2log(x)=A^2+2A=3$



    $displaystyleimplies(A+3)(A-1)=0, A=-3,1\therefore x=10^{-3},10$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you for taking the time!
      $endgroup$
      – VRT
      Dec 18 '18 at 21:30














    1












    1








    1





    $begingroup$

    Note that $x=1$ is always a solution since $1$ raised to any power is $1=1/x$.



    Take $log x=A$, you get $(log x)^2+log(x^2)=(log x)^2+2log(x)=A^2+2A=3$



    $displaystyleimplies(A+3)(A-1)=0, A=-3,1\therefore x=10^{-3},10$






    share|cite|improve this answer









    $endgroup$



    Note that $x=1$ is always a solution since $1$ raised to any power is $1=1/x$.



    Take $log x=A$, you get $(log x)^2+log(x^2)=(log x)^2+2log(x)=A^2+2A=3$



    $displaystyleimplies(A+3)(A-1)=0, A=-3,1\therefore x=10^{-3},10$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 18 '18 at 21:20









    Shubham JohriShubham Johri

    5,186717




    5,186717












    • $begingroup$
      Thank you for taking the time!
      $endgroup$
      – VRT
      Dec 18 '18 at 21:30


















    • $begingroup$
      Thank you for taking the time!
      $endgroup$
      – VRT
      Dec 18 '18 at 21:30
















    $begingroup$
    Thank you for taking the time!
    $endgroup$
    – VRT
    Dec 18 '18 at 21:30




    $begingroup$
    Thank you for taking the time!
    $endgroup$
    – VRT
    Dec 18 '18 at 21:30











    1












    $begingroup$

    Hint: clearly $x ne 0$ so multiplying by $x$ we get $x ^{3 - (lg x)^2-lg x^2} = 1$. So either $x=1$ or $3 - (lg x)^2-lg x^2=0$. The last equation you can solve by taking $y=lg x$ and solving $3-y^2-2y=0$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Hint: clearly $x ne 0$ so multiplying by $x$ we get $x ^{3 - (lg x)^2-lg x^2} = 1$. So either $x=1$ or $3 - (lg x)^2-lg x^2=0$. The last equation you can solve by taking $y=lg x$ and solving $3-y^2-2y=0$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Hint: clearly $x ne 0$ so multiplying by $x$ we get $x ^{3 - (lg x)^2-lg x^2} = 1$. So either $x=1$ or $3 - (lg x)^2-lg x^2=0$. The last equation you can solve by taking $y=lg x$ and solving $3-y^2-2y=0$.






        share|cite|improve this answer









        $endgroup$



        Hint: clearly $x ne 0$ so multiplying by $x$ we get $x ^{3 - (lg x)^2-lg x^2} = 1$. So either $x=1$ or $3 - (lg x)^2-lg x^2=0$. The last equation you can solve by taking $y=lg x$ and solving $3-y^2-2y=0$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 18 '18 at 21:17









        VasyaVasya

        3,3771516




        3,3771516























            1












            $begingroup$

            Recall that $log(x^2) = 2log x$; so, you have to solve
            $$
            3 - (log x)^2 - 2log x = 0
            $$
            for $x>0$. Set $y = log x$. Then, the original equation is equivalent to solving
            $$
            3 - y^2 -2y =0
            $$
            (and then, given a solution $y^ast$, $x^ast = e^{y^ast}$ will be a solution of the original equation). But this is now a quadratic equation, which you know how to handle.



            Important: when you "got rid of $x$ in the base", you implicitly simplified by $log x$. So you may have discarded a solution, $x=1$ (for which $log x=0$). You have to check separately whether this was a solution of the original equation.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thank you so so much for taking the time! This makes perfect sense!
              $endgroup$
              – VRT
              Dec 18 '18 at 21:30










            • $begingroup$
              @VRT You're welcome.
              $endgroup$
              – Clement C.
              Dec 18 '18 at 21:34
















            1












            $begingroup$

            Recall that $log(x^2) = 2log x$; so, you have to solve
            $$
            3 - (log x)^2 - 2log x = 0
            $$
            for $x>0$. Set $y = log x$. Then, the original equation is equivalent to solving
            $$
            3 - y^2 -2y =0
            $$
            (and then, given a solution $y^ast$, $x^ast = e^{y^ast}$ will be a solution of the original equation). But this is now a quadratic equation, which you know how to handle.



            Important: when you "got rid of $x$ in the base", you implicitly simplified by $log x$. So you may have discarded a solution, $x=1$ (for which $log x=0$). You have to check separately whether this was a solution of the original equation.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thank you so so much for taking the time! This makes perfect sense!
              $endgroup$
              – VRT
              Dec 18 '18 at 21:30










            • $begingroup$
              @VRT You're welcome.
              $endgroup$
              – Clement C.
              Dec 18 '18 at 21:34














            1












            1








            1





            $begingroup$

            Recall that $log(x^2) = 2log x$; so, you have to solve
            $$
            3 - (log x)^2 - 2log x = 0
            $$
            for $x>0$. Set $y = log x$. Then, the original equation is equivalent to solving
            $$
            3 - y^2 -2y =0
            $$
            (and then, given a solution $y^ast$, $x^ast = e^{y^ast}$ will be a solution of the original equation). But this is now a quadratic equation, which you know how to handle.



            Important: when you "got rid of $x$ in the base", you implicitly simplified by $log x$. So you may have discarded a solution, $x=1$ (for which $log x=0$). You have to check separately whether this was a solution of the original equation.






            share|cite|improve this answer









            $endgroup$



            Recall that $log(x^2) = 2log x$; so, you have to solve
            $$
            3 - (log x)^2 - 2log x = 0
            $$
            for $x>0$. Set $y = log x$. Then, the original equation is equivalent to solving
            $$
            3 - y^2 -2y =0
            $$
            (and then, given a solution $y^ast$, $x^ast = e^{y^ast}$ will be a solution of the original equation). But this is now a quadratic equation, which you know how to handle.



            Important: when you "got rid of $x$ in the base", you implicitly simplified by $log x$. So you may have discarded a solution, $x=1$ (for which $log x=0$). You have to check separately whether this was a solution of the original equation.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 18 '18 at 21:20









            Clement C.Clement C.

            50.6k33892




            50.6k33892












            • $begingroup$
              Thank you so so much for taking the time! This makes perfect sense!
              $endgroup$
              – VRT
              Dec 18 '18 at 21:30










            • $begingroup$
              @VRT You're welcome.
              $endgroup$
              – Clement C.
              Dec 18 '18 at 21:34


















            • $begingroup$
              Thank you so so much for taking the time! This makes perfect sense!
              $endgroup$
              – VRT
              Dec 18 '18 at 21:30










            • $begingroup$
              @VRT You're welcome.
              $endgroup$
              – Clement C.
              Dec 18 '18 at 21:34
















            $begingroup$
            Thank you so so much for taking the time! This makes perfect sense!
            $endgroup$
            – VRT
            Dec 18 '18 at 21:30




            $begingroup$
            Thank you so so much for taking the time! This makes perfect sense!
            $endgroup$
            – VRT
            Dec 18 '18 at 21:30












            $begingroup$
            @VRT You're welcome.
            $endgroup$
            – Clement C.
            Dec 18 '18 at 21:34




            $begingroup$
            @VRT You're welcome.
            $endgroup$
            – Clement C.
            Dec 18 '18 at 21:34


















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