Solve $log_3 x - 2log_x 3 = 1$ and find the larger value of $x$ out of the two












0












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I don't have the slightest idea about how to tackle this one. I could change $2log_x 3$ to $frac{2}{log_3 x}$ and deducting that from $log_3 x$ would give me $frac{(log_3x)^2-2}{log_3x}$, but I don't know how to proceed.






logarithms







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  • $begingroup$
    It looks like you are very close to a quadratic in $log_3x$... Can you write the full equation resulting from your work thus far? Also, I believe you get $frac 2{log_3x}$ as the changed value...
    $endgroup$
    – abiessu
    Dec 13 '18 at 13:05












  • $begingroup$
    Thanks! Fixed that...
    $endgroup$
    – Nameless King
    Dec 13 '18 at 13:15
















0












$begingroup$


I don't have the slightest idea about how to tackle this one. I could change $2log_x 3$ to $frac{2}{log_3 x}$ and deducting that from $log_3 x$ would give me $frac{(log_3x)^2-2}{log_3x}$, but I don't know how to proceed.






logarithms







share|cite|improve this question











$endgroup$












  • $begingroup$
    It looks like you are very close to a quadratic in $log_3x$... Can you write the full equation resulting from your work thus far? Also, I believe you get $frac 2{log_3x}$ as the changed value...
    $endgroup$
    – abiessu
    Dec 13 '18 at 13:05












  • $begingroup$
    Thanks! Fixed that...
    $endgroup$
    – Nameless King
    Dec 13 '18 at 13:15














0












0








0





$begingroup$


I don't have the slightest idea about how to tackle this one. I could change $2log_x 3$ to $frac{2}{log_3 x}$ and deducting that from $log_3 x$ would give me $frac{(log_3x)^2-2}{log_3x}$, but I don't know how to proceed.






logarithms







share|cite|improve this question











$endgroup$




I don't have the slightest idea about how to tackle this one. I could change $2log_x 3$ to $frac{2}{log_3 x}$ and deducting that from $log_3 x$ would give me $frac{(log_3x)^2-2}{log_3x}$, but I don't know how to proceed.






logarithms




logarithms






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 13 '18 at 13:13







Nameless King

















asked Dec 13 '18 at 13:01









Nameless KingNameless King

306




306












  • $begingroup$
    It looks like you are very close to a quadratic in $log_3x$... Can you write the full equation resulting from your work thus far? Also, I believe you get $frac 2{log_3x}$ as the changed value...
    $endgroup$
    – abiessu
    Dec 13 '18 at 13:05












  • $begingroup$
    Thanks! Fixed that...
    $endgroup$
    – Nameless King
    Dec 13 '18 at 13:15


















  • $begingroup$
    It looks like you are very close to a quadratic in $log_3x$... Can you write the full equation resulting from your work thus far? Also, I believe you get $frac 2{log_3x}$ as the changed value...
    $endgroup$
    – abiessu
    Dec 13 '18 at 13:05












  • $begingroup$
    Thanks! Fixed that...
    $endgroup$
    – Nameless King
    Dec 13 '18 at 13:15
















$begingroup$
It looks like you are very close to a quadratic in $log_3x$... Can you write the full equation resulting from your work thus far? Also, I believe you get $frac 2{log_3x}$ as the changed value...
$endgroup$
– abiessu
Dec 13 '18 at 13:05






$begingroup$
It looks like you are very close to a quadratic in $log_3x$... Can you write the full equation resulting from your work thus far? Also, I believe you get $frac 2{log_3x}$ as the changed value...
$endgroup$
– abiessu
Dec 13 '18 at 13:05














$begingroup$
Thanks! Fixed that...
$endgroup$
– Nameless King
Dec 13 '18 at 13:15




$begingroup$
Thanks! Fixed that...
$endgroup$
– Nameless King
Dec 13 '18 at 13:15










3 Answers
3






active

oldest

votes


















5












$begingroup$

$$log_3 x-2log_x 3 = 1$$



Here, there is a very good way to manipulate the expression. You have



$$log_a b = frac{1}{log_b a}$$



Applying it here, you get



$$frac{1}{log_x 3}-2log_x 3 = 1$$



Now, let $t = log_x 3$. You get



$$frac{1}{t}-2t = 1$$



which results in a quadratic equation. Can you work out the rest?



Of course, your way also works in a similar manner, except you made an error in your work: $2log_x 3 = frac{2}{log_3 x} color{red}{neq frac{1}{2log_3 x}}$.






share|cite|improve this answer











$endgroup$





















    6












    $begingroup$

    If $log_3x=yimplies x=3^y=?$



    we have $$y-dfrac2y=1iff0=y^2-y-2=(y-2)(y+1)$$



    $implies y=?$






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$


      Write $$log_3 x - 2log_x 3 = 1$$ as $$frac{log x}{log 3}-2frac{ log 3}{log x}=1$$ (Taking care that $xneq 1$)




      Take LCM: $$(log x) ^2 -2(log 3)^2 =log x cdot log 3$$
      Now substitute $$log x =X text { and } log 3=y$$
      You get:
      $$X^2-2y^2=Xy$$ or $$X^2-Xy-2y^2=0$$ Which factorises to:




      $$(X-2y)(X+y)$$




      So, $$X=2y text{ or } X=-y$$
      Substitute $x$ and $y$ back:
      $$log x=2log 3$$or$$ log x=-log 3$$
      So, $$log x= log 3^2=log 9$$ or $$log x=log 3^{-1}=log frac 13$$



      Giving you




      $$x=9 text{ or } x=frac 13$$







      share|cite|improve this answer











      $endgroup$













        Your Answer





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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        5












        $begingroup$

        $$log_3 x-2log_x 3 = 1$$



        Here, there is a very good way to manipulate the expression. You have



        $$log_a b = frac{1}{log_b a}$$



        Applying it here, you get



        $$frac{1}{log_x 3}-2log_x 3 = 1$$



        Now, let $t = log_x 3$. You get



        $$frac{1}{t}-2t = 1$$



        which results in a quadratic equation. Can you work out the rest?



        Of course, your way also works in a similar manner, except you made an error in your work: $2log_x 3 = frac{2}{log_3 x} color{red}{neq frac{1}{2log_3 x}}$.






        share|cite|improve this answer











        $endgroup$


















          5












          $begingroup$

          $$log_3 x-2log_x 3 = 1$$



          Here, there is a very good way to manipulate the expression. You have



          $$log_a b = frac{1}{log_b a}$$



          Applying it here, you get



          $$frac{1}{log_x 3}-2log_x 3 = 1$$



          Now, let $t = log_x 3$. You get



          $$frac{1}{t}-2t = 1$$



          which results in a quadratic equation. Can you work out the rest?



          Of course, your way also works in a similar manner, except you made an error in your work: $2log_x 3 = frac{2}{log_3 x} color{red}{neq frac{1}{2log_3 x}}$.






          share|cite|improve this answer











          $endgroup$
















            5












            5








            5





            $begingroup$

            $$log_3 x-2log_x 3 = 1$$



            Here, there is a very good way to manipulate the expression. You have



            $$log_a b = frac{1}{log_b a}$$



            Applying it here, you get



            $$frac{1}{log_x 3}-2log_x 3 = 1$$



            Now, let $t = log_x 3$. You get



            $$frac{1}{t}-2t = 1$$



            which results in a quadratic equation. Can you work out the rest?



            Of course, your way also works in a similar manner, except you made an error in your work: $2log_x 3 = frac{2}{log_3 x} color{red}{neq frac{1}{2log_3 x}}$.






            share|cite|improve this answer











            $endgroup$



            $$log_3 x-2log_x 3 = 1$$



            Here, there is a very good way to manipulate the expression. You have



            $$log_a b = frac{1}{log_b a}$$



            Applying it here, you get



            $$frac{1}{log_x 3}-2log_x 3 = 1$$



            Now, let $t = log_x 3$. You get



            $$frac{1}{t}-2t = 1$$



            which results in a quadratic equation. Can you work out the rest?



            Of course, your way also works in a similar manner, except you made an error in your work: $2log_x 3 = frac{2}{log_3 x} color{red}{neq frac{1}{2log_3 x}}$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 13 '18 at 13:15

























            answered Dec 13 '18 at 13:06









            KM101KM101

            5,9251524




            5,9251524























                6












                $begingroup$

                If $log_3x=yimplies x=3^y=?$



                we have $$y-dfrac2y=1iff0=y^2-y-2=(y-2)(y+1)$$



                $implies y=?$






                share|cite|improve this answer









                $endgroup$


















                  6












                  $begingroup$

                  If $log_3x=yimplies x=3^y=?$



                  we have $$y-dfrac2y=1iff0=y^2-y-2=(y-2)(y+1)$$



                  $implies y=?$






                  share|cite|improve this answer









                  $endgroup$
















                    6












                    6








                    6





                    $begingroup$

                    If $log_3x=yimplies x=3^y=?$



                    we have $$y-dfrac2y=1iff0=y^2-y-2=(y-2)(y+1)$$



                    $implies y=?$






                    share|cite|improve this answer









                    $endgroup$



                    If $log_3x=yimplies x=3^y=?$



                    we have $$y-dfrac2y=1iff0=y^2-y-2=(y-2)(y+1)$$



                    $implies y=?$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 13 '18 at 13:06









                    lab bhattacharjeelab bhattacharjee

                    225k15157275




                    225k15157275























                        1












                        $begingroup$


                        Write $$log_3 x - 2log_x 3 = 1$$ as $$frac{log x}{log 3}-2frac{ log 3}{log x}=1$$ (Taking care that $xneq 1$)




                        Take LCM: $$(log x) ^2 -2(log 3)^2 =log x cdot log 3$$
                        Now substitute $$log x =X text { and } log 3=y$$
                        You get:
                        $$X^2-2y^2=Xy$$ or $$X^2-Xy-2y^2=0$$ Which factorises to:




                        $$(X-2y)(X+y)$$




                        So, $$X=2y text{ or } X=-y$$
                        Substitute $x$ and $y$ back:
                        $$log x=2log 3$$or$$ log x=-log 3$$
                        So, $$log x= log 3^2=log 9$$ or $$log x=log 3^{-1}=log frac 13$$



                        Giving you




                        $$x=9 text{ or } x=frac 13$$







                        share|cite|improve this answer











                        $endgroup$


















                          1












                          $begingroup$


                          Write $$log_3 x - 2log_x 3 = 1$$ as $$frac{log x}{log 3}-2frac{ log 3}{log x}=1$$ (Taking care that $xneq 1$)




                          Take LCM: $$(log x) ^2 -2(log 3)^2 =log x cdot log 3$$
                          Now substitute $$log x =X text { and } log 3=y$$
                          You get:
                          $$X^2-2y^2=Xy$$ or $$X^2-Xy-2y^2=0$$ Which factorises to:




                          $$(X-2y)(X+y)$$




                          So, $$X=2y text{ or } X=-y$$
                          Substitute $x$ and $y$ back:
                          $$log x=2log 3$$or$$ log x=-log 3$$
                          So, $$log x= log 3^2=log 9$$ or $$log x=log 3^{-1}=log frac 13$$



                          Giving you




                          $$x=9 text{ or } x=frac 13$$







                          share|cite|improve this answer











                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$


                            Write $$log_3 x - 2log_x 3 = 1$$ as $$frac{log x}{log 3}-2frac{ log 3}{log x}=1$$ (Taking care that $xneq 1$)




                            Take LCM: $$(log x) ^2 -2(log 3)^2 =log x cdot log 3$$
                            Now substitute $$log x =X text { and } log 3=y$$
                            You get:
                            $$X^2-2y^2=Xy$$ or $$X^2-Xy-2y^2=0$$ Which factorises to:




                            $$(X-2y)(X+y)$$




                            So, $$X=2y text{ or } X=-y$$
                            Substitute $x$ and $y$ back:
                            $$log x=2log 3$$or$$ log x=-log 3$$
                            So, $$log x= log 3^2=log 9$$ or $$log x=log 3^{-1}=log frac 13$$



                            Giving you




                            $$x=9 text{ or } x=frac 13$$







                            share|cite|improve this answer











                            $endgroup$




                            Write $$log_3 x - 2log_x 3 = 1$$ as $$frac{log x}{log 3}-2frac{ log 3}{log x}=1$$ (Taking care that $xneq 1$)




                            Take LCM: $$(log x) ^2 -2(log 3)^2 =log x cdot log 3$$
                            Now substitute $$log x =X text { and } log 3=y$$
                            You get:
                            $$X^2-2y^2=Xy$$ or $$X^2-Xy-2y^2=0$$ Which factorises to:




                            $$(X-2y)(X+y)$$




                            So, $$X=2y text{ or } X=-y$$
                            Substitute $x$ and $y$ back:
                            $$log x=2log 3$$or$$ log x=-log 3$$
                            So, $$log x= log 3^2=log 9$$ or $$log x=log 3^{-1}=log frac 13$$



                            Giving you




                            $$x=9 text{ or } x=frac 13$$








                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Dec 13 '18 at 13:21

























                            answered Dec 13 '18 at 13:15









                            ideaidea

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