I am confused as to how the inverse of $f(x)=frac{3+ln x}{3-ln x}$ is found.












2












$begingroup$


The problem within my textbook is $f(x)=$$frac{3+ln{x}}{3-ln{x}}$ I checked the answer and it is $y=$$e^frac{3x-3}{x+1}$
I've tried many times to simplify the equation after switching the variables but I don't know how to separate the $y$ from the natural logarithm. If someone could show me the steps involved I would appreciate it.






algebra-precalculus







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  • 3




    $begingroup$
    Could you type out what you have tried so far so we can try to spot your error?
    $endgroup$
    – TM Gallagher
    Mar 16 at 16:09






  • 1




    $begingroup$
    If you have $ln y = ....something to do with(x)...$ then just raise $e$ to both sides of the equation. $y = e^{ln y} = e^{....something to do with (x)....}$.
    $endgroup$
    – fleablood
    Mar 16 at 16:27






  • 1




    $begingroup$
    .... but you need to get $ln y = .... somethingtodo with (x)...$ can you see how $x = frac {3+ln y}{3-ln y}implies ln y = frac {3x -3}{x+1}$?
    $endgroup$
    – fleablood
    Mar 16 at 16:28
















2












$begingroup$


The problem within my textbook is $f(x)=$$frac{3+ln{x}}{3-ln{x}}$ I checked the answer and it is $y=$$e^frac{3x-3}{x+1}$
I've tried many times to simplify the equation after switching the variables but I don't know how to separate the $y$ from the natural logarithm. If someone could show me the steps involved I would appreciate it.






algebra-precalculus







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Could you type out what you have tried so far so we can try to spot your error?
    $endgroup$
    – TM Gallagher
    Mar 16 at 16:09






  • 1




    $begingroup$
    If you have $ln y = ....something to do with(x)...$ then just raise $e$ to both sides of the equation. $y = e^{ln y} = e^{....something to do with (x)....}$.
    $endgroup$
    – fleablood
    Mar 16 at 16:27






  • 1




    $begingroup$
    .... but you need to get $ln y = .... somethingtodo with (x)...$ can you see how $x = frac {3+ln y}{3-ln y}implies ln y = frac {3x -3}{x+1}$?
    $endgroup$
    – fleablood
    Mar 16 at 16:28














2












2








2





$begingroup$


The problem within my textbook is $f(x)=$$frac{3+ln{x}}{3-ln{x}}$ I checked the answer and it is $y=$$e^frac{3x-3}{x+1}$
I've tried many times to simplify the equation after switching the variables but I don't know how to separate the $y$ from the natural logarithm. If someone could show me the steps involved I would appreciate it.






algebra-precalculus







share|cite|improve this question











$endgroup$




The problem within my textbook is $f(x)=$$frac{3+ln{x}}{3-ln{x}}$ I checked the answer and it is $y=$$e^frac{3x-3}{x+1}$
I've tried many times to simplify the equation after switching the variables but I don't know how to separate the $y$ from the natural logarithm. If someone could show me the steps involved I would appreciate it.






algebra-precalculus




algebra-precalculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 17 at 3:58









YuiTo Cheng

2,1362837




2,1362837










asked Mar 16 at 16:06









ShonellShonell

163




163








  • 3




    $begingroup$
    Could you type out what you have tried so far so we can try to spot your error?
    $endgroup$
    – TM Gallagher
    Mar 16 at 16:09






  • 1




    $begingroup$
    If you have $ln y = ....something to do with(x)...$ then just raise $e$ to both sides of the equation. $y = e^{ln y} = e^{....something to do with (x)....}$.
    $endgroup$
    – fleablood
    Mar 16 at 16:27






  • 1




    $begingroup$
    .... but you need to get $ln y = .... somethingtodo with (x)...$ can you see how $x = frac {3+ln y}{3-ln y}implies ln y = frac {3x -3}{x+1}$?
    $endgroup$
    – fleablood
    Mar 16 at 16:28














  • 3




    $begingroup$
    Could you type out what you have tried so far so we can try to spot your error?
    $endgroup$
    – TM Gallagher
    Mar 16 at 16:09






  • 1




    $begingroup$
    If you have $ln y = ....something to do with(x)...$ then just raise $e$ to both sides of the equation. $y = e^{ln y} = e^{....something to do with (x)....}$.
    $endgroup$
    – fleablood
    Mar 16 at 16:27






  • 1




    $begingroup$
    .... but you need to get $ln y = .... somethingtodo with (x)...$ can you see how $x = frac {3+ln y}{3-ln y}implies ln y = frac {3x -3}{x+1}$?
    $endgroup$
    – fleablood
    Mar 16 at 16:28








3




3




$begingroup$
Could you type out what you have tried so far so we can try to spot your error?
$endgroup$
– TM Gallagher
Mar 16 at 16:09




$begingroup$
Could you type out what you have tried so far so we can try to spot your error?
$endgroup$
– TM Gallagher
Mar 16 at 16:09




1




1




$begingroup$
If you have $ln y = ....something to do with(x)...$ then just raise $e$ to both sides of the equation. $y = e^{ln y} = e^{....something to do with (x)....}$.
$endgroup$
– fleablood
Mar 16 at 16:27




$begingroup$
If you have $ln y = ....something to do with(x)...$ then just raise $e$ to both sides of the equation. $y = e^{ln y} = e^{....something to do with (x)....}$.
$endgroup$
– fleablood
Mar 16 at 16:27




1




1




$begingroup$
.... but you need to get $ln y = .... somethingtodo with (x)...$ can you see how $x = frac {3+ln y}{3-ln y}implies ln y = frac {3x -3}{x+1}$?
$endgroup$
– fleablood
Mar 16 at 16:28




$begingroup$
.... but you need to get $ln y = .... somethingtodo with (x)...$ can you see how $x = frac {3+ln y}{3-ln y}implies ln y = frac {3x -3}{x+1}$?
$endgroup$
– fleablood
Mar 16 at 16:28










3 Answers
3






active

oldest

votes


















10












$begingroup$

Well, as always you want to solve for $y$ in



$x = frac {3+ln y}{3-ln y}$



$x(3-ln y) = (3 + ln y)$



$3x - 3 = ln y + xln y$



$(x+1) ln y = 3x-3$



$ln y = frac {3x-3}{x+1}$



$e^{ln y} = e^{frac {3x-3}{x+1}}$



So $y = e^{frac {3x-3}{x+1}}$




I don't know how to seperate the y from the natural logarithm




Then inverse of $ln y = K$ is $e^{ln y} = y = e^K$. $ln x$ and $e^x$ are inverses of each other. You "cancel" $ln y=k$ by raising $e$ to the value to get $y = e^k$, and you "cancel" $e^y = M$ by taking the natural log to get $y = ln M$.






share|cite|improve this answer











$endgroup$





















    2












    $begingroup$

    Steps:



    Start with:$$y=frac{3+ln x}{3-ln x}$$



    Rewrite the equality as: $$ln x=g(y)$$ and conclude that: $$x=e^{ln x}=e^{g(y)}$$



    Now switch $x$ and $y$ to get the final result.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      $$y=frac{3+ln x}{3-ln x}$$



      Swap $(x,y)$ to find inverse. This is the procedure!



      $$x=frac{3+ln y}{3-ln y}$$



      Componendo/Dividendo



      $$frac{x-1}{x+1}= dfrac{2 ln y}{6}$$



      $$ y rightarrow y_{,(inv-fn)} = e^{frac{3(x-1)}{(x+1)} } $$



      Also the graphs reflect each other with respect to mirror line $ y=x.$



      The basic rule or definition of log /exp is the same



      $$ A = ln_{base}C rightarrow base^{A}= C. $$






      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









        10












        $begingroup$

        Well, as always you want to solve for $y$ in



        $x = frac {3+ln y}{3-ln y}$



        $x(3-ln y) = (3 + ln y)$



        $3x - 3 = ln y + xln y$



        $(x+1) ln y = 3x-3$



        $ln y = frac {3x-3}{x+1}$



        $e^{ln y} = e^{frac {3x-3}{x+1}}$



        So $y = e^{frac {3x-3}{x+1}}$




        I don't know how to seperate the y from the natural logarithm




        Then inverse of $ln y = K$ is $e^{ln y} = y = e^K$. $ln x$ and $e^x$ are inverses of each other. You "cancel" $ln y=k$ by raising $e$ to the value to get $y = e^k$, and you "cancel" $e^y = M$ by taking the natural log to get $y = ln M$.






        share|cite|improve this answer











        $endgroup$


















          10












          $begingroup$

          Well, as always you want to solve for $y$ in



          $x = frac {3+ln y}{3-ln y}$



          $x(3-ln y) = (3 + ln y)$



          $3x - 3 = ln y + xln y$



          $(x+1) ln y = 3x-3$



          $ln y = frac {3x-3}{x+1}$



          $e^{ln y} = e^{frac {3x-3}{x+1}}$



          So $y = e^{frac {3x-3}{x+1}}$




          I don't know how to seperate the y from the natural logarithm




          Then inverse of $ln y = K$ is $e^{ln y} = y = e^K$. $ln x$ and $e^x$ are inverses of each other. You "cancel" $ln y=k$ by raising $e$ to the value to get $y = e^k$, and you "cancel" $e^y = M$ by taking the natural log to get $y = ln M$.






          share|cite|improve this answer











          $endgroup$
















            10












            10








            10





            $begingroup$

            Well, as always you want to solve for $y$ in



            $x = frac {3+ln y}{3-ln y}$



            $x(3-ln y) = (3 + ln y)$



            $3x - 3 = ln y + xln y$



            $(x+1) ln y = 3x-3$



            $ln y = frac {3x-3}{x+1}$



            $e^{ln y} = e^{frac {3x-3}{x+1}}$



            So $y = e^{frac {3x-3}{x+1}}$




            I don't know how to seperate the y from the natural logarithm




            Then inverse of $ln y = K$ is $e^{ln y} = y = e^K$. $ln x$ and $e^x$ are inverses of each other. You "cancel" $ln y=k$ by raising $e$ to the value to get $y = e^k$, and you "cancel" $e^y = M$ by taking the natural log to get $y = ln M$.






            share|cite|improve this answer











            $endgroup$



            Well, as always you want to solve for $y$ in



            $x = frac {3+ln y}{3-ln y}$



            $x(3-ln y) = (3 + ln y)$



            $3x - 3 = ln y + xln y$



            $(x+1) ln y = 3x-3$



            $ln y = frac {3x-3}{x+1}$



            $e^{ln y} = e^{frac {3x-3}{x+1}}$



            So $y = e^{frac {3x-3}{x+1}}$




            I don't know how to seperate the y from the natural logarithm




            Then inverse of $ln y = K$ is $e^{ln y} = y = e^K$. $ln x$ and $e^x$ are inverses of each other. You "cancel" $ln y=k$ by raising $e$ to the value to get $y = e^k$, and you "cancel" $e^y = M$ by taking the natural log to get $y = ln M$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Mar 16 at 20:29









            jpmc26

            25129




            25129










            answered Mar 16 at 16:25









            fleabloodfleablood

            73.4k22791




            73.4k22791























                2












                $begingroup$

                Steps:



                Start with:$$y=frac{3+ln x}{3-ln x}$$



                Rewrite the equality as: $$ln x=g(y)$$ and conclude that: $$x=e^{ln x}=e^{g(y)}$$



                Now switch $x$ and $y$ to get the final result.






                share|cite|improve this answer









                $endgroup$


















                  2












                  $begingroup$

                  Steps:



                  Start with:$$y=frac{3+ln x}{3-ln x}$$



                  Rewrite the equality as: $$ln x=g(y)$$ and conclude that: $$x=e^{ln x}=e^{g(y)}$$



                  Now switch $x$ and $y$ to get the final result.






                  share|cite|improve this answer









                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    Steps:



                    Start with:$$y=frac{3+ln x}{3-ln x}$$



                    Rewrite the equality as: $$ln x=g(y)$$ and conclude that: $$x=e^{ln x}=e^{g(y)}$$



                    Now switch $x$ and $y$ to get the final result.






                    share|cite|improve this answer









                    $endgroup$



                    Steps:



                    Start with:$$y=frac{3+ln x}{3-ln x}$$



                    Rewrite the equality as: $$ln x=g(y)$$ and conclude that: $$x=e^{ln x}=e^{g(y)}$$



                    Now switch $x$ and $y$ to get the final result.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Mar 16 at 16:16









                    drhabdrhab

                    103k545136




                    103k545136























                        0












                        $begingroup$

                        $$y=frac{3+ln x}{3-ln x}$$



                        Swap $(x,y)$ to find inverse. This is the procedure!



                        $$x=frac{3+ln y}{3-ln y}$$



                        Componendo/Dividendo



                        $$frac{x-1}{x+1}= dfrac{2 ln y}{6}$$



                        $$ y rightarrow y_{,(inv-fn)} = e^{frac{3(x-1)}{(x+1)} } $$



                        Also the graphs reflect each other with respect to mirror line $ y=x.$



                        The basic rule or definition of log /exp is the same



                        $$ A = ln_{base}C rightarrow base^{A}= C. $$






                        share|cite|improve this answer











                        $endgroup$


















                          0












                          $begingroup$

                          $$y=frac{3+ln x}{3-ln x}$$



                          Swap $(x,y)$ to find inverse. This is the procedure!



                          $$x=frac{3+ln y}{3-ln y}$$



                          Componendo/Dividendo



                          $$frac{x-1}{x+1}= dfrac{2 ln y}{6}$$



                          $$ y rightarrow y_{,(inv-fn)} = e^{frac{3(x-1)}{(x+1)} } $$



                          Also the graphs reflect each other with respect to mirror line $ y=x.$



                          The basic rule or definition of log /exp is the same



                          $$ A = ln_{base}C rightarrow base^{A}= C. $$






                          share|cite|improve this answer











                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            $$y=frac{3+ln x}{3-ln x}$$



                            Swap $(x,y)$ to find inverse. This is the procedure!



                            $$x=frac{3+ln y}{3-ln y}$$



                            Componendo/Dividendo



                            $$frac{x-1}{x+1}= dfrac{2 ln y}{6}$$



                            $$ y rightarrow y_{,(inv-fn)} = e^{frac{3(x-1)}{(x+1)} } $$



                            Also the graphs reflect each other with respect to mirror line $ y=x.$



                            The basic rule or definition of log /exp is the same



                            $$ A = ln_{base}C rightarrow base^{A}= C. $$






                            share|cite|improve this answer











                            $endgroup$



                            $$y=frac{3+ln x}{3-ln x}$$



                            Swap $(x,y)$ to find inverse. This is the procedure!



                            $$x=frac{3+ln y}{3-ln y}$$



                            Componendo/Dividendo



                            $$frac{x-1}{x+1}= dfrac{2 ln y}{6}$$



                            $$ y rightarrow y_{,(inv-fn)} = e^{frac{3(x-1)}{(x+1)} } $$



                            Also the graphs reflect each other with respect to mirror line $ y=x.$



                            The basic rule or definition of log /exp is the same



                            $$ A = ln_{base}C rightarrow base^{A}= C. $$







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Mar 17 at 8:24

























                            answered Mar 16 at 16:51









                            NarasimhamNarasimham

                            21.1k62158




                            21.1k62158






























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