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







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












$begingroup$


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





        StackExchange.ifUsing("editor", function () {
        return StackExchange.using("mathjaxEditing", function () {
        StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
        StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
        });
        });
        }, "mathjax-editing");

        StackExchange.ready(function() {
        var channelOptions = {
        tags: "".split(" "),
        id: "69"
        };
        initTagRenderer("".split(" "), "".split(" "), channelOptions);

        StackExchange.using("externalEditor", function() {
        // Have to fire editor after snippets, if snippets enabled
        if (StackExchange.settings.snippets.snippetsEnabled) {
        StackExchange.using("snippets", function() {
        createEditor();
        });
        }
        else {
        createEditor();
        }
        });

        function createEditor() {
        StackExchange.prepareEditor({
        heartbeatType: 'answer',
        autoActivateHeartbeat: false,
        convertImagesToLinks: true,
        noModals: true,
        showLowRepImageUploadWarning: true,
        reputationToPostImages: 10,
        bindNavPrevention: true,
        postfix: "",
        imageUploader: {
        brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
        contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
        allowUrls: true
        },
        noCode: true, onDemand: true,
        discardSelector: ".discard-answer"
        ,immediatelyShowMarkdownHelp:true
        });


        }
        });














        draft saved

        draft discarded


















        StackExchange.ready(
        function () {
        StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3150533%2fi-am-confused-as-to-how-the-inverse-of-fx-frac3-ln-x3-ln-x-is-found%23new-answer', 'question_page');
        }
        );

        Post as a guest















        Required, but never shown

























        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






























                                draft saved

                                draft discarded




















































                                Thanks for contributing an answer to Mathematics Stack Exchange!


                                • Please be sure to answer the question. Provide details and share your research!

                                But avoid …



                                • Asking for help, clarification, or responding to other answers.

                                • Making statements based on opinion; back them up with references or personal experience.


                                Use MathJax to format equations. MathJax reference.


                                To learn more, see our tips on writing great answers.




                                draft saved


                                draft discarded














                                StackExchange.ready(
                                function () {
                                StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3150533%2fi-am-confused-as-to-how-the-inverse-of-fx-frac3-ln-x3-ln-x-is-found%23new-answer', 'question_page');
                                }
                                );

                                Post as a guest















                                Required, but never shown





















































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown

































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown







                                -

                                Popular posts from this blog

                                How to change which sound is reproduced for terminal bell?

                                Image
                                3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question ...
                                Read more

                                Can I use Tabulator js library in my java Spring + Thymeleaf project?

                                Image
                                0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"...
                                Read more

                                Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

                                Image
                                up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx...
                                Read more